Differentiation | An Introduction to Real Analysis

The Derivative

We begin with the definition of the derivative of a function.

Let

I \subset R

be an interval and let

c \in I

. We say that

f : I \to R

is differentiable at $c$ or has a derivative at $c$ if

lim_{x \to c} \frac{f (x) - f (c)}{x - c}

exists. We say that

f

is differentiable on

I

f

is differentiable at every point in

I

By definition,

f

has a derivative at

c

if there exists a number

L \in R

such that for every

ε > 0

there exists

δ > 0

such that if

| x - c | < δ

then

| \frac{f (x) - f (c)}{x - c} - L | < ε .

f

is differentiable at

c

, we will denote

lim_{x \to c} \frac{f (x) - f (c)}{x - c}

f^{'} (c)

, that is,

f^{'} (c) = lim_{x \to c} \frac{f (x) - f (c)}{x - c} .

The rule that sends

c

to the number

f^{'} (c)

defines a function on a possibly smaller subset

J \subset I

. The function

f^{'} : J \to R

is called the derivative of

f

Let

f (x) = 1 / x

for

x \in (0, \infty)

. Prove that

f^{'} (x) = - \frac{1}{x^{2}}

Let

f (x) = \sin (x)

for

x \in R

. Prove that

f^{'} (x) = \cos (x)

Recall that

\sin (x) - \sin (c) = 2 \sin (\frac{x - c}{2}) \cos (\frac{x + c}{2})

and that

lim_{x \to 0} \frac{\sin (x)}{x} = 1

. Therefore,

\begin{aligned} lim_{x \to c} \frac{\sin (x) - \sin (c)}{x - c} & = lim_{x \to c} \frac{2 \sin (\frac{x - c}{2}) \cos (\frac{x + c}{2})}{x - c} \\ = lim_{x \to c} (\frac{\sin (\frac{x - c}{2})}{\frac{x - c}{2}}) \cos (\frac{x + c}{2}) \\ = 1 \cdot \cos (c) = \cos (c) . \end{aligned}

Hence

f^{'} (c) = \cos (c)

for all

c

and thus

f^{'} (x) = \cos (x)

Prove by definition that

f (x) = \frac{x}{1 + x^{2}}

is differentiable on

R

We have that

\begin{aligned} \frac{f (x) - f (c)}{x - c} & = \frac{\frac{x}{1 + x^{2}} - \frac{c}{1 + c^{2}}}{x - c} \\ = \frac{x (1 + c^{2}) - c (1 + x^{2})}{(1 + x^{2}) (1 + c^{2}) (x - c)} \\ = \frac{1 - c x}{(1 + c^{2}) (1 + x^{2})} . \end{aligned}

Now

lim_{x \to c} \frac{f (x) - f (c)}{x - c} = \frac{1 - c^{2}}{(1 + c^{2})^{2}} .

Hence,

f^{'} (c)

exists for all

c \in R

and the derivative function of

f

f^{'} (x) = \frac{1 - x^{2}}{(1 + x^{2})^{2}} .

Prove that

f^{'} (x) = α

f (x) = α x + b

We have that

f (x) - f (c) = α x - α c = α (x - c)

. Therefore,

lim_{x \to c} \frac{f (x) - f (c)}{x - c} = α

. This proves that

f^{'} (x) = α

for all

x \in R

Compute the derivative function of

f (x) = | x |

for

x \in R

x > 0

then

f (x) = x

and thus

f^{'} (x) = 1

for

x > 0

. If

x < 0

then

f (x) = - x

and therefore

f^{'} (x) = - 1

for

x < 0

. Now consider

c = 0

. We have that

\frac{f (x) - f (c)}{x - c} = \frac{| x |}{x} .

We claim that the limit

lim_{x \to 0} \frac{| x |}{x}

does not exist and thus

f^{'} (0)

does not exist. To see this, consider

x_{n} = 1 / n

. Then

(x_{n}) \to 0

and

f (x_{n}) = 1

for all

n

. On the other hand, consider

y_{n} = - 1 / n

. Then

(y_{n}) \to 0

and

f (y_{n}) = - 1

. Hence,

lim_{n \to \infty} f (x_{n}) \neq lim_{n \to \infty} f (y_{n})

, and thus the claim holds by the Sequential criterion for limits. The derivative function

f^{'}

f

is therefore defined on

A = R ∖ {0}

and is given by

f^{'} (x) = {\begin{cases} 1, & x > 0 \\ - 1, & x > 0. \end{cases}

Hence, even though

f

is continuous at every point in its domain

R

, it is not differentiable at every point in its domain. In other words, continuity is not a sufficient condition for differentiability.

Suppose that

f : I \to R

is differentiable at

c

. Then

f

is continuous at

c

To prove that

f

is continuous at

c

we must show that

lim_{x \to c} f (x) = f (c)

. By assumption

lim_{x \to c} \frac{f (x) - f (c)}{x - c} = f^{'} (c)

exists, and clearly

lim_{x \to c} (x - c) = 0

. Hence we can apply the Limits laws and compute

\begin{aligned} lim_{x \to c} f (x) & = lim_{x \to c} (f (x) - f (c) + f (c)) \\ = lim_{x \to c} (\frac{f (x) - f (c)}{(x - c)} (x - c) + f (c)) \\ = f^{'} (c) \cdot 0 + f (c) \\ = f (c) \end{aligned}

and the proof is complete.

Let

f : I \to R

and

g : I \to R

be differentiable at

c \in I

. The following hold:

If $α \in R$ then $(α f)$ is differentiable and $(α f)^{'} (c) = α f^{'} (c)$ .
is differentiable at and .
is differentiable at and .
If then is differentiable at and

Parts (i) and (ii) are straightforward. We will prove only (iii) and (iv). For (iii), we have that

Now

because

is differentiable at

. Therefore,

To prove part (iv), since

, then there exist a

-neighborhood

such that

for all

. If

then

Since

, it follows that

and the proof is complete.

We now prove the Chain Rule.

Let

and

be functions such that

and let

. If

exists and

exists then

exists and

Suppose that there exists a neighborhood of

where

. Otherwise, the composite function

is constant in a neighborhood of

, and then clearly differentiable at

. Consider the function

defined by

Now

Hence,

is differentiable at

and therefore

is at

. Now,

and therefore

Therefore,

as claimed.

Compute

Where is

continuous?

When

is the composition and product of differentiable functions at

, and therefore

is differentiable at

. For instance, on

, the functions

and

are differentiable at every

. Hence, if

we have that

Consider now

. If

exists it is equal to

Using the Squeeze Theorem, we deduce that

. Therefore,

From the above formula obtained for

, we observe that when

is continuous since it is the product/difference/composition of continuous functions. To determine continuity of

consider

. Consider the sequence

, which clearly converges to

. Now,

for all

and therefore

. The sequence

does not converge and therefore

does not exist. Thus,

is not continuous at

Compute

Where is

continuous?

When

is the composition and product of differentiable functions, and therefore

is differentiable at

. For instance, on

, the functions

and

are differentiable on

. Hence, if

we have that

Consider now

. If

exists it is equal to

and using the Squeeze Theorem we deduce that

. Therefore,

When

is continuous since it is the product/difference/composition of continuous functions. To determine continuity of

we consider the limit

. Now

using the Squeeze Theorem, and similarly

using the Squeeze Theorem. Therefore,

exists and is equal to

, which is equal to

. Hence,

is continuous at

, and thus continuous everywhere.

Consider the function

Show that

Exercises

Use the definition of the derivative of a function to find

. Clearly state the domain of

Use the definition of the derivative of a function to find

. Clearly state the domain of

Let

be defined by

Show that is differentiable at and find .
Prove that if then is not differentiable at .

Let

for

. Determine whether

exist and if yes find them. Hint: Consider writing

as a piecewise function and use the definition of the derivative.

is differentiable at

, explain why

Give an example of a function

and a number

such that

exists but

does not exist.

The Mean Value Theorem

Let

be a function and let

We say that has a relative maximum at if there exists such that for all .
We say that has a relative minimum at if there exists such that for all .

A point

is called a critical point of

. The next theorem says that a relative maximum/minimum of a differentiable function can only occur at a critical point.

Let

be a function and let

be an interior point of

. Suppose that

has a relative maximum (or minimum) at

. If

is differentiable at

then

is a critical point of

, that is,

Suppose that

has a relative maximum at

; the relative minimum case is similar. Then for

, it holds that

for

and some

. Consider the function

defined by

for

and

. Then the function

is continuous at

because

. Now for

it holds that

and therefore

. Similarly, for

it holds that

and therefore

. Thus

has a relative maximum (or minimum) at

then either

does not exist.

The function

has a relative minimum at

, however,

is not differentiable at

Let

be continuous on

and differentiable on

. If

then there exists

such that

Since

is continuous on

it achieves its maximum and minimum at some point

and

, respectively, that is

for all

. If

is constant then

for all

. If

is not constant then

. Since

it follows that at least one of

and

is not contained in

, and hence by Theorem 6.2.2 there exists

such that

We now state and prove the main result of this section.

Let

be continuous on

and differentiable on

. Then there exists

such that

then the result follows from Rolle's Theorem (

for some

). Let

be the line from

, that is,

and define the function

for

. Then

and

, and thus

. Clearly,

is continuous on

and differentiable on

, and it is straightforward to verify that

. By Rolle's Theorem, there exists

such that

, and therefore

Let

be continuous on

and differentiable on

. If

for all

then

is constant on

Let

. Now

restricted to

satisfies all the assumptions needed in the Mean Value Theorem. Therefore, there exists

such that

. But

and thus

. This holds for all

and thus

is constant on

Show by example that Theorem 6.2.7 is not true for a function

is not a closed and bounded interval.

are continuous and differentiable on

and

for all

then

for some constant

Use the Mean Value theorem to show that

for all

Suppose that

and let

so that

. By the MVT, there exists

such that

, that is

. Now

and therefore

. Therefore,

. The case

can be treated similarly.

The function

is increasing if

whenever

. Similarly,

is decreasing if

whenever

. In either case, we say that

is monotone.

The sign of the derivative

determines where

is increasing/decreasing.

Suppose that

is differentiable.

Then is increasing if and only if for all .
Then is decreasing if and only if for all .

Suppose that

is increasing. Then for all

with

it holds that

and therefore

. Hence, this proves that

for all

. Now suppose that

for all

. Suppose that

. Then by the Mean Value Theorem, there exists

such that

. Therefore, since

it follows that

. Part (ii) is proved similarly.

Exercises

Use the Mean Value Theorem to show that

In general, suppose that

is such that

exists on

and

is continuous on

. Prove that

is Lipschitz on

Give an example of a uniformly continuous function on

that is differentiable on

but whose derivative is not bounded on

. Justify your answer.

Let

be an interval and let

be differentiable on

. Prove that if

for

then

is strictly increasing on

Let

be continuous on

and differentiable on

. Show that if

then

exists and equals

. Hint: Use the definition of

, the Mean Value Theorem, and the Sequential Criterion for limits.

Let

be continuous and suppose that

exists on

. Prove that if

for

then

is strictly increasing on

Suppose that

is continuous on

and differentiable on

. We proved that if

for all

then

is constant on

. Give an example of a function

such that

for all

but

is not constant on

Let

be differentiable. Prove that if

then

is injective.

Taylor's Theorem

Taylor's theorem is a higher-order version of the Mean Value Theorem and it has abundant applications in numerical analysis. Taylor's theorem involves Taylor polynomials which you are familiar with from calculus.

Let

and suppose that

is such that the derivatives

exist for some positive integer

. Then the polynomial

is called the th order Taylor polynomial of based at . Using summation convention,

can be written as

By construction, the derivatives of

and

up to order

are identical at

(verify this!):

It is reasonable then to suspect that

is a good approximation to

for points

near

. If

then the difference between

and

and we call

the th order remainder based at . Hence, for each

, the remainder

is the error in approximating

with

. You may be asking yourself why we would need to approximate

if the function

is known and given. For example, if say

then why would we need to approximate say

since any basic calculator could easily compute

? Well, what your calculator is actually computing is an approximation to

using a (rational) number such as

and using a large value of

for accuracy (although modern numerical algorithms for computing trigonometric functions have superseded Taylor approximations but Taylor approximations are a good start). Taylor's theorem provides an expression for the remainder term

using the derivative

Let

be a function such that for some

the functions

are continuous on

and

exists on

. Fix

. Then for any

there exists

between

and

such that

where

then

and then

can be chosen arbitrarily. Thus, suppose that

, let

, and define the function

Since

exists on

then

exists on

. Moreover, since

for

then

for

. Now

and therefore since

by Rolle's theorem there exists

in between

and

such that

. Now we can apply Rolle's theorem to

since

and

, and therefore there exists

in between

and

such that

. By applying this same argument repeatedly, there exists

in between

and

such that

. Now,

and since

then

from which we conclude that

and the proof is complete.

Consider the function

given by

. Use

based at

to estimate

and give a bound on the error with your estimation.

Note that

and so the estimate of

using

. To determine

we need

. We compute

Therefore,

Now

and therefore

The error is

which is unknown but we can approximate it using Taylor's theorem. To that end, by Taylor's theorem, for any

there exists

in between

and

such that

Therefore, for

, there exists

such that

Therefore, a bound for the error is

since

Let

be the sine function, that is,

Approximate using centered at and give a bound on the error.
Restrict to a closed and bounded interval of the form . Show that for any there exists such that if then for all .

(a) It is straightforward to compute that

and

. Thus, by Taylor's theorem for any

there exists

in between

and

such that

The estimate for

By Taylor's theorem, there exists

such that

and

Now since

for all

, we have

(b) Since

has derivatives of all orders, for any

we have by Taylor's theorem that

where

is in between

and

. Now, the derivative of

of any order is one of

, and therefore

. Since

then

and therefore

. Therefore, for all

we have

Consider the sequence

. Applying the Ratio test we obtain

Therefore, by the Ratio test

. Hence, for any

there exists

such that

for all

. Therefore, for all

we have that

for all

Taylor's theorem can be used to derive useful inequalities.

Prove that for all

it holds that

Let

. Applying Taylor's theorem to

we obtain

where

and

is in between

and

. Now, if

then

and then

, from which it follows that

. If on the other hand

then

and then

, from which it follows that

. Hence, the inequality holds for

. Now if

then

Hence the inequality holds for all

Exercises

Use Taylor's theorem to prove that if

then

Then use these inequalities to approximate

and

, and for each case determine a bound on the error of your approximation.

Let

be such that

exists for all

and for all

(such a function is called infinitely differentiable on ). Suppose further that there exists

such that

for all

and all

. Let

be the

th order Taylor polynomial of

centered at

. Let

, where

. Prove that for any fixed

there exists

such that for

it holds that

for all

. Hint:

is continuous on

for every

Euler's number is approximately

. Use Taylor's theorem at

and the estimate

to show that, for all

Let

be the cosine function

. Approximate

using

centered at

and give a bound on the error of your estimation.