MATH 108 - ELEMENTARY AND
INTERMEDIATE ALGEBRA -
Differentiation of Trigonometric
Functions
Question Bank - Set 5
Liberty University
Question 1
Question
Find the derivative of the function f(x) = cos2(2x)−sin(3x).
Solution
Step 1: Apply the chain rule to differentiate cos2(2x) with respect to x.
Let u= cos(2x) and v= cos(2x).
f′(x) = 2 cos(2x)(−sin(2x)) −cos2(2x)·2 sin(2x)
=−2 cos(2x) sin(2x)−2 cos2(2x) sin(2x)
=−2 sin(4x)−2 cos2(2x) sin(2x)
Step 2: Differentiate sin(3x) with respect to x.
f′(x) = −2 sin(4x)−2 cos2(2x) sin(2x)−3 cos(3x).
Therefore, the derivative of the function f(x) = cos2(2x)−sin(3x) is f′(x) =
−2 sin(4x)−2 cos2(2x) sin(2x)−3 cos(3x).
Question 2
Question
Compute the derivative of f(x) = 4 sin(x) cos(x)
3x.
Solution
Step 1: Apply the quotient rule to differentiate the function f(x) = 4 sin(x) cos(x)
3x.
Step 2: Let u(x) = 4 sin(x) cos(x) and v(x) = 3x. Then, the derivative of
f(x) is given by
f′(x) = u′v−uv′
v2
=(4 cos(x) cos(x)−4 sin(x)(−sin(x))) ·3x−4 sin(x) cos(x)·3
(3x)2
=(4 cos2(x) + 4 sin2(x)) ·3x+ 4 sin(x) cos(x)·3
9x2
=12x
9x2
=4
3x.
Therefore, the derivative of f(x) = 4 sin(x) cos(x)
3xis f′(x) = 4
3x.
Question 3
Question
Find the derivative of the function f(x) = sin2(x) cos(3x).
Solution
Step 1: Use the product rule to differentiate the function. The product rule
states that if uand vare differentiable functions of x, then the derivative of
their product is given by (uv)′=u′v+uv′. Step 2: Let u= sin2(x) and
v= cos(3x). Step 3: Compute u′and v′:
u′= (sin(x))2)′= 2 sin(x) cos(x)
v′= (cos(3x))′=−3 sin(3x)
Step 4: Apply the product rule formula to find f′(x):
f′(x) = u′v+uv′
f′(x) = 2 sin(x) cos(x)·cos(3x) + sin2(x)·(−3 sin(3x))
Step 5: Simplify the expression:
f′(x) = 2 sin(x) cos(x) cos(3x)−3 sin2(x) sin(3x)
Step 6: Apply trigonometric identities to simplify further:
f′(x) = sin(2x) cos(3x)−
3
2sin(2x)
2
Step 7: Finally, simplify the expression:
f′(x) = 1
2sin(2x)(2 cos(3x)−3)
Therefore, the derivative of the function f(x) = sin2(x) cos(3x) is f′(x) =
1
2sin(2x)(2 cos(3x)−3).
Question 4
Question
Find the derivative of the function f(x) = sin(2x)−cos(3x).
Solution
Step 1: Use the differentiation rules for trigonometric functions to find the
derivative of f(x).
f(x) = sin(2x)−cos(3x)
f′(x) = d
dx (sin(2x)) −
d
dx (cos(3x))
Step 2: Apply the chain rule and the derivative of sine function to sin(2x).
d
dx (sin(2x)) = cos(2x)·
d
dx (2x)
= 2 cos(2x)
Step 3: Apply the chain rule and the derivative of cosine function to cos(3x).
d
dx (cos(3x)) = −sin(3x)·
d
dx (3x)
=−3 sin(3x)
Step 4: Put the derivatives of sin(2x) and cos(3x) back into the expression
for f′(x).
f′(x) = 2 cos(2x)−3 sin(3x)
= 2 cos(2x)−3 sin(3x)
Therefore, the derivative of the function f(x) = sin(2x)−cos(3x) is f′(x) =
2 cos(2x)−3 sin(3x).
Question 5
Question
Find the derivative of f(x) = sin2(3x) + cos(2x).
3
Solution
To find the derivative of f(x), we will differentiate each term separately using
the rules of differentiation.
f(x) = sin2(3x) + cos(2x)
Step 1: Differentiate sin2(3x) using the chain rule.
d
dx (sin2(3x)) = 2 sin(3x) cos(3x)·3
= 6 sin(3x) cos(3x)
Step 2: Differentiate cos(2x).
d
dx (cos(2x)) = −sin(2x)·2
=−2 sin(2x)
Step 3: Combine the derivatives to find f′(x).
f′(x) = 6 sin(3x) cos(3x)−2 sin(2x)
Therefore, the derivative of f(x) = sin2(3x)+cos(2x) is f′(x) = 6 sin(3x) cos(3x)−
2 sin(2x).
Question 6
Question
Find the derivative of the following function with respect to x:f(x) = tan2(3x)+
sin(2x) cos(4x).
Solution
To find the derivative of f(x) with respect to x, we will differentiate each term
separately using the rules of differentiation. Step 1: Apply the chain rule to dif-
ferentiate tan2(3x). Step 2: Differentiate sin(2x). Step 3: Differentiate cos(4x).
Step 4: Add up the derivatives to find the derivative of f(x).
Step 1: Let u= tan(3x). Using the chain rule, we have:
d
dx (tan2(3x)) = d
dx (u2)=2u·
du
dx
Step 2:
d
dx (sin(2x)) = 2 cos(2x)
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Step 3:
d
dx (cos(4x)) = −4 sin(4x)
Step 4: Putting it all together, we have:
f′(x) = 2 tan(3x)·3 sec2(3x) + 2 cos(2x) cos(4x)−4 sin(2x) sin(4x)
Therefore, the derivative of f(x) with respect to xis:
f′(x) = 6 tan(3x) sec2(3x) + 2 cos(2x) cos(4x)−4 sin(2x) sin(4x)
Question 7
Question
Find the derivative of the function f(x) = sin(3x) cos(2x).
Solution
To find the derivative of f(x) = sin(3x) cos(2x), we will use the product rule.
Step 1: Apply the product rule:
f′(x) = (sin(3x))′
·cos(2x) + sin(3x)·(cos(2x))′
Step 2: Find the derivatives of sin(3x) and cos(2x).
(sin(3x))′= 3 cos(3x)
(cos(2x))′=−2 sin(2x)
Step 3: Substitute the derivatives back into the product rule formula:
f′(x) = 3 cos(3x)·cos(2x) + sin(3x)·(−2 sin(2x))
Step 4: Simplify the expression:
f′(x) = 3 cos(3x) cos(2x)−2 sin(3x) sin(2x)
So, the derivative of the function f(x) = sin(3x) cos(2x) is 3 cos(3x) cos(2x)−
2 sin(3x) sin(2x).
Question 8
Question
Find the derivative of the following function: f(x) = sin2(3x) + cos(2x).
5
Solution
To find the derivative of f(x), we will use the power rule, chain rule, and sum
rule for differentiation.
Step 1: Start by finding the derivative of sin2(3x).
d
dx (sin2(3x)) = 2 sin(3x) cos(3x)·3
= 6 sin(3x) cos(3x)
Step 2: Next, find the derivative of cos(2x).
d
dx (cos(2x)) = −sin(2x)·2
=−2 sin(2x)
Step 3: Now, add the derivatives of sin2(3x) and cos(2x) to find the deriva-
tive of f(x).
f′(x) = 6 sin(3x) cos(3x)−2 sin(2x)
= 6 ·
1
2
·sin(6x)−2 sin(2x)
= 3 sin(6x)−2 sin(2x)
Therefore, the derivative of f(x) = sin2(3x) + cos(2x) is f′(x) = 3 sin(6x)−
2 sin(2x).
Question 9
Question
Find the derivative of the function f(x) = sin(x) cos(x).
Solution
Step 1: To find the derivative of f(x), we will use the product rule, which states
that the derivative of the product of two functions u(x) and v(x) is given by:
(uv)′=u′v+uv′.
Step 2: Let u(x) = sin(x) and v(x) = cos(x). Then, we have:
u′(x) = cos(x) and v′(x) = −sin(x)
Step 3: Using the product rule, we can find the derivative of f(x):
f′(x) = (sin(x))(cos(x))′+ (sin(x))′(cos(x))
f′(x) = cos(x) cos(x) + sin(x)(−sin(x))
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f′(x) = cos2(x)−sin2(x)
Step 4: Recall the Pythagorean trigonometric identity: cos2(x)−sin2(x) =
cos(2x). Substituting this into f′(x) gives:
f′(x) = cos(2x)
Therefore, the derivative of f(x) = sin(x) cos(x) is f′(x) = cos(2x).
Question 10
Question
Find the derivative of the function f(x) = sin2(x) + cos2(x).
Solution
Step 1: Recall the trigonometric identity sin2(x) + cos2(x) = 1. Step 2: Rewrite
f(x) in terms of sin2(x). Step 3: Differentiate f(x) using the chain rule. Step
4: Simplify the derivative to obtain the final answer.
Question 11
Question
Find the derivative of the function f(x) = sin(3x) cos(2x).
Solution
To find the derivative of the given function f(x) = sin(3x) cos(2x), we will use
the product rule of differentiation.
Step 1: Apply the product rule, which states that if f(x) = u(x)v(x), then
f′(x) = u′v+uv′. Let u(x) = sin(3x) and v(x) = cos(2x).
Step 2: Find u′(x) and v′(x): - u′(x) = 3 cos(3x) (derivative of sin(3x)
using the chain rule) - v′(x) = −2 sin(2x) (derivative of cos(2x) using the chain
rule)
Step 3: Apply the product rule to find f′(x):
f′(x) = u′v+uv′= (3 cos(3x))(cos(2x)) + (sin(3x))(−2 sin(2x))
Step 4: Simplify the expression:
f′(x) = 3 cos(3x) cos(2x)−2 sin(3x) sin(2x)
Thus, the derivative of the function f(x) = sin(3x) cos(2x) is 3 cos(3x) cos(2x)−
2 sin(3x) sin(2x).
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Question 12
Question
Find the derivative of the function f(x) = sin3(2x) + cos3(3x).
Solution
Step 1: To find the derivative of f(x), we will use the chain rule and the power
rule for differentiation.
Step 2: Let’s start by finding the derivative of sin3(2x). Using the chain
rule, we have
d
dx (sin3(2x)) = 3 sin2(2x)·cos(2x)·2.
Step 3: Simplifying the above expression, we get
d
dx (sin3(2x)) = 6 sin2(2x) cos(2x).
Step 4: Next, let’s find the derivative of cos3(3x). Using the chain rule, we
have d
dx (cos3(3x)) = 3 cos2(3x)·(−sin(3x)) ·3.
Step 5: Simplifying the above expression, we get
d
dx (cos3(3x)) = −9 cos2(3x) sin(3x).
Step 6: Therefore, the derivative of the function f(x) = sin3(2x) + cos3(3x)
is
f′(x) = 6 sin2(2x) cos(2x)−9 cos2(3x) sin(3x).
Question 13
Question
Find the derivative of f(x) = sin2(3x) with respect to x.
Solution
Step 1: Apply the chain rule. Step 2: Identify the inner function and the
outer function. Step 3: Let u= 3x, then the function can be rewritten as
g(u) = sin2u. Step 4: Find the derivative of the outer function. Step 5: Find
the derivative of the inner function. Step 6: Apply the chain rule formula to
find the derivative of the composite function.
Therefore, the derivative of f(x) = sin2(3x) with respect to xis 6 sin(3x) cos(3x) .
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Question 14
Question
Find the derivative of the function f(x) = cos(2x)·sin(3x).
Solution
Step 1: Apply the product rule to find the derivative of the function f(x) =
cos(2x)·sin(3x).
Let u(x) = cos(2x) and v(x) = sin(3x).
The product rule states that (uv)′=u′v+uv′.
Step 2: Find u′(x) and v′(x).
Using the chain rule, we have u′(x) = −2 sin(2x) and v′(x) = 3 cos(3x).
Step 3: Apply the product rule to find f′(x).
f′(x)=(u(x)v(x))′
=u′(x)v(x) + u(x)v′(x)
= (−2 sin(2x)) sin(3x) + cos(2x)(3 cos(3x))
=−2 sin(2x) sin(3x) + 3 cos(2x) cos(3x).
Therefore, the derivative of the function f(x) = cos(2x)·sin(3x) is f′(x) =
−2 sin(2x) sin(3x) + 3 cos(2x) cos(3x).
Question 15
Question
Find the derivative of the function f(x) = sin2(2x)
1+cos(2x).
Solution
Step 1: To find the derivative of f(x), we will first rewrite it using trigonometric
identities. Step 2: Recall the double angle identity: sin(2x) = 2 sin(x) cos(x)
and cos(2x) = cos2(x)−sin2(x). Step 3: Substituting these identities into f(x),
we get:
f(x) = (2 sin(x) cos(x))2
1 + (cos2(x)−sin2(x))
Step 4: Simplifying the expression, we have:
f(x) = 4 sin2(x) cos2(x)
2 cos2(x)
Step 5: Further simplifying, we obtain:
f(x) = 2 sin2(x)
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Step 6: Now, differentiate f(x) with respect to xusing the power rule and
chain rule for trigonometric functions. Step 7: The derivative of 2 sin2(x) is
4 sin(x) cos(x) by the chain rule. Step 8: Therefore, the derivative of f(x) is
4 sin(x) cos(x) .
Question 16
Question
Find the derivative of the function f(x) = sin(x) cos(x).
Solution
To find the derivative of f(x) = sin(x) cos(x), we will use the product rule of
differentiation.
Step 1: Apply the product rule. Let u= sin(x) and v= cos(x). The
product rule states that (uv)′=u′v+uv′.
Step 2: Find u′and v′.
u′= cos(x) (derivative of sin(x))
v′=−sin(x) (derivative of cos(x))
Step 3: Substitute u,v,u′, and v′into the product rule formula.
f′(x) = u′v+uv′
= (cos(x))(cos(x)) + (sin(x))(−sin(x))
= cos2(x)−sin2(x)
Step 4: Use the trigonometric identity cos2(x)−sin2(x) = cos(2x).
f′(x) = cos(2x)
So, the derivative of f(x) = sin(x) cos(x) is f′(x) = cos(2x).
Question 17
Question
Find the derivative of the function f(x) = sin2(4x) cos(3x).
Solution
To find the derivative of the given function, we will use the product rule and
chain rule for differentiation.
Step 1: Apply the product rule: (uv)′=u′v+uv′, where u= sin2(4x) and
v= cos(3x).
10
Let u= sin2(4x), then u′= 2 sin(4x) cos(4x) by the chain rule.
Let v= cos(3x), then v′=−3 sin(3x) by the chain rule.
Step 2: Substitute u′,v′,u, and vinto the product rule formula.
(f(x))′= (u′v+uv′)
= (2 sin(4x) cos(4x))(cos(3x)) + (sin2(4x))(−3 sin(3x))
= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin2(4x)
Step 3: Simplify the expression if possible.
(f(x))′= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin2(4x)
= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin(4x) sin(4x)
= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin(4x) cos(4x)
Therefore, the derivative of the function f(x) = sin2(4x) cos(3x) is 2 sin(4x) cos(4x) cos(3x)−
3 sin(3x) sin(4x) cos(4x).
Question 18
Question
Find the derivative of y= sin2(3x) with respect to x.
Solution
Step 1: Apply the chain rule to differentiate y= sin2(3x). Step 2: Let u=
sin(3x) and y=u2. Step 3: Find dy
du and du
dx . Step 4: Use the chain rule:
dy
dx =dy
du
·du
dx .
Step 1: Apply the chain rule to differentiate y= sin2(3x). Step 2: Let
u= sin(3x) and y=u2. Step 3: Find dy
du and du
dx .
dy
du = 2u
du
dx = 3 cos(3x)
Step 4: Use the chain rule:
dy
dx =dy
du
·
du
dx = 2u·3 cos(3x) = 6 sin(3x) cos(3x)
Therefore, the derivative of y= sin2(3x) with respect to xis 6 sin(3x) cos(3x).
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Question 19
Question
Find the derivative of f(x) = sin(2x) cos(3x).
Solution
Step 1: Apply the product rule, (uv)′=u′v+uv′, where u= sin(2x) and
v= cos(3x). Step 2: Find u′and v′. Step 3: Calculate u′:
u= sin(2x)
u′=d
dx (sin(2x)) = 2 cos(2x)
Step 4: Calculate v′:
v= cos(3x)
v′=d
dx (cos(3x)) = −3 sin(3x)
Step 5: Apply the product rule:
f′(x) = u′v+uv′
f′(x) = (2 cos(2x))(cos(3x)) + (sin(2x))(−3 sin(3x))
Step 6: Simplify the expression:
f′(x) = 2 cos(2x) cos(3x)−3 sin(2x) sin(3x)
f′(x) = 2 cos(2x) cos(3x)−3 cosπ
2
−3x
f′(x) = 2 cos(2x) cos(3x)−3 cosπ
2cos(3x) + 3 sinπ
2sin(3x)
f′(x) = 2 cos(2x) cos(3x)−3 cos(3x)
Step 7: Therefore, the derivative of f(x) = sin(2x) cos(3x) is f′(x) = 2 cos(2x) cos(3x)−
3 cos(3x).
Question 20
Question
Let f(x) = cos(5x) sin(3x). Find f′(x).
12
Solution
Step 1: Apply the product rule to differentiate f(x). Step 2: Let u= cos(5x)
and v= sin(3x). Step 3: Find u′and v′. Step 4: Calculate u′v+uv′to find
f′(x).
Step 1: Apply the product rule:
f′(x) = d
dx (cos(5x)) sin(3x) + cos(5x)d
dx (sin(3x))
Step 2: Let u= cos(5x) and v= sin(3x).
Step 3: Calculate u′and v′:
u′=−5 sin(5x)
v′= 3 cos(3x)
Step 4: Substitute u′and v′back into the product rule formula:
f′(x) = −5 sin(5x) sin(3x) + cos(5x)3 cos(3x)
Therefore, f′(x) = −5 sin(5x) sin(3x) + 3 cos(5x) cos(3x).
Question 21
Question
Find the derivative of the function f(x) = cos(3x) sin(4x).
Solution
Step 1: Apply the product rule to differentiate the product of cos(3x) and
sin(4x). Step 2: Recall that the derivative of cos(u) is −sin(u)·u′and the
derivative of sin(u) is cos(u)·u′. Step 3: Let u= 3xand v= sin(4x). Step
4: Calculate u′and v′. Step 5: Find the derivative of the function f(x) =
cos(3x) sin(4x) by applying the product rule. Step 6: Combine the results to
obtain the final expression for f′(x).
Question 22
Question
Find the derivative of the function f(x) = sin2(3x) + cos(2x).
13
Solution
Step 1: Apply the chain rule to differentiate sin2(3x).
d
dx (sin2(3x)) = 2 sin(3x) cos(3x)·3 = 6 sin(3x) cos(3x)
Step 2: Apply the chain rule to differentiate cos(2x).
d
dx (cos(2x)) = −sin(2x)·2 = −2 sin(2x)
Step 3: Add the derivatives of the two parts together.
f′(x) = 6 sin(3x) cos(3x)−2 sin(2x)
Question 23
Question
Find the derivative of the function y= sin(x) cos(x).
Solution
We can use the product rule to find the derivative of the given function.
Step 1: Apply the product rule. Let f(x) = sin(x) and g(x) = cos(x). The
product rule states that the derivative of the product of two functions is given
by
(f·g)′=f′
·g+f·g′.
Therefore, the derivative of y= sin(x) cos(x) is
y′= (sin(x))′cos(x) + sin(x)(cos(x))′.
Step 2: Find the derivatives of sin(x) and cos(x). The derivative of sin(x)
is cos(x) and the derivative of cos(x) is −sin(x).
Step 3: Substitute the derivatives into the product rule. Substitute cos(x)
for (sin(x))′and −sin(x) for (cos(x))′into the derivative expression:
y′= cos(x) cos(x) + sin(x)(−sin(x)).
Step 4: Simplify the expression. Simplify the expression to get the final
answer:
y′= cos2(x)−sin2(x).
Therefore, the derivative of y= sin(x) cos(x) is y′= cos2(x)−sin2(x).
Question 24
Question
Find the derivative of the function f(x) = sin(2x) cos(3x).
14
Solution
Step 1: Apply the product rule to differentiate f(x) = sin(2x) cos(3x), which
states that the derivative of the product of two functions is the first function
times the derivative of the second, plus the second function times the derivative
of the first. Step 2: Let u= sin(2x) and v= cos(3x). Then we have f(x) = u·v.
Step 3: Find u′and v′. Step 4: Calculate u′:u′=d
dx [sin(2x)] Step 5: Apply
the chain rule to differentiate sin(2x): u′= 2 cos(2x) Step 6: Calculate v′:
v′=d
dx [cos(3x)] Step 7: Apply the chain rule to differentiate cos(3x): v′=
−3 sin(3x) Step 8: Now, apply the product rule to find f′(x): f′(x) = u′v+uv′
Step 9: Substitute u= sin(2x), u′= 2 cos(2x), v= cos(3x), and v′=−3 sin(3x)
into the formula for f′(x). Step 10: Simplify the expression for f′(x) to get the
final answer.
Question 25
Question
Find the derivative of the function f(x) = sin2(x) cos(x).
Solution
Step 1: Apply the product rule by differentiating the first function sin2(x) and
keeping the second function cos(x) unchanged.
f′(x) = (2 sin(x) cos(x)) cos(x) + sin2(x)(−sin(x))
= 2 sin(x) cos2(x)−sin3(x)
Therefore, the derivative of f(x) = sin2(x) cos(x) is f′(x) = 2 sin(x) cos2(x)−
sin3(x).
Question 26
Question
Find the derivative of the function y= sec2(3x) with respect to x.
Solution
Step 1: Recall the derivative of sec(x).
d
dx (sec(x)) = sec(x) tan(x)
Step 2: Rewrite the given function using basic trigonometric identities.
y= sec2(3x) = 1
cos(3x)2
=1
cos2(3x)
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Step 3: Differentiate using the chain rule.
dy
dx =d
dx 1
cos2(3x)
Step 4: Apply the chain rule to differentiate the function.
dy
dx =−21
cos(3x)
·
d
dx (cos(3x)) = −2 sec(3x) tan(3x)
Step 5: Substitute back the definition of sec(x).
dy
dx =−21
cos(3x)(tan(3x)) = −2 sec(3x) tan(3x)
Therefore, the derivative of y= sec2(3x) with respect to xis −2 sec(3x) tan(3x) .
Question 27
Question
Find the derivative of the function f(x) = sin2(x) cos(x).
Solution
Step 1: Apply the product rule. Step 2: Recall that the product rule states
that if f(x) = u(x)v(x), then f′(x) = u′(x)v(x) + u(x)v′(x). Step 3: Let
u(x) = sin2(x) and v(x) = cos(x). Step 4: Find u′(x) and v′(x).
Step 4: u′(x) = 2 sin(x) cos(x) and v′(x) = −sin(x)
Step 5: Apply the product rule to find f′(x).
Step 5: f′(x) = u′(x)v(x) + u(x)v′(x)
f′(x) = (2 sin(x) cos(x))(cos(x)) + (sin2(x))(−sin(x))
Step 6: Simplify the expression.
Step 6: f′(x) = 2 sin(x) cos2(x)−sin3(x)
Therefore, the derivative of the function f(x) = sin2(x) cos(x) is 2 sin(x) cos2(x)−
sin3(x).
Question 28
Question
Find the derivative of the function f(x) = sin2(x) tan(x).
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Solution
Step 1: Use the product rule to differentiate the function f(x) = sin2(x) tan(x).
The product rule states that if f(x) = u(x)v(x), then f′(x) = u(x)v′(x) +
v(x)u′(x). Step 2: Let u(x) = sin2(x) and v(x) = tan(x). Then, u′(x) =
2 sin(x) cos(x) and v′(x) = sec2(x). Step 3: Apply the product rule to find
f′(x).
f′(x) = u(x)v′(x) + v(x)u′(x) = sin2(x) sec2(x) + tan(x)·2 sin(x) cos(x)
Step 4: Simplify the expression by using trigonometric identities. Step 5: Recall
that tan(x) = sin(x)
cos(x)and sec(x) = 1
cos(x). Step 6: Substitute these identities into
the expression for f′(x).
f′(x) = sin2(x)
cos2(x)+2 sin(x) cos(x) sin(x)
cos(x)
Step 7: Simplify the expression further. Step 8: Combine the terms to get the
final derivative.
f′(x) = sin2(x) sec2(x) + 2 sin2(x) = sin2(x)(1 + 2 sec2(x))
Question 29
Question
Find the derivative of y= sin2(3x) with respect to x.
Solution
Step 1: To find the derivative of y= sin2(3x), we will use the chain rule. Let
u= 3xand v= sin(u). Then y=v2.
Step 2: Find dy
du by differentiating y=v2with respect to uusing the power
rule: dy
du = 2v.
Step 3: Find du
dx to apply the chain rule. Since u= 3x, we have du
dx = 3.
Step 4: Combine the results from Step 2 and Step 3 using the chain rule:
dy
dx =dy
du
·du
dx = 2v·3.
Step 5: Substitute back v= sin(u), we get dy
dx = 2 sin(3x)·3.
Step 6: Simplify to obtain the final answer: dy
dx = 6 sin(3x).
Question 30
Question
Find the derivative of the function f(x) = cos3(2x).
17
Solution
Step 1: Apply the chain rule for differentiation. Step 2: Recall that the deriva-
tive of cos(u) is −sin(u)·u′. Step 3: Let u= 2x, then u′= 2. Step 4:
Substitute u= 2xand u′= 2 into the derivative formula for cos(u). Step 5:
Find the derivative of f(x) = cos3(2x).
Step 1: Apply the chain rule for differentiation:
f′(x) = −3 cos2(2x) sin(2x)·2
Step 2: Recall that d
dx [cos(u)] = −sin(u)·u′:
f′(x) = −3 (cos(2x))2
·sin(2x)·2
Step 3: Let u= 2x, then u′=2:
f′(x) = −3 (cos(u))2
·sin(u)·2
Step 4: Substitute u= 2x, and u′= 2 into the derivative formula for cos(u) :
f′(x) = −3 cos2(2x) sin(2x)·2
Step 5: Simplify the derivative of f(x) :
f′(x) = −6 cos2(2x) sin(2x)
Question 31
Question
Find the derivative of the function f(x) = cos(2x) sin(3x).
Solution
Step 1: Apply the product rule to differentiate the function f(x). Step 2: Let
u= cos(2x) and v= sin(3x). Step 3: Find u′and v′. Step 4: Use the product
rule: f′(x) = u′
·v+u·v′.
Step 1: Apply the product rule to differentiate f(x):
f′(x) = d
dx (cos(2x)·sin(3x))
Step 2: Let u= cos(2x) and v= sin(3x), so u′=−2 sin(2x) and v′=
3 cos(3x).
Step 3: Find u′and v′:
u′=−2 sin(2x)
v′= 3 cos(3x)
Step 4: Use the product rule:
f′(x) = u′
·v+u·v′
18
f′(x) = (−2 sin(2x)) ·sin(3x) + cos(2x)·(3 cos(3x))
f′(x) = −2 sin(2x) sin(3x) + 3 cos(2x) cos(3x)
Question 32
Question
Differentiate the function f(x) = sinx2cos(x) with respect to x.
Solution
Step 1: Apply the product rule, which states that if f(x) = u(x)v(x), then
f′(x) = u′(x)v(x) + u(x)v′(x).
Step 2: Let u(x) = sinx2and v(x) = cos(x). Then, we have u′(x) =
cosx2·2xand v′(x) = −sin(x).
Step 3: Using the product rule, we can find f′(x) as follows:
f′(x) = u′(x)v(x) + u(x)v′(x)
= (cosx2·2x)·(cos(x)) + (sinx2)·(−sin(x))
= 2xcosx2cos(x)−sinx2sin(x).
Therefore, the derivative of f(x) = sinx2cos(x) with respect to xis f′(x) =
2xcosx2cos(x)−sinx2sin(x).
Question 33
Question
Find the derivative of y= sin2(x) cos3(x) with respect to x.
Solution
Step 1: Apply the product rule to differentiate y= sin2(x) cos3(x).
Step 2: Let u= sin2(x) and v= cos3(x). Then,
u′= 2 sin(x) cos(x) and v′=−3 cos2(x) sin(x).
Step 3: The product rule states that (uv)′=u′v+uv′, so
dy
dx =u′v+uv′.
Step 4: Substitute u,v,u′, and v′into the equation to find the derivative:
dy
dx = (2 sin(x) cos(x))(cos3(x)) + (sin2(x))(−3 cos2(x) sin(x)).
19
Step 5: Simplify the expression by distributing and combining like terms:
dy
dx = 2 sin(x) cos(x) cos3(x)−3 sin2(x) cos2(x) sin(x).
Step 6: Finally, the derivative of y= sin2(x) cos3(x) with respect to xis
dy
dx = 2 sin(x) cos4(x)−3 sin2(x) cos2(x) sin(x).
Question 34
Question
Find the derivative of y=sin2x+cos2x
sin xcos x.
Solution
Step 1: Simplify the expression.
sin2x+ cos2x= 1.
y=1
sin xcos x.
Step 2: Rewrite the expression using trigonometric identities.
y=1
1
2sin 2x= 2 csc 2x.
Step 3: Differentiate ywith respect to x.
d
dx (2 csc 2x) = −2 csc 2xcot 2x.
Therefore, the derivative of ywith respect to xis −2 csc 2xcot 2x.
Question 35
Question
Find the derivative of f(x) = sin3x2−2xwith respect to x.
Solution
Step 1: Apply the chain rule, which states that if g(x) = sin(u(x)), then g′(x) =
u′(x) cos(u(x)).
f′(x) = d
dx [sin3x2−2x] = cos3x2−2x·
d
dx (3x2−2x)
20
Solution
Step 1: Apply the quotient rule to differentiate the function f(x) = 4 sin(x) cos(x)
3x.
Step 2: Let u(x) = 4 sin(x) cos(x) and v(x) = 3x. Then, the derivative of
f(x) is given by
f′(x) = u′v−uv′
v2
=(4 cos(x) cos(x)−4 sin(x)(−sin(x))) ·3x−4 sin(x) cos(x)·3
(3x)2
=(4 cos2(x) + 4 sin2(x)) ·3x+ 4 sin(x) cos(x)·3
9x2
=12x
9x2
=4
3x.
Therefore, the derivative of f(x) = 4 sin(x) cos(x)
3xis f′(x) = 4
3x.
Question 3
Question
Find the derivative of the function f(x) = sin2(x) cos(3x).
Solution
Step 1: Use the product rule to differentiate the function. The product rule
states that if uand vare differentiable functions of x, then the derivative of
their product is given by (uv)′=u′v+uv′. Step 2: Let u= sin2(x) and
v= cos(3x). Step 3: Compute u′and v′:
u′= (sin(x))2)′= 2 sin(x) cos(x)
v′= (cos(3x))′=−3 sin(3x)
Step 4: Apply the product rule formula to find f′(x):
f′(x) = u′v+uv′
f′(x) = 2 sin(x) cos(x)·cos(3x) + sin2(x)·(−3 sin(3x))
Step 5: Simplify the expression:
f′(x) = 2 sin(x) cos(x) cos(3x)−3 sin2(x) sin(3x)
Step 6: Apply trigonometric identities to simplify further:
f′(x) = sin(2x) cos(3x)−
3
2sin(2x)
2
Step 7: Finally, simplify the expression:
f′(x) = 1
2sin(2x)(2 cos(3x)−3)
Therefore, the derivative of the function f(x) = sin2(x) cos(3x) is f′(x) =
1
2sin(2x)(2 cos(3x)−3).
Question 4
Question
Find the derivative of the function f(x) = sin(2x)−cos(3x).
Solution
Step 1: Use the differentiation rules for trigonometric functions to find the
derivative of f(x).
f(x) = sin(2x)−cos(3x)
f′(x) = d
dx (sin(2x)) −
d
dx (cos(3x))
Step 2: Apply the chain rule and the derivative of sine function to sin(2x).
d
dx (sin(2x)) = cos(2x)·
d
dx (2x)
= 2 cos(2x)
Step 3: Apply the chain rule and the derivative of cosine function to cos(3x).
d
dx (cos(3x)) = −sin(3x)·
d
dx (3x)
=−3 sin(3x)
Step 4: Put the derivatives of sin(2x) and cos(3x) back into the expression
for f′(x).
f′(x) = 2 cos(2x)−3 sin(3x)
= 2 cos(2x)−3 sin(3x)
Therefore, the derivative of the function f(x) = sin(2x)−cos(3x) is f′(x) =
2 cos(2x)−3 sin(3x).
Question 5
Question
Find the derivative of f(x) = sin2(3x) + cos(2x).
3
Solution
To find the derivative of f(x), we will differentiate each term separately using
the rules of differentiation.
f(x) = sin2(3x) + cos(2x)
Step 1: Differentiate sin2(3x) using the chain rule.
d
dx (sin2(3x)) = 2 sin(3x) cos(3x)·3
= 6 sin(3x) cos(3x)
Step 2: Differentiate cos(2x).
d
dx (cos(2x)) = −sin(2x)·2
=−2 sin(2x)
Step 3: Combine the derivatives to find f′(x).
f′(x) = 6 sin(3x) cos(3x)−2 sin(2x)
Therefore, the derivative of f(x) = sin2(3x)+cos(2x) is f′(x) = 6 sin(3x) cos(3x)−
2 sin(2x).
Question 6
Question
Find the derivative of the following function with respect to x:f(x) = tan2(3x)+
sin(2x) cos(4x).
Solution
To find the derivative of f(x) with respect to x, we will differentiate each term
separately using the rules of differentiation. Step 1: Apply the chain rule to dif-
ferentiate tan2(3x). Step 2: Differentiate sin(2x). Step 3: Differentiate cos(4x).
Step 4: Add up the derivatives to find the derivative of f(x).
Step 1: Let u= tan(3x). Using the chain rule, we have:
d
dx (tan2(3x)) = d
dx (u2)=2u·
du
dx
Step 2:
d
dx (sin(2x)) = 2 cos(2x)
4
Step 3:
d
dx (cos(4x)) = −4 sin(4x)
Step 4: Putting it all together, we have:
f′(x) = 2 tan(3x)·3 sec2(3x) + 2 cos(2x) cos(4x)−4 sin(2x) sin(4x)
Therefore, the derivative of f(x) with respect to xis:
f′(x) = 6 tan(3x) sec2(3x) + 2 cos(2x) cos(4x)−4 sin(2x) sin(4x)
Question 7
Question
Find the derivative of the function f(x) = sin(3x) cos(2x).
Solution
To find the derivative of f(x) = sin(3x) cos(2x), we will use the product rule.
Step 1: Apply the product rule:
f′(x) = (sin(3x))′
·cos(2x) + sin(3x)·(cos(2x))′
Step 2: Find the derivatives of sin(3x) and cos(2x).
(sin(3x))′= 3 cos(3x)
(cos(2x))′=−2 sin(2x)
Step 3: Substitute the derivatives back into the product rule formula:
f′(x) = 3 cos(3x)·cos(2x) + sin(3x)·(−2 sin(2x))
Step 4: Simplify the expression:
f′(x) = 3 cos(3x) cos(2x)−2 sin(3x) sin(2x)
So, the derivative of the function f(x) = sin(3x) cos(2x) is 3 cos(3x) cos(2x)−
2 sin(3x) sin(2x).
Question 8
Question
Find the derivative of the following function: f(x) = sin2(3x) + cos(2x).
5
Solution
To find the derivative of f(x), we will use the power rule, chain rule, and sum
rule for differentiation.
Step 1: Start by finding the derivative of sin2(3x).
d
dx (sin2(3x)) = 2 sin(3x) cos(3x)·3
= 6 sin(3x) cos(3x)
Step 2: Next, find the derivative of cos(2x).
d
dx (cos(2x)) = −sin(2x)·2
=−2 sin(2x)
Step 3: Now, add the derivatives of sin2(3x) and cos(2x) to find the deriva-
tive of f(x).
f′(x) = 6 sin(3x) cos(3x)−2 sin(2x)
= 6 ·
1
2
·sin(6x)−2 sin(2x)
= 3 sin(6x)−2 sin(2x)
Therefore, the derivative of f(x) = sin2(3x) + cos(2x) is f′(x) = 3 sin(6x)−
2 sin(2x).
Question 9
Question
Find the derivative of the function f(x) = sin(x) cos(x).
Solution
Step 1: To find the derivative of f(x), we will use the product rule, which states
that the derivative of the product of two functions u(x) and v(x) is given by:
(uv)′=u′v+uv′.
Step 2: Let u(x) = sin(x) and v(x) = cos(x). Then, we have:
u′(x) = cos(x) and v′(x) = −sin(x)
Step 3: Using the product rule, we can find the derivative of f(x):
f′(x) = (sin(x))(cos(x))′+ (sin(x))′(cos(x))
f′(x) = cos(x) cos(x) + sin(x)(−sin(x))
6
f′(x) = cos2(x)−sin2(x)
Step 4: Recall the Pythagorean trigonometric identity: cos2(x)−sin2(x) =
cos(2x). Substituting this into f′(x) gives:
f′(x) = cos(2x)
Therefore, the derivative of f(x) = sin(x) cos(x) is f′(x) = cos(2x).
Question 10
Question
Find the derivative of the function f(x) = sin2(x) + cos2(x).
Solution
Step 1: Recall the trigonometric identity sin2(x) + cos2(x) = 1. Step 2: Rewrite
f(x) in terms of sin2(x). Step 3: Differentiate f(x) using the chain rule. Step
4: Simplify the derivative to obtain the final answer.
Question 11
Question
Find the derivative of the function f(x) = sin(3x) cos(2x).
Solution
To find the derivative of the given function f(x) = sin(3x) cos(2x), we will use
the product rule of differentiation.
Step 1: Apply the product rule, which states that if f(x) = u(x)v(x), then
f′(x) = u′v+uv′. Let u(x) = sin(3x) and v(x) = cos(2x).
Step 2: Find u′(x) and v′(x): - u′(x) = 3 cos(3x) (derivative of sin(3x)
using the chain rule) - v′(x) = −2 sin(2x) (derivative of cos(2x) using the chain
rule)
Step 3: Apply the product rule to find f′(x):
f′(x) = u′v+uv′= (3 cos(3x))(cos(2x)) + (sin(3x))(−2 sin(2x))
Step 4: Simplify the expression:
f′(x) = 3 cos(3x) cos(2x)−2 sin(3x) sin(2x)
Thus, the derivative of the function f(x) = sin(3x) cos(2x) is 3 cos(3x) cos(2x)−
2 sin(3x) sin(2x).
7
Question 12
Question
Find the derivative of the function f(x) = sin3(2x) + cos3(3x).
Solution
Step 1: To find the derivative of f(x), we will use the chain rule and the power
rule for differentiation.
Step 2: Let’s start by finding the derivative of sin3(2x). Using the chain
rule, we have
d
dx (sin3(2x)) = 3 sin2(2x)·cos(2x)·2.
Step 3: Simplifying the above expression, we get
d
dx (sin3(2x)) = 6 sin2(2x) cos(2x).
Step 4: Next, let’s find the derivative of cos3(3x). Using the chain rule, we
have d
dx (cos3(3x)) = 3 cos2(3x)·(−sin(3x)) ·3.
Step 5: Simplifying the above expression, we get
d
dx (cos3(3x)) = −9 cos2(3x) sin(3x).
Step 6: Therefore, the derivative of the function f(x) = sin3(2x) + cos3(3x)
is
f′(x) = 6 sin2(2x) cos(2x)−9 cos2(3x) sin(3x).
Question 13
Question
Find the derivative of f(x) = sin2(3x) with respect to x.
Solution
Step 1: Apply the chain rule. Step 2: Identify the inner function and the
outer function. Step 3: Let u= 3x, then the function can be rewritten as
g(u) = sin2u. Step 4: Find the derivative of the outer function. Step 5: Find
the derivative of the inner function. Step 6: Apply the chain rule formula to
find the derivative of the composite function.
Therefore, the derivative of f(x) = sin2(3x) with respect to xis 6 sin(3x) cos(3x) .
8
Question 14
Question
Find the derivative of the function f(x) = cos(2x)·sin(3x).
Solution
Step 1: Apply the product rule to find the derivative of the function f(x) =
cos(2x)·sin(3x).
Let u(x) = cos(2x) and v(x) = sin(3x).
The product rule states that (uv)′=u′v+uv′.
Step 2: Find u′(x) and v′(x).
Using the chain rule, we have u′(x) = −2 sin(2x) and v′(x) = 3 cos(3x).
Step 3: Apply the product rule to find f′(x).
f′(x)=(u(x)v(x))′
=u′(x)v(x) + u(x)v′(x)
= (−2 sin(2x)) sin(3x) + cos(2x)(3 cos(3x))
=−2 sin(2x) sin(3x) + 3 cos(2x) cos(3x).
Therefore, the derivative of the function f(x) = cos(2x)·sin(3x) is f′(x) =
−2 sin(2x) sin(3x) + 3 cos(2x) cos(3x).
Question 15
Question
Find the derivative of the function f(x) = sin2(2x)
1+cos(2x).
Solution
Step 1: To find the derivative of f(x), we will first rewrite it using trigonometric
identities. Step 2: Recall the double angle identity: sin(2x) = 2 sin(x) cos(x)
and cos(2x) = cos2(x)−sin2(x). Step 3: Substituting these identities into f(x),
we get:
f(x) = (2 sin(x) cos(x))2
1 + (cos2(x)−sin2(x))
Step 4: Simplifying the expression, we have:
f(x) = 4 sin2(x) cos2(x)
2 cos2(x)
Step 5: Further simplifying, we obtain:
f(x) = 2 sin2(x)
9
Step 6: Now, differentiate f(x) with respect to xusing the power rule and
chain rule for trigonometric functions. Step 7: The derivative of 2 sin2(x) is
4 sin(x) cos(x) by the chain rule. Step 8: Therefore, the derivative of f(x) is
4 sin(x) cos(x) .
Question 16
Question
Find the derivative of the function f(x) = sin(x) cos(x).
Solution
To find the derivative of f(x) = sin(x) cos(x), we will use the product rule of
differentiation.
Step 1: Apply the product rule. Let u= sin(x) and v= cos(x). The
product rule states that (uv)′=u′v+uv′.
Step 2: Find u′and v′.
u′= cos(x) (derivative of sin(x))
v′=−sin(x) (derivative of cos(x))
Step 3: Substitute u,v,u′, and v′into the product rule formula.
f′(x) = u′v+uv′
= (cos(x))(cos(x)) + (sin(x))(−sin(x))
= cos2(x)−sin2(x)
Step 4: Use the trigonometric identity cos2(x)−sin2(x) = cos(2x).
f′(x) = cos(2x)
So, the derivative of f(x) = sin(x) cos(x) is f′(x) = cos(2x).
Question 17
Question
Find the derivative of the function f(x) = sin2(4x) cos(3x).
Solution
To find the derivative of the given function, we will use the product rule and
chain rule for differentiation.
Step 1: Apply the product rule: (uv)′=u′v+uv′, where u= sin2(4x) and
v= cos(3x).
10
Let u= sin2(4x), then u′= 2 sin(4x) cos(4x) by the chain rule.
Let v= cos(3x), then v′=−3 sin(3x) by the chain rule.
Step 2: Substitute u′,v′,u, and vinto the product rule formula.
(f(x))′= (u′v+uv′)
= (2 sin(4x) cos(4x))(cos(3x)) + (sin2(4x))(−3 sin(3x))
= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin2(4x)
Step 3: Simplify the expression if possible.
(f(x))′= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin2(4x)
= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin(4x) sin(4x)
= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin(4x) cos(4x)
Therefore, the derivative of the function f(x) = sin2(4x) cos(3x) is 2 sin(4x) cos(4x) cos(3x)−
3 sin(3x) sin(4x) cos(4x).
Question 18
Question
Find the derivative of y= sin2(3x) with respect to x.
Solution
Step 1: Apply the chain rule to differentiate y= sin2(3x). Step 2: Let u=
sin(3x) and y=u2. Step 3: Find dy
du and du
dx . Step 4: Use the chain rule:
dy
dx =dy
du
·du
dx .
Step 1: Apply the chain rule to differentiate y= sin2(3x). Step 2: Let
u= sin(3x) and y=u2. Step 3: Find dy
du and du
dx .
dy
du = 2u
du
dx = 3 cos(3x)
Step 4: Use the chain rule:
dy
dx =dy
du
·
du
dx = 2u·3 cos(3x) = 6 sin(3x) cos(3x)
Therefore, the derivative of y= sin2(3x) with respect to xis 6 sin(3x) cos(3x).
11
Question 19
Question
Find the derivative of f(x) = sin(2x) cos(3x).
Solution
Step 1: Apply the product rule, (uv)′=u′v+uv′, where u= sin(2x) and
v= cos(3x). Step 2: Find u′and v′. Step 3: Calculate u′:
u= sin(2x)
u′=d
dx (sin(2x)) = 2 cos(2x)
Step 4: Calculate v′:
v= cos(3x)
v′=d
dx (cos(3x)) = −3 sin(3x)
Step 5: Apply the product rule:
f′(x) = u′v+uv′
f′(x) = (2 cos(2x))(cos(3x)) + (sin(2x))(−3 sin(3x))
Step 6: Simplify the expression:
f′(x) = 2 cos(2x) cos(3x)−3 sin(2x) sin(3x)
f′(x) = 2 cos(2x) cos(3x)−3 cosπ
2
−3x
f′(x) = 2 cos(2x) cos(3x)−3 cosπ
2cos(3x) + 3 sinπ
2sin(3x)
f′(x) = 2 cos(2x) cos(3x)−3 cos(3x)
Step 7: Therefore, the derivative of f(x) = sin(2x) cos(3x) is f′(x) = 2 cos(2x) cos(3x)−
3 cos(3x).
Question 20
Question
Let f(x) = cos(5x) sin(3x). Find f′(x).
12
Solution
Step 1: Apply the product rule to differentiate f(x). Step 2: Let u= cos(5x)
and v= sin(3x). Step 3: Find u′and v′. Step 4: Calculate u′v+uv′to find
f′(x).
Step 1: Apply the product rule:
f′(x) = d
dx (cos(5x)) sin(3x) + cos(5x)d
dx (sin(3x))
Step 2: Let u= cos(5x) and v= sin(3x).
Step 3: Calculate u′and v′:
u′=−5 sin(5x)
v′= 3 cos(3x)
Step 4: Substitute u′and v′back into the product rule formula:
f′(x) = −5 sin(5x) sin(3x) + cos(5x)3 cos(3x)
Therefore, f′(x) = −5 sin(5x) sin(3x) + 3 cos(5x) cos(3x).
Question 21
Question
Find the derivative of the function f(x) = cos(3x) sin(4x).
Solution
Step 1: Apply the product rule to differentiate the product of cos(3x) and
sin(4x). Step 2: Recall that the derivative of cos(u) is −sin(u)·u′and the
derivative of sin(u) is cos(u)·u′. Step 3: Let u= 3xand v= sin(4x). Step
4: Calculate u′and v′. Step 5: Find the derivative of the function f(x) =
cos(3x) sin(4x) by applying the product rule. Step 6: Combine the results to
obtain the final expression for f′(x).
Question 22
Question
Find the derivative of the function f(x) = sin2(3x) + cos(2x).
13
Solution
Step 1: Apply the chain rule to differentiate sin2(3x).
d
dx (sin2(3x)) = 2 sin(3x) cos(3x)·3 = 6 sin(3x) cos(3x)
Step 2: Apply the chain rule to differentiate cos(2x).
d
dx (cos(2x)) = −sin(2x)·2 = −2 sin(2x)
Step 3: Add the derivatives of the two parts together.
f′(x) = 6 sin(3x) cos(3x)−2 sin(2x)
Question 23
Question
Find the derivative of the function y= sin(x) cos(x).
Solution
We can use the product rule to find the derivative of the given function.
Step 1: Apply the product rule. Let f(x) = sin(x) and g(x) = cos(x). The
product rule states that the derivative of the product of two functions is given
by
(f·g)′=f′
·g+f·g′.
Therefore, the derivative of y= sin(x) cos(x) is
y′= (sin(x))′cos(x) + sin(x)(cos(x))′.
Step 2: Find the derivatives of sin(x) and cos(x). The derivative of sin(x)
is cos(x) and the derivative of cos(x) is −sin(x).
Step 3: Substitute the derivatives into the product rule. Substitute cos(x)
for (sin(x))′and −sin(x) for (cos(x))′into the derivative expression:
y′= cos(x) cos(x) + sin(x)(−sin(x)).
Step 4: Simplify the expression. Simplify the expression to get the final
answer:
y′= cos2(x)−sin2(x).
Therefore, the derivative of y= sin(x) cos(x) is y′= cos2(x)−sin2(x).
Question 24
Question
Find the derivative of the function f(x) = sin(2x) cos(3x).
14
Solution
Step 1: Apply the product rule to differentiate f(x) = sin(2x) cos(3x), which
states that the derivative of the product of two functions is the first function
times the derivative of the second, plus the second function times the derivative
of the first. Step 2: Let u= sin(2x) and v= cos(3x). Then we have f(x) = u·v.
Step 3: Find u′and v′. Step 4: Calculate u′:u′=d
dx [sin(2x)] Step 5: Apply
the chain rule to differentiate sin(2x): u′= 2 cos(2x) Step 6: Calculate v′:
v′=d
dx [cos(3x)] Step 7: Apply the chain rule to differentiate cos(3x): v′=
−3 sin(3x) Step 8: Now, apply the product rule to find f′(x): f′(x) = u′v+uv′
Step 9: Substitute u= sin(2x), u′= 2 cos(2x), v= cos(3x), and v′=−3 sin(3x)
into the formula for f′(x). Step 10: Simplify the expression for f′(x) to get the
final answer.
Question 25
Question
Find the derivative of the function f(x) = sin2(x) cos(x).
Solution
Step 1: Apply the product rule by differentiating the first function sin2(x) and
keeping the second function cos(x) unchanged.
f′(x) = (2 sin(x) cos(x)) cos(x) + sin2(x)(−sin(x))
= 2 sin(x) cos2(x)−sin3(x)
Therefore, the derivative of f(x) = sin2(x) cos(x) is f′(x) = 2 sin(x) cos2(x)−
sin3(x).
Question 26
Question
Find the derivative of the function y= sec2(3x) with respect to x.
Solution
Step 1: Recall the derivative of sec(x).
d
dx (sec(x)) = sec(x) tan(x)
Step 2: Rewrite the given function using basic trigonometric identities.
y= sec2(3x) = 1
cos(3x)2
=1
cos2(3x)
15
Step 3: Differentiate using the chain rule.
dy
dx =d
dx 1
cos2(3x)
Step 4: Apply the chain rule to differentiate the function.
dy
dx =−21
cos(3x)
·
d
dx (cos(3x)) = −2 sec(3x) tan(3x)
Step 5: Substitute back the definition of sec(x).
dy
dx =−21
cos(3x)(tan(3x)) = −2 sec(3x) tan(3x)
Therefore, the derivative of y= sec2(3x) with respect to xis −2 sec(3x) tan(3x) .
Question 27
Question
Find the derivative of the function f(x) = sin2(x) cos(x).
Solution
Step 1: Apply the product rule. Step 2: Recall that the product rule states
that if f(x) = u(x)v(x), then f′(x) = u′(x)v(x) + u(x)v′(x). Step 3: Let
u(x) = sin2(x) and v(x) = cos(x). Step 4: Find u′(x) and v′(x).
Step 4: u′(x) = 2 sin(x) cos(x) and v′(x) = −sin(x)
Step 5: Apply the product rule to find f′(x).
Step 5: f′(x) = u′(x)v(x) + u(x)v′(x)
f′(x) = (2 sin(x) cos(x))(cos(x)) + (sin2(x))(−sin(x))
Step 6: Simplify the expression.
Step 6: f′(x) = 2 sin(x) cos2(x)−sin3(x)
Therefore, the derivative of the function f(x) = sin2(x) cos(x) is 2 sin(x) cos2(x)−
sin3(x).
Question 28
Question
Find the derivative of the function f(x) = sin2(x) tan(x).
16
Solution
Step 1: Use the product rule to differentiate the function f(x) = sin2(x) tan(x).
The product rule states that if f(x) = u(x)v(x), then f′(x) = u(x)v′(x) +
v(x)u′(x). Step 2: Let u(x) = sin2(x) and v(x) = tan(x). Then, u′(x) =
2 sin(x) cos(x) and v′(x) = sec2(x). Step 3: Apply the product rule to find
f′(x).
f′(x) = u(x)v′(x) + v(x)u′(x) = sin2(x) sec2(x) + tan(x)·2 sin(x) cos(x)
Step 4: Simplify the expression by using trigonometric identities. Step 5: Recall
that tan(x) = sin(x)
cos(x)and sec(x) = 1
cos(x). Step 6: Substitute these identities into
the expression for f′(x).
f′(x) = sin2(x)
cos2(x)+2 sin(x) cos(x) sin(x)
cos(x)
Step 7: Simplify the expression further. Step 8: Combine the terms to get the
final derivative.
f′(x) = sin2(x) sec2(x) + 2 sin2(x) = sin2(x)(1 + 2 sec2(x))
Question 29
Question
Find the derivative of y= sin2(3x) with respect to x.
Solution
Step 1: To find the derivative of y= sin2(3x), we will use the chain rule. Let
u= 3xand v= sin(u). Then y=v2.
Step 2: Find dy
du by differentiating y=v2with respect to uusing the power
rule: dy
du = 2v.
Step 3: Find du
dx to apply the chain rule. Since u= 3x, we have du
dx = 3.
Step 4: Combine the results from Step 2 and Step 3 using the chain rule:
dy
dx =dy
du
·du
dx = 2v·3.
Step 5: Substitute back v= sin(u), we get dy
dx = 2 sin(3x)·3.
Step 6: Simplify to obtain the final answer: dy
dx = 6 sin(3x).
Question 30
Question
Find the derivative of the function f(x) = cos3(2x).
17
Solution
Step 1: Apply the chain rule for differentiation. Step 2: Recall that the deriva-
tive of cos(u) is −sin(u)·u′. Step 3: Let u= 2x, then u′= 2. Step 4:
Substitute u= 2xand u′= 2 into the derivative formula for cos(u). Step 5:
Find the derivative of f(x) = cos3(2x).
Step 1: Apply the chain rule for differentiation:
f′(x) = −3 cos2(2x) sin(2x)·2
Step 2: Recall that d
dx [cos(u)] = −sin(u)·u′:
f′(x) = −3 (cos(2x))2
·sin(2x)·2
Step 3: Let u= 2x, then u′=2:
f′(x) = −3 (cos(u))2
·sin(u)·2
Step 4: Substitute u= 2x, and u′= 2 into the derivative formula for cos(u) :
f′(x) = −3 cos2(2x) sin(2x)·2
Step 5: Simplify the derivative of f(x) :
f′(x) = −6 cos2(2x) sin(2x)
Question 31
Question
Find the derivative of the function f(x) = cos(2x) sin(3x).
Solution
Step 1: Apply the product rule to differentiate the function f(x). Step 2: Let
u= cos(2x) and v= sin(3x). Step 3: Find u′and v′. Step 4: Use the product
rule: f′(x) = u′
·v+u·v′.
Step 1: Apply the product rule to differentiate f(x):
f′(x) = d
dx (cos(2x)·sin(3x))
Step 2: Let u= cos(2x) and v= sin(3x), so u′=−2 sin(2x) and v′=
3 cos(3x).
Step 3: Find u′and v′:
u′=−2 sin(2x)
v′= 3 cos(3x)
Step 4: Use the product rule:
f′(x) = u′
·v+u·v′
18
f′(x) = (−2 sin(2x)) ·sin(3x) + cos(2x)·(3 cos(3x))
f′(x) = −2 sin(2x) sin(3x) + 3 cos(2x) cos(3x)
Question 32
Question
Differentiate the function f(x) = sinx2cos(x) with respect to x.
Solution
Step 1: Apply the product rule, which states that if f(x) = u(x)v(x), then
f′(x) = u′(x)v(x) + u(x)v′(x).
Step 2: Let u(x) = sinx2and v(x) = cos(x). Then, we have u′(x) =
cosx2·2xand v′(x) = −sin(x).
Step 3: Using the product rule, we can find f′(x) as follows:
f′(x) = u′(x)v(x) + u(x)v′(x)
= (cosx2·2x)·(cos(x)) + (sinx2)·(−sin(x))
= 2xcosx2cos(x)−sinx2sin(x).
Therefore, the derivative of f(x) = sinx2cos(x) with respect to xis f′(x) =
2xcosx2cos(x)−sinx2sin(x).
Question 33
Question
Find the derivative of y= sin2(x) cos3(x) with respect to x.
Solution
Step 1: Apply the product rule to differentiate y= sin2(x) cos3(x).
Step 2: Let u= sin2(x) and v= cos3(x). Then,
u′= 2 sin(x) cos(x) and v′=−3 cos2(x) sin(x).
Step 3: The product rule states that (uv)′=u′v+uv′, so
dy
dx =u′v+uv′.
Step 4: Substitute u,v,u′, and v′into the equation to find the derivative:
dy
dx = (2 sin(x) cos(x))(cos3(x)) + (sin2(x))(−3 cos2(x) sin(x)).
19
Step 5: Simplify the expression by distributing and combining like terms:
dy
dx = 2 sin(x) cos(x) cos3(x)−3 sin2(x) cos2(x) sin(x).
Step 6: Finally, the derivative of y= sin2(x) cos3(x) with respect to xis
dy
dx = 2 sin(x) cos4(x)−3 sin2(x) cos2(x) sin(x).
Question 34
Question
Find the derivative of y=sin2x+cos2x
sin xcos x.
Solution
Step 1: Simplify the expression.
sin2x+ cos2x= 1.
y=1
sin xcos x.
Step 2: Rewrite the expression using trigonometric identities.
y=1
1
2sin 2x= 2 csc 2x.
Step 3: Differentiate ywith respect to x.
d
dx (2 csc 2x) = −2 csc 2xcot 2x.
Therefore, the derivative of ywith respect to xis −2 csc 2xcot 2x.
Question 35
Question
Find the derivative of f(x) = sin3x2−2xwith respect to x.
Solution
Step 1: Apply the chain rule, which states that if g(x) = sin(u(x)), then g′(x) =
u′(x) cos(u(x)).
f′(x) = d
dx [sin3x2−2x] = cos3x2−2x·
d
dx (3x2−2x)
20
Solution
Step 1: Apply the quotient rule to differentiate the function f(x) = 4 sin(x) cos(x)
3x.
Step 2: Let u(x) = 4 sin(x) cos(x) and v(x) = 3x. Then, the derivative of
f(x) is given by
f′(x) = u′v−uv′
v2
=(4 cos(x) cos(x)−4 sin(x)(−sin(x))) ·3x−4 sin(x) cos(x)·3
(3x)2
=(4 cos2(x) + 4 sin2(x)) ·3x+ 4 sin(x) cos(x)·3
9x2
=12x
9x2
=4
3x.
Therefore, the derivative of f(x) = 4 sin(x) cos(x)
3xis f′(x) = 4
3x.
Question 3
Question
Find the derivative of the function f(x) = sin2(x) cos(3x).
Solution
Step 1: Use the product rule to differentiate the function. The product rule
states that if uand vare differentiable functions of x, then the derivative of
their product is given by (uv)′=u′v+uv′. Step 2: Let u= sin2(x) and
v= cos(3x). Step 3: Compute u′and v′:
u′= (sin(x))2)′= 2 sin(x) cos(x)
v′= (cos(3x))′=−3 sin(3x)
Step 4: Apply the product rule formula to find f′(x):
f′(x) = u′v+uv′
f′(x) = 2 sin(x) cos(x)·cos(3x) + sin2(x)·(−3 sin(3x))
Step 5: Simplify the expression:
f′(x) = 2 sin(x) cos(x) cos(3x)−3 sin2(x) sin(3x)
Step 6: Apply trigonometric identities to simplify further:
f′(x) = sin(2x) cos(3x)−
3
2sin(2x)
2
Step 7: Finally, simplify the expression:
f′(x) = 1
2sin(2x)(2 cos(3x)−3)
Therefore, the derivative of the function f(x) = sin2(x) cos(3x) is f′(x) =
1
2sin(2x)(2 cos(3x)−3).
Question 4
Question
Find the derivative of the function f(x) = sin(2x)−cos(3x).
Solution
Step 1: Use the differentiation rules for trigonometric functions to find the
derivative of f(x).
f(x) = sin(2x)−cos(3x)
f′(x) = d
dx (sin(2x)) −
d
dx (cos(3x))
Step 2: Apply the chain rule and the derivative of sine function to sin(2x).
d
dx (sin(2x)) = cos(2x)·
d
dx (2x)
= 2 cos(2x)
Step 3: Apply the chain rule and the derivative of cosine function to cos(3x).
d
dx (cos(3x)) = −sin(3x)·
d
dx (3x)
=−3 sin(3x)
Step 4: Put the derivatives of sin(2x) and cos(3x) back into the expression
for f′(x).
f′(x) = 2 cos(2x)−3 sin(3x)
= 2 cos(2x)−3 sin(3x)
Therefore, the derivative of the function f(x) = sin(2x)−cos(3x) is f′(x) =
2 cos(2x)−3 sin(3x).
Question 5
Question
Find the derivative of f(x) = sin2(3x) + cos(2x).
3
Solution
To find the derivative of f(x), we will differentiate each term separately using
the rules of differentiation.
f(x) = sin2(3x) + cos(2x)
Step 1: Differentiate sin2(3x) using the chain rule.
d
dx (sin2(3x)) = 2 sin(3x) cos(3x)·3
= 6 sin(3x) cos(3x)
Step 2: Differentiate cos(2x).
d
dx (cos(2x)) = −sin(2x)·2
=−2 sin(2x)
Step 3: Combine the derivatives to find f′(x).
f′(x) = 6 sin(3x) cos(3x)−2 sin(2x)
Therefore, the derivative of f(x) = sin2(3x)+cos(2x) is f′(x) = 6 sin(3x) cos(3x)−
2 sin(2x).
Question 6
Question
Find the derivative of the following function with respect to x:f(x) = tan2(3x)+
sin(2x) cos(4x).
Solution
To find the derivative of f(x) with respect to x, we will differentiate each term
separately using the rules of differentiation. Step 1: Apply the chain rule to dif-
ferentiate tan2(3x). Step 2: Differentiate sin(2x). Step 3: Differentiate cos(4x).
Step 4: Add up the derivatives to find the derivative of f(x).
Step 1: Let u= tan(3x). Using the chain rule, we have:
d
dx (tan2(3x)) = d
dx (u2)=2u·
du
dx
Step 2:
d
dx (sin(2x)) = 2 cos(2x)
4
Step 3:
d
dx (cos(4x)) = −4 sin(4x)
Step 4: Putting it all together, we have:
f′(x) = 2 tan(3x)·3 sec2(3x) + 2 cos(2x) cos(4x)−4 sin(2x) sin(4x)
Therefore, the derivative of f(x) with respect to xis:
f′(x) = 6 tan(3x) sec2(3x) + 2 cos(2x) cos(4x)−4 sin(2x) sin(4x)
Question 7
Question
Find the derivative of the function f(x) = sin(3x) cos(2x).
Solution
To find the derivative of f(x) = sin(3x) cos(2x), we will use the product rule.
Step 1: Apply the product rule:
f′(x) = (sin(3x))′
·cos(2x) + sin(3x)·(cos(2x))′
Step 2: Find the derivatives of sin(3x) and cos(2x).
(sin(3x))′= 3 cos(3x)
(cos(2x))′=−2 sin(2x)
Step 3: Substitute the derivatives back into the product rule formula:
f′(x) = 3 cos(3x)·cos(2x) + sin(3x)·(−2 sin(2x))
Step 4: Simplify the expression:
f′(x) = 3 cos(3x) cos(2x)−2 sin(3x) sin(2x)
So, the derivative of the function f(x) = sin(3x) cos(2x) is 3 cos(3x) cos(2x)−
2 sin(3x) sin(2x).
Question 8
Question
Find the derivative of the following function: f(x) = sin2(3x) + cos(2x).
5
Solution
To find the derivative of f(x), we will use the power rule, chain rule, and sum
rule for differentiation.
Step 1: Start by finding the derivative of sin2(3x).
d
dx (sin2(3x)) = 2 sin(3x) cos(3x)·3
= 6 sin(3x) cos(3x)
Step 2: Next, find the derivative of cos(2x).
d
dx (cos(2x)) = −sin(2x)·2
=−2 sin(2x)
Step 3: Now, add the derivatives of sin2(3x) and cos(2x) to find the deriva-
tive of f(x).
f′(x) = 6 sin(3x) cos(3x)−2 sin(2x)
= 6 ·
1
2
·sin(6x)−2 sin(2x)
= 3 sin(6x)−2 sin(2x)
Therefore, the derivative of f(x) = sin2(3x) + cos(2x) is f′(x) = 3 sin(6x)−
2 sin(2x).
Question 9
Question
Find the derivative of the function f(x) = sin(x) cos(x).
Solution
Step 1: To find the derivative of f(x), we will use the product rule, which states
that the derivative of the product of two functions u(x) and v(x) is given by:
(uv)′=u′v+uv′.
Step 2: Let u(x) = sin(x) and v(x) = cos(x). Then, we have:
u′(x) = cos(x) and v′(x) = −sin(x)
Step 3: Using the product rule, we can find the derivative of f(x):
f′(x) = (sin(x))(cos(x))′+ (sin(x))′(cos(x))
f′(x) = cos(x) cos(x) + sin(x)(−sin(x))
6
f′(x) = cos2(x)−sin2(x)
Step 4: Recall the Pythagorean trigonometric identity: cos2(x)−sin2(x) =
cos(2x). Substituting this into f′(x) gives:
f′(x) = cos(2x)
Therefore, the derivative of f(x) = sin(x) cos(x) is f′(x) = cos(2x).
Question 10
Question
Find the derivative of the function f(x) = sin2(x) + cos2(x).
Solution
Step 1: Recall the trigonometric identity sin2(x) + cos2(x) = 1. Step 2: Rewrite
f(x) in terms of sin2(x). Step 3: Differentiate f(x) using the chain rule. Step
4: Simplify the derivative to obtain the final answer.
Question 11
Question
Find the derivative of the function f(x) = sin(3x) cos(2x).
Solution
To find the derivative of the given function f(x) = sin(3x) cos(2x), we will use
the product rule of differentiation.
Step 1: Apply the product rule, which states that if f(x) = u(x)v(x), then
f′(x) = u′v+uv′. Let u(x) = sin(3x) and v(x) = cos(2x).
Step 2: Find u′(x) and v′(x): - u′(x) = 3 cos(3x) (derivative of sin(3x)
using the chain rule) - v′(x) = −2 sin(2x) (derivative of cos(2x) using the chain
rule)
Step 3: Apply the product rule to find f′(x):
f′(x) = u′v+uv′= (3 cos(3x))(cos(2x)) + (sin(3x))(−2 sin(2x))
Step 4: Simplify the expression:
f′(x) = 3 cos(3x) cos(2x)−2 sin(3x) sin(2x)
Thus, the derivative of the function f(x) = sin(3x) cos(2x) is 3 cos(3x) cos(2x)−
2 sin(3x) sin(2x).
7
Question 12
Question
Find the derivative of the function f(x) = sin3(2x) + cos3(3x).
Solution
Step 1: To find the derivative of f(x), we will use the chain rule and the power
rule for differentiation.
Step 2: Let’s start by finding the derivative of sin3(2x). Using the chain
rule, we have
d
dx (sin3(2x)) = 3 sin2(2x)·cos(2x)·2.
Step 3: Simplifying the above expression, we get
d
dx (sin3(2x)) = 6 sin2(2x) cos(2x).
Step 4: Next, let’s find the derivative of cos3(3x). Using the chain rule, we
have d
dx (cos3(3x)) = 3 cos2(3x)·(−sin(3x)) ·3.
Step 5: Simplifying the above expression, we get
d
dx (cos3(3x)) = −9 cos2(3x) sin(3x).
Step 6: Therefore, the derivative of the function f(x) = sin3(2x) + cos3(3x)
is
f′(x) = 6 sin2(2x) cos(2x)−9 cos2(3x) sin(3x).
Question 13
Question
Find the derivative of f(x) = sin2(3x) with respect to x.
Solution
Step 1: Apply the chain rule. Step 2: Identify the inner function and the
outer function. Step 3: Let u= 3x, then the function can be rewritten as
g(u) = sin2u. Step 4: Find the derivative of the outer function. Step 5: Find
the derivative of the inner function. Step 6: Apply the chain rule formula to
find the derivative of the composite function.
Therefore, the derivative of f(x) = sin2(3x) with respect to xis 6 sin(3x) cos(3x) .
8
Question 14
Question
Find the derivative of the function f(x) = cos(2x)·sin(3x).
Solution
Step 1: Apply the product rule to find the derivative of the function f(x) =
cos(2x)·sin(3x).
Let u(x) = cos(2x) and v(x) = sin(3x).
The product rule states that (uv)′=u′v+uv′.
Step 2: Find u′(x) and v′(x).
Using the chain rule, we have u′(x) = −2 sin(2x) and v′(x) = 3 cos(3x).
Step 3: Apply the product rule to find f′(x).
f′(x)=(u(x)v(x))′
=u′(x)v(x) + u(x)v′(x)
= (−2 sin(2x)) sin(3x) + cos(2x)(3 cos(3x))
=−2 sin(2x) sin(3x) + 3 cos(2x) cos(3x).
Therefore, the derivative of the function f(x) = cos(2x)·sin(3x) is f′(x) =
−2 sin(2x) sin(3x) + 3 cos(2x) cos(3x).
Question 15
Question
Find the derivative of the function f(x) = sin2(2x)
1+cos(2x).
Solution
Step 1: To find the derivative of f(x), we will first rewrite it using trigonometric
identities. Step 2: Recall the double angle identity: sin(2x) = 2 sin(x) cos(x)
and cos(2x) = cos2(x)−sin2(x). Step 3: Substituting these identities into f(x),
we get:
f(x) = (2 sin(x) cos(x))2
1 + (cos2(x)−sin2(x))
Step 4: Simplifying the expression, we have:
f(x) = 4 sin2(x) cos2(x)
2 cos2(x)
Step 5: Further simplifying, we obtain:
f(x) = 2 sin2(x)
9
Step 6: Now, differentiate f(x) with respect to xusing the power rule and
chain rule for trigonometric functions. Step 7: The derivative of 2 sin2(x) is
4 sin(x) cos(x) by the chain rule. Step 8: Therefore, the derivative of f(x) is
4 sin(x) cos(x) .
Question 16
Question
Find the derivative of the function f(x) = sin(x) cos(x).
Solution
To find the derivative of f(x) = sin(x) cos(x), we will use the product rule of
differentiation.
Step 1: Apply the product rule. Let u= sin(x) and v= cos(x). The
product rule states that (uv)′=u′v+uv′.
Step 2: Find u′and v′.
u′= cos(x) (derivative of sin(x))
v′=−sin(x) (derivative of cos(x))
Step 3: Substitute u,v,u′, and v′into the product rule formula.
f′(x) = u′v+uv′
= (cos(x))(cos(x)) + (sin(x))(−sin(x))
= cos2(x)−sin2(x)
Step 4: Use the trigonometric identity cos2(x)−sin2(x) = cos(2x).
f′(x) = cos(2x)
So, the derivative of f(x) = sin(x) cos(x) is f′(x) = cos(2x).
Question 17
Question
Find the derivative of the function f(x) = sin2(4x) cos(3x).
Solution
To find the derivative of the given function, we will use the product rule and
chain rule for differentiation.
Step 1: Apply the product rule: (uv)′=u′v+uv′, where u= sin2(4x) and
v= cos(3x).
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Let u= sin2(4x), then u′= 2 sin(4x) cos(4x) by the chain rule.
Let v= cos(3x), then v′=−3 sin(3x) by the chain rule.
Step 2: Substitute u′,v′,u, and vinto the product rule formula.
(f(x))′= (u′v+uv′)
= (2 sin(4x) cos(4x))(cos(3x)) + (sin2(4x))(−3 sin(3x))
= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin2(4x)
Step 3: Simplify the expression if possible.
(f(x))′= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin2(4x)
= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin(4x) sin(4x)
= 2 sin(4x) cos(4x) cos(3x)−3 sin(3x) sin(4x) cos(4x)
Therefore, the derivative of the function f(x) = sin2(4x) cos(3x) is 2 sin(4x) cos(4x) cos(3x)−
3 sin(3x) sin(4x) cos(4x).
Question 18
Question
Find the derivative of y= sin2(3x) with respect to x.
Solution
Step 1: Apply the chain rule to differentiate y= sin2(3x). Step 2: Let u=
sin(3x) and y=u2. Step 3: Find dy
du and du
dx . Step 4: Use the chain rule:
dy
dx =dy
du
·du
dx .
Step 1: Apply the chain rule to differentiate y= sin2(3x). Step 2: Let
u= sin(3x) and y=u2. Step 3: Find dy
du and du
dx .
dy
du = 2u
du
dx = 3 cos(3x)
Step 4: Use the chain rule:
dy
dx =dy
du
·
du
dx = 2u·3 cos(3x) = 6 sin(3x) cos(3x)
Therefore, the derivative of y= sin2(3x) with respect to xis 6 sin(3x) cos(3x).
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Question 19
Question
Find the derivative of f(x) = sin(2x) cos(3x).
Solution
Step 1: Apply the product rule, (uv)′=u′v+uv′, where u= sin(2x) and
v= cos(3x). Step 2: Find u′and v′. Step 3: Calculate u′:
u= sin(2x)
u′=d
dx (sin(2x)) = 2 cos(2x)
Step 4: Calculate v′:
v= cos(3x)
v′=d
dx (cos(3x)) = −3 sin(3x)
Step 5: Apply the product rule:
f′(x) = u′v+uv′
f′(x) = (2 cos(2x))(cos(3x)) + (sin(2x))(−3 sin(3x))
Step 6: Simplify the expression:
f′(x) = 2 cos(2x) cos(3x)−3 sin(2x) sin(3x)
f′(x) = 2 cos(2x) cos(3x)−3 cosπ
2
−3x
f′(x) = 2 cos(2x) cos(3x)−3 cosπ
2cos(3x) + 3 sinπ
2sin(3x)
f′(x) = 2 cos(2x) cos(3x)−3 cos(3x)
Step 7: Therefore, the derivative of f(x) = sin(2x) cos(3x) is f′(x) = 2 cos(2x) cos(3x)−
3 cos(3x).
Question 20
Question
Let f(x) = cos(5x) sin(3x). Find f′(x).
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Solution
Step 1: Apply the product rule to differentiate f(x). Step 2: Let u= cos(5x)
and v= sin(3x). Step 3: Find u′and v′. Step 4: Calculate u′v+uv′to find
f′(x).
Step 1: Apply the product rule:
f′(x) = d
dx (cos(5x)) sin(3x) + cos(5x)d
dx (sin(3x))
Step 2: Let u= cos(5x) and v= sin(3x).
Step 3: Calculate u′and v′:
u′=−5 sin(5x)
v′= 3 cos(3x)
Step 4: Substitute u′and v′back into the product rule formula:
f′(x) = −5 sin(5x) sin(3x) + cos(5x)3 cos(3x)
Therefore, f′(x) = −5 sin(5x) sin(3x) + 3 cos(5x) cos(3x).
Question 21
Question
Find the derivative of the function f(x) = cos(3x) sin(4x).
Solution
Step 1: Apply the product rule to differentiate the product of cos(3x) and
sin(4x). Step 2: Recall that the derivative of cos(u) is −sin(u)·u′and the
derivative of sin(u) is cos(u)·u′. Step 3: Let u= 3xand v= sin(4x). Step
4: Calculate u′and v′. Step 5: Find the derivative of the function f(x) =
cos(3x) sin(4x) by applying the product rule. Step 6: Combine the results to
obtain the final expression for f′(x).
Question 22
Question
Find the derivative of the function f(x) = sin2(3x) + cos(2x).
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Solution
Step 1: Apply the chain rule to differentiate sin2(3x).
d
dx (sin2(3x)) = 2 sin(3x) cos(3x)·3 = 6 sin(3x) cos(3x)
Step 2: Apply the chain rule to differentiate cos(2x).
d
dx (cos(2x)) = −sin(2x)·2 = −2 sin(2x)
Step 3: Add the derivatives of the two parts together.
f′(x) = 6 sin(3x) cos(3x)−2 sin(2x)
Question 23
Question
Find the derivative of the function y= sin(x) cos(x).
Solution
We can use the product rule to find the derivative of the given function.
Step 1: Apply the product rule. Let f(x) = sin(x) and g(x) = cos(x). The
product rule states that the derivative of the product of two functions is given
by
(f·g)′=f′
·g+f·g′.
Therefore, the derivative of y= sin(x) cos(x) is
y′= (sin(x))′cos(x) + sin(x)(cos(x))′.
Step 2: Find the derivatives of sin(x) and cos(x). The derivative of sin(x)
is cos(x) and the derivative of cos(x) is −sin(x).
Step 3: Substitute the derivatives into the product rule. Substitute cos(x)
for (sin(x))′and −sin(x) for (cos(x))′into the derivative expression:
y′= cos(x) cos(x) + sin(x)(−sin(x)).
Step 4: Simplify the expression. Simplify the expression to get the final
answer:
y′= cos2(x)−sin2(x).
Therefore, the derivative of y= sin(x) cos(x) is y′= cos2(x)−sin2(x).
Question 24
Question
Find the derivative of the function f(x) = sin(2x) cos(3x).
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Solution
Step 1: Apply the product rule to differentiate f(x) = sin(2x) cos(3x), which
states that the derivative of the product of two functions is the first function
times the derivative of the second, plus the second function times the derivative
of the first. Step 2: Let u= sin(2x) and v= cos(3x). Then we have f(x) = u·v.
Step 3: Find u′and v′. Step 4: Calculate u′:u′=d
dx [sin(2x)] Step 5: Apply
the chain rule to differentiate sin(2x): u′= 2 cos(2x) Step 6: Calculate v′:
v′=d
dx [cos(3x)] Step 7: Apply the chain rule to differentiate cos(3x): v′=
−3 sin(3x) Step 8: Now, apply the product rule to find f′(x): f′(x) = u′v+uv′
Step 9: Substitute u= sin(2x), u′= 2 cos(2x), v= cos(3x), and v′=−3 sin(3x)
into the formula for f′(x). Step 10: Simplify the expression for f′(x) to get the
final answer.
Question 25
Question
Find the derivative of the function f(x) = sin2(x) cos(x).
Solution
Step 1: Apply the product rule by differentiating the first function sin2(x) and
keeping the second function cos(x) unchanged.
f′(x) = (2 sin(x) cos(x)) cos(x) + sin2(x)(−sin(x))
= 2 sin(x) cos2(x)−sin3(x)
Therefore, the derivative of f(x) = sin2(x) cos(x) is f′(x) = 2 sin(x) cos2(x)−
sin3(x).
Question 26
Question
Find the derivative of the function y= sec2(3x) with respect to x.
Solution
Step 1: Recall the derivative of sec(x).
d
dx (sec(x)) = sec(x) tan(x)
Step 2: Rewrite the given function using basic trigonometric identities.
y= sec2(3x) = 1
cos(3x)2
=1
cos2(3x)
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Step 3: Differentiate using the chain rule.
dy
dx =d
dx 1
cos2(3x)
Step 4: Apply the chain rule to differentiate the function.
dy
dx =−21
cos(3x)
·
d
dx (cos(3x)) = −2 sec(3x) tan(3x)
Step 5: Substitute back the definition of sec(x).
dy
dx =−21
cos(3x)(tan(3x)) = −2 sec(3x) tan(3x)
Therefore, the derivative of y= sec2(3x) with respect to xis −2 sec(3x) tan(3x) .
Question 27
Question
Find the derivative of the function f(x) = sin2(x) cos(x).
Solution
Step 1: Apply the product rule. Step 2: Recall that the product rule states
that if f(x) = u(x)v(x), then f′(x) = u′(x)v(x) + u(x)v′(x). Step 3: Let
u(x) = sin2(x) and v(x) = cos(x). Step 4: Find u′(x) and v′(x).
Step 4: u′(x) = 2 sin(x) cos(x) and v′(x) = −sin(x)
Step 5: Apply the product rule to find f′(x).
Step 5: f′(x) = u′(x)v(x) + u(x)v′(x)
f′(x) = (2 sin(x) cos(x))(cos(x)) + (sin2(x))(−sin(x))
Step 6: Simplify the expression.
Step 6: f′(x) = 2 sin(x) cos2(x)−sin3(x)
Therefore, the derivative of the function f(x) = sin2(x) cos(x) is 2 sin(x) cos2(x)−
sin3(x).
Question 28
Question
Find the derivative of the function f(x) = sin2(x) tan(x).
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Solution
Step 1: Use the product rule to differentiate the function f(x) = sin2(x) tan(x).
The product rule states that if f(x) = u(x)v(x), then f′(x) = u(x)v′(x) +
v(x)u′(x). Step 2: Let u(x) = sin2(x) and v(x) = tan(x). Then, u′(x) =
2 sin(x) cos(x) and v′(x) = sec2(x). Step 3: Apply the product rule to find
f′(x).
f′(x) = u(x)v′(x) + v(x)u′(x) = sin2(x) sec2(x) + tan(x)·2 sin(x) cos(x)
Step 4: Simplify the expression by using trigonometric identities. Step 5: Recall
that tan(x) = sin(x)
cos(x)and sec(x) = 1
cos(x). Step 6: Substitute these identities into
the expression for f′(x).
f′(x) = sin2(x)
cos2(x)+2 sin(x) cos(x) sin(x)
cos(x)
Step 7: Simplify the expression further. Step 8: Combine the terms to get the
final derivative.
f′(x) = sin2(x) sec2(x) + 2 sin2(x) = sin2(x)(1 + 2 sec2(x))
Question 29
Question
Find the derivative of y= sin2(3x) with respect to x.
Solution
Step 1: To find the derivative of y= sin2(3x), we will use the chain rule. Let
u= 3xand v= sin(u). Then y=v2.
Step 2: Find dy
du by differentiating y=v2with respect to uusing the power
rule: dy
du = 2v.
Step 3: Find du
dx to apply the chain rule. Since u= 3x, we have du
dx = 3.
Step 4: Combine the results from Step 2 and Step 3 using the chain rule:
dy
dx =dy
du
·du
dx = 2v·3.
Step 5: Substitute back v= sin(u), we get dy
dx = 2 sin(3x)·3.
Step 6: Simplify to obtain the final answer: dy
dx = 6 sin(3x).
Question 30
Question
Find the derivative of the function f(x) = cos3(2x).
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Solution
Step 1: Apply the chain rule for differentiation. Step 2: Recall that the deriva-
tive of cos(u) is −sin(u)·u′. Step 3: Let u= 2x, then u′= 2. Step 4:
Substitute u= 2xand u′= 2 into the derivative formula for cos(u). Step 5:
Find the derivative of f(x) = cos3(2x).
Step 1: Apply the chain rule for differentiation:
f′(x) = −3 cos2(2x) sin(2x)·2
Step 2: Recall that d
dx [cos(u)] = −sin(u)·u′:
f′(x) = −3 (cos(2x))2
·sin(2x)·2
Step 3: Let u= 2x, then u′=2:
f′(x) = −3 (cos(u))2
·sin(u)·2
Step 4: Substitute u= 2x, and u′= 2 into the derivative formula for cos(u) :
f′(x) = −3 cos2(2x) sin(2x)·2
Step 5: Simplify the derivative of f(x) :
f′(x) = −6 cos2(2x) sin(2x)
Question 31
Question
Find the derivative of the function f(x) = cos(2x) sin(3x).
Solution
Step 1: Apply the product rule to differentiate the function f(x). Step 2: Let
u= cos(2x) and v= sin(3x). Step 3: Find u′and v′. Step 4: Use the product
rule: f′(x) = u′
·v+u·v′.
Step 1: Apply the product rule to differentiate f(x):
f′(x) = d
dx (cos(2x)·sin(3x))
Step 2: Let u= cos(2x) and v= sin(3x), so u′=−2 sin(2x) and v′=
3 cos(3x).
Step 3: Find u′and v′:
u′=−2 sin(2x)
v′= 3 cos(3x)
Step 4: Use the product rule:
f′(x) = u′
·v+u·v′
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f′(x) = (−2 sin(2x)) ·sin(3x) + cos(2x)·(3 cos(3x))
f′(x) = −2 sin(2x) sin(3x) + 3 cos(2x) cos(3x)
Question 32
Question
Differentiate the function f(x) = sinx2cos(x) with respect to x.
Solution
Step 1: Apply the product rule, which states that if f(x) = u(x)v(x), then
f′(x) = u′(x)v(x) + u(x)v′(x).
Step 2: Let u(x) = sinx2and v(x) = cos(x). Then, we have u′(x) =
cosx2·2xand v′(x) = −sin(x).
Step 3: Using the product rule, we can find f′(x) as follows:
f′(x) = u′(x)v(x) + u(x)v′(x)
= (cosx2·2x)·(cos(x)) + (sinx2)·(−sin(x))
= 2xcosx2cos(x)−sinx2sin(x).
Therefore, the derivative of f(x) = sinx2cos(x) with respect to xis f′(x) =
2xcosx2cos(x)−sinx2sin(x).
Question 33
Question
Find the derivative of y= sin2(x) cos3(x) with respect to x.
Solution
Step 1: Apply the product rule to differentiate y= sin2(x) cos3(x).
Step 2: Let u= sin2(x) and v= cos3(x). Then,
u′= 2 sin(x) cos(x) and v′=−3 cos2(x) sin(x).
Step 3: The product rule states that (uv)′=u′v+uv′, so
dy
dx =u′v+uv′.
Step 4: Substitute u,v,u′, and v′into the equation to find the derivative:
dy
dx = (2 sin(x) cos(x))(cos3(x)) + (sin2(x))(−3 cos2(x) sin(x)).
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Step 5: Simplify the expression by distributing and combining like terms:
dy
dx = 2 sin(x) cos(x) cos3(x)−3 sin2(x) cos2(x) sin(x).
Step 6: Finally, the derivative of y= sin2(x) cos3(x) with respect to xis
dy
dx = 2 sin(x) cos4(x)−3 sin2(x) cos2(x) sin(x).
Question 34
Question
Find the derivative of y=sin2x+cos2x
sin xcos x.
Solution
Step 1: Simplify the expression.
sin2x+ cos2x= 1.
y=1
sin xcos x.
Step 2: Rewrite the expression using trigonometric identities.
y=1
1
2sin 2x= 2 csc 2x.
Step 3: Differentiate ywith respect to x.
d
dx (2 csc 2x) = −2 csc 2xcot 2x.
Therefore, the derivative of ywith respect to xis −2 csc 2xcot 2x.
Question 35
Question
Find the derivative of f(x) = sin3x2−2xwith respect to x.
Solution
Step 1: Apply the chain rule, which states that if g(x) = sin(u(x)), then g′(x) =
u′(x) cos(u(x)).
f′(x) = d
dx [sin3x2−2x] = cos3x2−2x·
d
dx (3x2−2x)
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Step 2: Find the derivative of 3x2−2xusing the power rule.
f′(x) = cos3x2−2x·(6x−2)
Step 3: Simplify the expression.
f′(x) = cos3x2−2x·(6x−2) = (6x−2) cos3x2−2x
Therefore, the derivative of f(x) = sin3x2−2xwith respect to xis f′(x) =
(6x−2) cos3x2−2x.
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