Exercises: Derivatives 1 – Serlo

Computing derivatives with differential quotients

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Exrecises: derivative and differentiability

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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  .

  1.  
  2.  

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.  
  2.   for  
  3.  

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

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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

  1.  
  2.  
  3.  

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

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Applying the computation rules

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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

  1.  
  2.  
  3.  
  4.  
  5.  

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

  1.  
  2.  

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 ( )

  1.  
  2.  
  3.  
  4.  
  5.  

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.  
  2.  
  3.   with  
  4.  

Proof of computational laws

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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  
  3.