Exercises: Derivatives 2 – Serlo

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Derivative of the inverse function[Bearbeiten]

Exercise (Derivative of the inverse function)

Consider the function

Justify that the following derivatives and limits exist and calculate them:

  1. and
  2. and

Solution (Derivative of the inverse function)

Part 1:

Proof step: Existence and computation of

is differentiable on as a sum of the differentiable functions and with derivative

So .

Proof step: Existence and computation of

is differentiable on as a sum of the differentiable functions and with derivative

Further, , since for . So is strictly monotonously increasing by the monotony criterion and hence injective. Further, is bijective. And we have , so , and there is . Hence is differentiable at with

Part 2:

Proof step: Computation of

There is

So

Proof step: Computation of

There is . According to the intermediate value theorem, is bijective. For all we thus have

So

Derivative of a general logarithm function [Bearbeiten]

Exercise (Derivative of a general logarithm function)

Show that the function

with base is bijective and differentiable on . Show further that the inverse function

is differentiable on all of with derivative .

Solution (Derivative of a general logarithm function)

Proof step: is bijective and differentiable

Fall 1:

is continuous on as a composition of continuous functions. Further , since , and there is:

as well as

By the intermediate value theorem, and is therefore surjective. Moreover, is differentiable according to the chain rule as a composition of differentiable functions, and

for all . By the monotonicity criterion, is strictly monotonously increasing, and hence injective. So we have shown the bijectivity of .

Fall 2:

is continuous on as a composition of continuous functions. Further , since , and there is:

as well as

By the intermediate value theorem, and is therefore surjective. Moreover, is differentiable according to the chain rule as a composition of differentiable functions, and

for all . By the monotonicity criterion, is strictly monotonously decreasing, and hence injective. So we have shown the bijectivity of .

Proof step: exists and is differentiable

is bijective on and hence invertible. Further, there is

So

Since in addition for all , the theorem about inverse function derivatives yields that is differentiable on with

Exercise (derivatives on and area-functions)

Show that the functions

are differentiable, and determine their derivative.

Proof (derivatives on and area-functions)

Differentiability of :

For the cotangent function there is:. Thus the function is differentiable and strictly monotonically decreasing, and thus injective. Further, . Thus is bijective. The inverse function

is differentiable according to the theorem on the derivative of the inverse function, and for there is:

Differentiability of :

The hyperbolic sine function is differentiable with . Thus it is strictly monotonously increasing, and hence injective. Further, . So it is also surjective. The inverse function

is differentiable by the theorem on the derivative of the inverse function , and for there is:

Differentiability of :

The hyperbolic cosine function is differentiable with on . Thus it is strictly monotonically increasing, and hence injective. Further, . So it is also surjective. The inverse function

is differentiable by to the theorem on the derivative of the inverse function, and for there is:

Differentiability of :

For the cotangent function there is: . Thus the function is strictly monotonically increasing, and thus injective. Further, . Thus is bijective. The inverse function

is differentiable by the theorem on the derivative of the inverse function, and for there is:

Exercise (Non-differentiable functions at zero)

Let . Show that:

  1. Let with and for all . Then and are not simultaneously differentiable at zero.
  2. Let , and let be differentiable at zero. Further let and for all . Then is not differentiable at zero.

Solution (Non-differentiable functions at zero)

Part 1: Assuming and are differentiable at zero. Then according to the product rule, is also differentiable at zero, and from we get for all that:

 ?

So and cannot both be differentiable at zero.

Part 2: Assume that is differentiable at zero. Then according to the chain rule is also differentiable at zero, and from we get for all that:

 ?

So cannot be differentiable at zero.

Derivatives of higher order[Bearbeiten]

Exercise 1[Bearbeiten]

Exercise (Arbitrarily often differentiable function)

Show that the function is arbitrarily often differentiable and for all there is:

Proof (Arbitrarily often differentiable function)

The proof goes by induction over :

Theorem whose validity shall be proven for the :

1. Base case:

1. inductive step:

2a. inductive hypothesis:

2b. induction theorem:

2b. proof of induction step:

Exercise 2[Bearbeiten]

Exercise (Exactly one/two/three times differentiable functions)

Provide an example of a

  1. function
  2. function that is differentiable, but not continuously differentiable on
  3. function

Solution (Exactly one/two/three times differentiable functions)

Solution sub-exercise 1:

or or

Solution sub-exercise 2:

or

Solution sub-exercise 3:

or or

Solution sub-exercise 4:

or or

Hint

We can successively proceed with the construction of functions and obtain some for all , with

or or

Exercise 3[Bearbeiten]

Exercise (Application of the Leibniz rule)

Determine the following derivatives using the Leibniz rule

  1. for

Solution (Application of the Leibniz rule)

Solution sub-exercise 1:

The functions and are arbitrarily often differentiable on . Hence there is

Solution sub-exercise 2:

The functions and are arbitrarily often differentiable on . Hence there is