Exercises: Derivatives 1 – Serlo

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Computing derivatives with differential quotients[Bearbeiten]

Exrecises: derivative and differentiability[Bearbeiten]

Exercise (Differentiable power function)

Show that the power function is differentiable at and compute the derivative. What is the derivative of at any ?

Solution (Differentiable power function)

The differential quotient of at is given by

So is differentiable at , with derivative . For a general there is

Exercise (Derivative of a product function)

Let defined by

Determine .

Solution (Derivative of a product function)

There is

For we have used that is continuous as a product of the continuous functions for .

Exercise (Derivative of a function with case distinction)

Check whether the following functions are differentiable at .

Solution (Derivative of a function with case distinction)

Part 1: Since oscillates very quickly between and , just like for , it is to be expected that at is not continuous. For this purpose we consider the null sequences and . For these sequences

and

So does not exist. According to the sequence criterion is therefore not continuous at zero and thus not differentiable.

Part 2: The function is continuous at zero by the sequence criterion, as . So we can consider the differential quotient:

In Part 1 we have shown that this limit value does not exist. Therefore is also not differentiable at zero.

Exercise (Criterion for non-differentiability of a general function at zero)

Let . Show that: If for some and , then is not differentiable at zero.

Solution (Criterion for non-differentiability of a general function at zero)

There is

since

So does not exist.

Exercise (Determining limits with the differential quotient)

Let be differentiable at . Show that the following limits exist

  1. for

How to get to the proof? (Determining limits with the differential quotient)

Since in is differentiable, there is

and

In addition, from the exercises within the article "derivatives", we know that

The idea is to transform the limits so that we can calculate them using the differential quotient.

Solution (Determining limits with the differential quotient)

Part 1: Because there is also . So

Part 2: With and there is also and . Hence

Part 3: Here we need the "original" differential quotient :

Exercise (Implication of differentiability)

Let be differentiable at . Further let and be sequences with for all , as well as . Show that then, there is

Additional question: Does the converse statement also hold? I.e. does the limit value with sequences and as above exist, so is differentiable at , and is equal to this limit?

Hint: Show first that

Solution (Implication of differentiability)

There is

Since now the product of a bounded sequence and a null sequence converges to zero, there is with the calculation rules for sequences

Concerning the additional question: The converse is false. Let us consider the following function, which is not continuous at (and therefore not differentiable):

Then, there is for all null sequences and with :

Exercises: examples for derivatives[Bearbeiten]

Exercise (Derivatives of linear and quadratic functions)

Determine (using the definition) the derivative of a linear function

and of a quadratic function

with .

Solution (Derivatives of linear and quadratic functions)

1. linear function: For there is

2. quadratic function: For there is

Exercise (Derivative of the logarithm function)

Compute the derivative of the natural logarithm function

directly, using the differential quotient.

Solution (Derivative of the logarithm function)

1st way:

For there is

Nun for we have the inequality

If we swap the roles of and , then there is

Since the left and right-hand sides of the inequality for converge towards , the squeeze theorem implies

2nd way: -method

For there is

Exercise (Computing the derivatives of hyperbolic functions and )

Determine die derivatives of the following functions using the differential quotient

Solution (Computing the derivatives of hyperbolic functions and )

Part 1: Let . Then, there is

Alternative proof:

Part 2: Let . Then, there is

Alternative proof:

Part 3: Let . Then, there is

So

Alternative proof:

Computation rules for derivatives[Bearbeiten]

Applying the computation rules[Bearbeiten]

Exercise (derivatives of a power function)

Show by induction in , that the power function

is differentiable with

Proof (derivatives of a power function)

Induction base:

If , then there is

Induction assumption:

For with , there is

Induction step:

Let . Then is differentiable by induction assumption and the product rule. For there is

Exercise (Derivatives of secant and cosecant)

The functions (secant) and (cosecant) are defined as follows:

as well as

Determine their domain of definition and all derivatives.

Solution (Derivatives of secant and cosecant)

Part 1: secant

domain of definition: is well-defined

Derivative: Since is differentiable on , there is with the chain rule

Part 2: cosecant

domain of definition: is well-defined

Derivative: Since is differentiable on , there is with the chain rule

Exercise (Computing derivatives)

Determine the domain of definition of the following functions, as well as their derivatives

Solution (Computing derivatives)

Part 1:

domain of definition:

Derivative: For there is with the product rule

Part 2:

domain of definition: , as

Derivative: For there is with the chain rule

Part 3:

domain of definition:

Derivative: For there is with the chain- and product rule

Part 4:

domain of definition: , since there must be

Derivative: For there is with the quotient rule

Part 5:

domain of definition:

Derivative:

For there is

For there is

Further there is

as well as

So we have

Concluding all three cases, we get for

Exercise (derivatives of exponential functions)

Determine the derivatives of the following functions on their domain of definition ()

For the function we may leave out the bracket, since in general dis well-defined.

Solution (derivatives of exponential functions)

Part 1: There is . The function is differentiable with . Hence is differentiable by the chain- and product rule and for there is

Part 2: There is . Hence is differentiable by the chain- and product rule and for there is

Part 3: There is . Hence is differentiable by the chain- and product rule and for there is

Part 4: There is . Hence is differentiable by the chain- and product rule and for there is

Part 5: There is . The function is differentiable with . Hence is differentiable by the chain- and product rule and for there is

Exercise (Proof of sum formulas using the derivative)

Proofs by means of binomial theorem (missing) that for all :

Use the binomial theorem and set . Then take the derivative on both sides.

Proof (Proof of sum formulas using the derivative)

For the binomial theorem reads

for and . Now the left-hand side of the equation is a polynomial and the right-hand side is a power function . Both sides are therefore differentiable on with

and

Since there is also . So in particular

and

Exercise (Logarithmic derivatives)

Determine the logarithmic derivatives of the following functions

  1. with

Proof of computational laws[Bearbeiten]

Exercise (Alternative proof of the product rule)

Prove that for differentiable the product rule

holds - by using the chain rule.

Hint: There is:

Proof (Alternative proof of the product rule)

The function is differentiable on with

By der chain rule, we hence have that is differentiable with

for all . Using the hint, we get with the factor- and sum rule

Exercise (Special case of the chain rule)

Derive a general derivative formula for the following function:

If are differentiable.

Solution (Special case of the chain rule)

There is

with and for all . The function is differentiable by the product rule with

By the chain rule, also is differentiable, and there is

Theorem (Computational laws for logarithmic derivatives)

For two differentiable functions and without zeros there is

  1. for and
  2. for and