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4/23/2018 Pre Test 3 (Chain Rule and Implicit)

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Student: _____________________ Date: _____________________

Instructor: Miftahul Qorib Course: Calculus Business - Spring 2018

Assignment: Pre Test 3 (Chain Rule and Implicit) Make Up

Use implicit differentiation to find and evaluate at ( , ).y′ y′ 4 − 2

08xy + y + 7 =

y′ =

Evaluate at ( , ).y′ 4 − 2

y′ (4, − 2) = (Simplify your answer.)

The total cost (in hundreds of dollars) of producing x calculators per day is given by the equation.

C(x) = 8 + 0 ≤ x ≤ 502x + 32

Perform the following calculations and interpret the results.

10 20 30 40 50 0

5

10

15

20

x

C o

st (

h u

n d

re d

d o

ll a

rs )

Production

(x)C′ =

(16)C′ =

(34)C′ =

Interpret the results. Choose the correct answer below.

A. As production increases, the cost of producing the next calculator decreases.

B. As production increases, the cost of producing the next calculator increases.

Find (x) and find the equation of the line tangent to the graph of f at the indicated value of x. Find the value(s) of x

where the tangent line is horizontal.

f′

f(x) ; x 1= ln 4 − x + 2x 2 4

=

(x)f′ =

The equation of the line tangent to the graph of f at x 1 is y .= =

Select the correct choice below and fill in any answer boxes in your choice.

A. The value(s) of x where the tangent line is horizontal is/are .

(Simplify your answer. Use a comma to separate answers as needed.)

B. The tangent line is never horizontal.

4/23/2018 Pre Test 3 (Chain Rule and Implicit)

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Find the derivative.

(x) if F(x)F′ = e x 7

+ 9 3

(x)F′ =

Use implicit differentiation to find , and evaluate for 0 at the point ( , ).y′ y′ x y − 4x + 162 3 = 2 4

y′ =

(Simplify your answer.)y′ ( )2,4 =

Find (x).f′

f(x) = 8 ln 7 + 6x 2

(x)f′ =

Use implicit differentiation to find for the equation below and evaluate at the indicated point.y′ y′

y x 0; ( , )y 2

+ 6 + 7 = − 1 1

Use implicit differentiation to find .y′

y′ =

Evaluate at the point ( , ).y′ − 1 1

y′ ( − 1,1) = (Simplify your answer.)

The demand x is the number of items that can be sold at a price of $p. For find the rate of

change of p with respect to x by differentiating implicitly.

x = , 3

15,000 − p 2

The rate of change of the price p with respect to the demand x is .

4/23/2018 Pre Test 3 (Chain Rule and Implicit)

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Find the derivative and simplify.

d

dx

ln (6 + x)

x 3

Choose the correct answer below.

A. 3(6 + x) ln (6 + x) − x

x 4 (6 + x)

B. 1 − 18x

2 (6 + x) ln (6 + x)

6x 3 (6 + x)

C. x − 3(6 + x) ln (6 + x)

x 4 (6 + x)

D. 18x

2 (6 + x) ln (6 + x) − 1

6x 3 (6 + x)

Find for x(t) defined implicitly by . Then evaluate at the point x′ x + t x + t − 9 = 03 3 4 x′ ( − 2,1).

x′ =

x′ ( − 2,1) =

The demand x is the number of items that can be sold at a price of $p. For find the rate of change of p with respect to x by differentiating implicitly.

x = p − 4p + 700, 4 3

The rate of change of the price p with respect to the demand x is .

Find .(x)f ′

f(x) = (1 + ln x) 6

(x)f ′ =

Find if .y′ y = ln x + 16 3 / 2

y′ =

Use implicit differentiation to find . Then evaluate at ( ,1).y′ y′ 1

x 5

− y 2

= ln y

y′ =

y′ (1,1) =