prooving limits and differentials

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Chapter 5 Derivatives

Example 1 Consider where . Recall the identity

For any ,

Example 2 Consider . The function is not differentiable at 0, since

which does not exist, as the limit is from the right and from the left.

Therefore, a continuous function is not always differentiable; that is,

continuous ⇏ differentiable.

A continuous function is not always differentiable; that is,

continuous ⇏ differentiable.

Proof: We are given that exists.

Thus, by the Algebraic Limit Theorem for Functions (Corollary 4.2.4). This implies that , i.e., is continuous at .

Algebraic combinations of continuous functions are continuous by the Algebraic Limit Theorem for Functions (Corollary 4.2.4).

Somewhat similar conclusions hold for derivatives.

Product Rule

Proof: First, rewrite the difference quotient as

Taking limits as and using the Algebraic Limit Theorem gives

Since is differentiable at implies is continuous at , Thus,

Idea: Write out the difference quotient and write the numerator as a single fraction. Within that single fraction subtract and add in its numerator. Proceed as before.

Quotient Rule

Idea: Essentially the idea is the following, except the nuisance of dealing with the possibility that for values of close to

Proof: Assume attains its maximum at some point . Since is in the open interval , there exists sequences

such that . Since is a maximum,

By the Order Limit Theorem,

Similarly, Therefore, .

The derivative possesses an intermediate value property:

To prove the derivatives for the trig functions, we require

Start with a unit circle with center .

Suppose and are points on the circle (points between and ).

Let be the perpendicular bisector of .

Proof :

Note: you can prove a similar inequality for .

Proof :

Notice in the last proof that we used cosine is continuous on R. Using geometry, one can prove that for all .

Use this, the definition of continuity and the trig identities

to prove the next theorem.

Theorem All trig functions are continuous on their domains.

Proof: Let , and show is continuous at . Let . Choose . Thus,

Thus, is continuous on R. Using trig identities and the Algebraic Limit Theorem, one can show the other trig functions are continuous on their domains.

Using the special limit

We can deduce another special limit involving cosine:

By replacing with the derivative can also be written as

Show if , then .

DERIVS. OF TRIG. FUNCTIONS

Equation 1

So, we have proved the formula for the derivative of the sine function:

Using the same methods, we can prove:

The tangent function can also be differentiated by using the definition of a derivative, but it is easier to use the Quotient Rule.

DERIV. OF SINE FUNCTION

Formula 4

NOTE: these are valid only when x is measured in radians.

DERIVS. OF TRIG. FUNCTIONS

Calculate the derivatives of , , and .

Recall we used radian measure to find limit of sin(theta)/theta as theta goes to 0.

For , consider

Since for all near , is continuous at 0.

Notice that which does not exist.

Thus, is continuous at 0, but not differentiable at 0.

Similar calculations show is continuous and differentiable at 0. But,

and

All together, is continuous and differentiable everywhere on R.

Thus, the derivative of is defined everywhere on R. But, has a discontinuity at 0. Hence, in general, derivatives need not be continuous!

g_1 is cont. at 0 but not diff. at 0 g_2 is cont. and diff. at 0

g_2 is cont. and diff. on R

g_2’ is not cont. at 0

Recall the theorems:

Extreme Value Theorem (Theorem 4.4.2) states that continuous functions on compact sets always attain maximum and minimum values.

Proof: Since is continuous on a compact set, attains a maximum and a minimum. If both the max and min occur at the endpoints, then it must be that is a constant function and on all of . Thus, there exists such that . On the other hand, if either the max or min occurs at some point in the interior , then by the Interior Extremum Theorem (Theorem 5.2.6).

Proof: The equation of the line through and is

Consider the difference between this line and the function ,

and observe that is continuous on , differentiable on , and satisfies . By Rolle’s Theorem, there exists a point such that . Since

we get the desired

sin

that

have

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show

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for 0<q<p/2,

we have that

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tan

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arc

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sin

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Theorem,

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By the

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lim

cos1cos1

lim

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lim

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⎝ ⎜

⎞

⎠ ⎟

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1+1

lim θ→0

cosθ −1 θ

= 0

lim

q®0

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(sin)cos

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sin

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cos

cos(sin)sin(cos)

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sec

coscos

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===

(sin)cos(csc)csccot

(cos)sin(sec)sectan

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==-

=-=

==-