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Computing derivatives - special – Serlo

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Special cases of the chain rule

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Now we want to list a few special cases of the chain rule, which occur frequently in practice. For the derivation of the derivatives of , , , , etc. we refer to the following chapter Examples for derivatives (missing).

Case: is linear

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Let and let be differentiable. Then also is differentiable ad at there is

Proof

is differentiable with for all . The chain rule implies differentiablilty of , where

Example

Let with . Then for all so

Case: is a power function

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Let be differentiable. The also is differentiable for all , where at there is

Proof

is differentiable with for all . The chain rule implies differentiablility of , where

Example (Deriving a power function)

Let with . Then and for all we have

Case: is a root function

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Let be differentiable. then with is differentiable as well and for all there is

Proof

is differentiable with for all . The chain rule implies differentiability of where

Example (Deriving a root function)

Let with . Then and for all there is

Case:

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Let be differentiable. Then is differentiable as well and for all there is

Proof

Let , which is differentiable with . Since is differentiable by assumption, we also get differentiability of . By the chain rule,

Example (Deriving exponential functions)

1. Let with . Then, for all and we have

2. Let with . Then, for all and we have

Special case: Differentiating "function to the power of a function"

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Consider the function

which is a special case of an exponential function. The inner function is . We may again just use the chain rule.

Example (Deriving exponential functions 2)

1. Let with . Then, for all and by the chain rule

2. Let with . Then, for all and by the chain rule

3. Let with . Then, for all and by the chain rule

Case:

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Let be and by the chain rule. Then, is and by the chain rule as well and for all there is

(logarithmic derivative)

Proof

Let , which is and by the chain rule with for all . Since is and by the chain rule by assumption, the chain rule implies differentiability of and

Example (Logarithmic derivatives)

1. Let with . Then, for all and by the chain rule

2. Let with . Then, for all and by the chain rule

Questions for understanding: Answer the questions:

  1. Why is the domain of only ?
  2. What of domain of ?

Solutions:

  1. There is is well-defined
  2. There is is well-defined. So . For the derivative of there is

Hint

Below we will see how we can use the logarithmic derivative to calculate easily the derivatives of product, quotient or power functions. This makes sense especially if the function consists of several products, for example. ()

Linear combinations of functions

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The factor and sum rule state that the derivative is linear. If we apply this linearity to functions, we get:

Theorem (Differentiating linear combinations of functions)

Let , be differentiable and . Then,

is differentiable as well and for all there is

Proof (Differentiating linear combinations of functions)

We show the assertion by induction over :

Induction base: . For there is

Induction assumption:

shall hold for some

Induction step: .

Example (Differentiability of polynomial functions)

The power function is differentiable for all where

The theorem from above applied to polynomial functions yields

for and differentiable with

Application: Deriving sum formulas

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We can use the linearity of the derivative to obtain new sum formulas from already known ones. Let us consider as an example the geometric sum formula (missing) for and :

Both sides of the equation can be understood as differentiable functions or or :

Since is a polynomial, we have for :

Furthermore, by the quotient rule

Since now , we also have . So for there is:

Additional question: Which sum formulas do we get for and ?

For we get

and for

Generalized product rule

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The product rule can also be applied to more than two differentiable functions by first combining several functions and then applying the product rule several times in succession. For three functions we get

For four functions we get analogously

We now recognize a clear formation law for derivatives: the product of the functions is added up, whereby in each summand the derivative "moves forward" by one position. In general, the derivative of a product function of functions is:

Theorem (Generalized product rule)

Let and let be differentiable. The the product function is also differentiable with

Exercise (Proof of the generalized product rule)

Prove the generalized product rule by induction over .

Proof (Proof of the generalized product rule)

Induction base: . Es gilt

Induction assumption:

is assumed for some

Induction step: .

Example (Generalized product rule)

The function

is differentiable, since , and are differentiable for all . In addition

, and

The generalized product rule yields for :

Exercise (Generalized product rule)

Determine the domain of definition and the derivative of

Solution (Generalized product rule)

Domain of definition: The functions , and are defined on all of . by contrast, is only defined on . Hence

Derivative: is differentiable, as the functions , , and are differentiable. In addition, for all there is:

, , and

The generalized product rule yields:

Hint

If additionally for all , we can divide both sides by this product, and thus obtain the form

The advantage of this representation is that the sum on the right side is much clearer. This is already the idea behind the logarithmic derivative, which we present in the next section.

Logarithmic derivatives

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The logarithmic derivative is a very elegant tool to calculate the derivative of some functions of a special form. For a differentiable function without zeros, the logarithmic derivative is defined by

We have already shown above that the chain rule yields:

The following table lists some standard examples of logarithmic derivatives:

Domain of definition
,

Exercise (Computing logarithmic derivatives)

Determine the logarithmic derivative (with domain of definition) of the following functions

  1. with

Solution (Computing logarithmic derivatives)

Part 1: There is for all . So

Since for all , the domain of definition for our logarithmic derivative of is equal to .

Part 2: The quotient rule yields

for all

So

Since for all , the domain of definition for our logarithmic derivative of is equal to .

Part 3: For there is

Since for all , the domain of definition for our logarithmic derivative of is equal to .

By direct computation we obtain the following rules for the logarithmic derivative:

Theorem (Computation rules for logarithmic derivatives)

For two differentiable functions and without zeros, there is

  1. for
  2. for

Note: The rules are analogous to the computation rules for the logarithm function.

Proof (Computation rules for logarithmic derivatives)

We will only prove rule 1 and rule 4, the other three we leave to you "as an exercise" (don't worry, there is a solution, here).

Rule 1: Since and are differentiable and free of zeros, is also differentiable and free of zeros. Thus the following applies

Rule 4: Since is differentiable and zero-free, is also differentiable and zero-free for . Further above we have already shown by the chain rule. So

Exercise (Computation rules for logarithmic derivatives)

Prove rules 2, 3 and 5 of the previous theorem

Proof (Computation rules for logarithmic derivatives)

Rule 2: Since are differentiable and zero-free, is also differentiable and zero-free. By the chain rule, . Thus, there is

Rule 3: Since and are differentiable and zero-free, is also differentiable and zero-free. Using rules 1 and 2 we get

Alternatively, the rule can be proved by using the quotient rule.

Rule 5: Since are differentiable and positive, is also differentiable and positive. With the chain rule, applies. Thus,

Hint

The summation rule can still be generalized to zero-free and differentiable ( as

Using those rules, we can now easily calculate derivatives. The transition to logarithmic derivatives does not usually require less computational effort, but it is much clearer than calculating with the usual rules, and therefore less susceptible to errors!

Example (Logarithmic derivatives 1)

First we differentiate the following product function by means of the logarithmic derivation

First we determine the domain of definition: there is , and . In order to be able to form the logarithmic derivative, and must be zero-free. Because of we will choose .


Now take the logarithmic derivative of : There is

Finally, we multiply the equation by and obtain

Example (Logarithmic derivatives 2)

Next, we differentiate the following quotient function:

Concerning the domain: The denominator is always different from zero. In order for to be free of zeros, the numerator must be unequal to zero. Therefore:

Hence, the domain of definition is .

With and we have for the logarithmic derivative of :

Multiplication by yields:

Example (Logarithmic derivatives 3)

Finally, we differentiate with the logarithmic derivative

Concerning the domain: In order for to be defined, must hold. The function is zero-free on all of . So .

The logarithmic derivative of is

Multiplication by yields:

Exercise (Logarithmic derivatives)

Using the logarithmic derivatives, differentiate the following functions on their domain of definition:

Generalized chain rule

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Just like the sum and product rule, the chain rule can be generalized to the composition of more than two functions. For two differentiable functions and the chain rule reads

If we have three functions , and , then by applying the rule twice we obtain

If we now take a closer look, we can see a law of formation: First the outermost function is differentiated and the two inner ones are inserted into the derivative function. Then the second function is differentiated and the innermost function is inserted, and the whole thing is multiplied by the first derivative. Finally, the innermost function is differentiated and multiplied. If we now generalize this to functions, we get:

Theorem (Generalized chain rule)

Let be differentiable for all , and for all . Then is also differentiable, and its derivative at is given by

Proof (Generalized chain rule)

We prove the theorem by induction over :

Induction base: . There is

. The chain rule yields

Induction assumption:

is assumed for all and some

Induction step: . For there is