Exercises: Derivatives 4 – Serlo

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Criterion for constancy[Bearbeiten]

Exercise (Simple application)

Let and . For the function we assume

for all . Show that then, is constant.

Solution (Simple application)

By assumption there is

Further there is , since . The squeeze theorem then yields

for all . By the computation rule for limits we then get a zero differential quotient

for all . Hence is constant.

Exercise (Proof of identities)

Show that

  1. for all

Proof (Proof of identities)

Part 1: The function

is differentiable by the chain- and difference rule with

So is constant. Since further there is

we have .

Part 2:

is differentiable according to the sum rule, since the arcus-functions are differentiable. Further there is

Hence is constant. Further there is

since and . So establishing the assertion.

Exercise (Logarithm representations of and )

Show that

  1. for
  2. for

Proof (Logarithm representations of and )

Part 1:

The function is differentiable, see examples for derivatives, with

By the chain- and sum rule also is differentiable with

So we get . But now,

since , and there is

So , and hence .

Part 2:

is differentiable, as well, with

By the factor-, chain- and quotient rule, also is differentiable with

So we have . Since

by , as well as

there is again , and hence .

Exercise (Extension of the identity theorem)

Let be twice differentiable with . Then and differ only by a linear function with .

Solution (Extension of the identity theorem)

Because according to the identity theorem there is a with

if we now set , then there is

the identity theorem again provides us with a such that

Exercise (General solution of a differential equation)

Let be differentiable and . Further let

Show that:

  1. and satisfy the differential equations.
  2. If two functions and satisfy the differential equations, then there is and .
  3. Furthermore, if and there is and , then and .

Proof (General solution of a differential equation)

Part 1: There is

and

So and satisfy the differential equations.

Part 2:We define (as given in the hint) the auxiliary functions

These are differentiable by the product-, sum- and difference rule with

and

by the criterion for constancy, there is now

with . Further there is

So and .

Part 3: If further and

then

Hint

Since and analogously both the functions and satisfy the differential equation (or ).

Monotony criterion[Bearbeiten]

Exercise (Monotony of the exponential function)

Show using the monotony criterion that:

  1. For all there is
  2. is strictly monotonously increasing.

Hint: use 1. in order to prove 2.

Solution (Monotony of the exponential function)

Part 1: For the differentiable auxiliary function there is

So is strictly monotonously decreasing by the monotony criterion. Further

Graph of the function

and

Since is continuous and strictly monotonously decreasing, there must be .

Part 2:

There is . Since is strictly monotonously increasing, the function is strictly monotonously increasing, if and only if the "inner function" is. This function in turn is differentiable on all of by the product rule and there is

By the monotony criterion, the function , and hence also is strictly monotonously increasing.

Exercise (Condition for monotonicity of a cubic function)

Let . Provide a condition for such that

is strictly monotonously increasing on all of .

Hint: Distinguish the cases , and

Solution (Condition for monotonicity of a cubic function)

For being strictly monotonously increasing on all of , we need by the monotony criterion that

holds for all .

Fall 1:

Then . For to be strictly monotonously increasing, must hold. For this is never possible for any .


However, if , then there is . So is strictly monotonously increasing for and .

Fall 2:

By completing the square, we get

So is strictly monotonously increasing whenever there is

This is satisfied for all if and only if the right-hand side is negative. This in turn is exactly the case for

Hence, is strictly monotonously increasing for and .

Fall 3:

Here, we have

However, this is never fulfilled for all . So in this case is never strictly monotonous increasing.

Hint

Similarly, we can show that is strictly monotonously decreasing in the cases and , as well as for and .

Exercise (Applying the monotony criterion)

Let be differentiable with . Further let for some (fixed) and all . Show that there is

for all

Hint: Consider the auxiliary function .

Proof (Applying the monotony criterion)

As stated in the hint we consider

is differentiable according to the product rule with

But now and by assumption . So there is

for all

by the monotony criterion, is monotonously decreasing. Since further there is we have

for all

Therefore, we also have

for all

Derivative and extrema[Bearbeiten]

Exercise (Extrema of functions 1)

Investigate whether the following functions have local/global extrema. Determine and characterise these if they exist.

Solution (Extrema of functions 1)

Part 1:

Part 1: Local extrema of

is differentiable on according to the quotient rule with

According to the sufficient criterion for the existence of an extremum , there must be . Now

So and are the candidates for local extrema in . Now there is

The case and is not possible. Thus holds on .

Further there is

So holds on and on .

By the sufficient criterion, is a (strict) local minimum and is a (strict) local maximum of .

Part 2: Global extrema of

For global extrema we first have to determine the limits and .

Since and there is

Graph of the function

Thus is unbounded from above, and therefore has no local extremum. Further, for every power of grows slower than . Thus

(As numerator and denominator are positive.) Now . Thus is a global minimum of .

Part 2:

Part 1: Local extrema of

is differentiable on by the chain rule with

Graph of the function

Since now and there is, . By the necessary criterion for extrema, has no local extrema on .

Since is continuous on , it follows from for all that is strictly monotonously decreasing on . Therefore has a local maximum at .

Part 2: Global extrema of

Using the same argument as in part 1, it follows that is even a global maximum of .

Exercise (Extrema of functions 2)

Investigate whether the following functions are continuous, differentiability and/or have local/global extrema:

Solution (Extrema of functions 2)

Part 1: continuity and differentiability

Continuity:

On the function is continuous as a polynomial. The function is continuous on as a composition of the continuous functions , and . At there is

In addition, since and by continuity of the exponential function

So is continuous at zero and hence on all of .

Differentiability:

On the function is differentiable as a polynomial, with

On the function is differentiable by the chain- and product rule as it is a composition of differentiable functions , and . There is

At there is, according to L'Hospital's rule,

So is not differentiable at zero.

Part 2: Local and global extrema

Local extrema:

On there is . So cannot have local extrema there.

On however

So is a candidate for a possible extremum. Further

So has a strict local minimum in .

Now we still have to examine . Since is not differentiable there, our necessary and sufficient criteria are not applicable. However, there is

for all

and

for all

Thus is strictly monotonously increasing on , and strictly monotonously decreasing on . Since is continuous at zero, it follows that

for all

So has a strict local maximum in .

Global extrema:

There is

Graph of the function

and

Therefore is unbounded from above and below, and has no global extrema.

Exercise (Extrema of functions 3)

Show that the function

has exactly two local extrema, and determine their type.

Solution (Extrema of functions 3)

Candidates for the extreme values are obtained from our necessary condition

Graph of the auxiliary function

Since the zeros of cannot be calculated explicitly, we need to examine this function more closely. There is

  1. , and

Because of continuity and 2. with the intermediate value theorem has (at least) two zeros and .

Because of 1., the function is strictly monotonously increasing on and strictly monotonously decreasing on . Thus is respectively injective on and and thus has exactly the two zeros and .

For the derivative of we now have

According to our first sufficient criterion, has a strict local maximum at and a strict local minimum at .

Computing limits via L'Hospital[Bearbeiten]

Exercise (L'Hospital 1)

Compute the following limits

  1. with

Solution (L'Hospital 1)

Part 1:

Part 2:

L'Hospital's rule is not applicable here. However, the function is continuous at the point , and therefore there is

Part 3:

Part 4:

This limit value does not exist. First it can be decomposed into

For the left-hand limit there is now with Part 1:

Analogously, however, for the right-hand limit:

So , and hence does not exist.

Part 5:

Part 6:

L'Hospital can be applied here, but it is useless:

Instead, it makes sense to use the definitions of and , and then transform the quotient:

Part 7:

L'Hospital cannot be applied here because the enumerator for diverges (improperly). Instead, the fraction can be estimated as follows:

Using the squeeze theorem, it follows .

Part 8:

Part 9:

Part 10:

Exercise (L'Hospital 2)

Compute the following limits:

  1. for
  2. for

Solution (L'Hospital 2)

Part 1:

and since the limit exists, the application of L'Hospital is justified.

Part 2:

and since the limit exists, the application of L'Hospital is justified.

Part 3:

and since the limit exists, the application of L'Hospital is justified.

Part 4:
Part 5:

For the expression in the exponent there is now

and since the limit exists, the application of L'Hospital is justified.

By continuity of at we now get

Part 6: First we have: If the limit exists, then the sequence limit also exists.

Further:

For the expression in the exponent there is now

and since the limit exists, the application of L'Hospital is justified.

By continuity of at we now get

And now, we also have .

Part 7:

For the expression in the exponent there is now

and since the limit exists, the application of L'Hospital is justified.

By continuity of at we now get

Part 8:

For the expression in the exponent there is now

and since the limit exists, the application of L'Hospital is justified.

By continuity of at we now get

Part 9:

and since the limit exists, the application of L'Hospital is justified.

Part 10:

and since the limit exists, the application of L'Hospital is justified.

Exercise (Differentiability at a point)

Let

  1. Show that is continuous at zero.
  2. Show that is differentiable on and compute the derivative.
  3. Determine by 1. and 2. the derivative .

Solution (Differentiability at a point)

Part 1: By L'Hospital there is

So is continuous at zero.

Part 2: Since , and are differentiable on , the quotient rule yields that is differentiable. Further there is for :

Part 3: We use the criterion from the theorem above. There is

Using the criterion, is differentiable at zero with .