Computing derivatives - special – Serlo
Special cases of the chain rule
BearbeitenNow 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
BearbeitenLet 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
BearbeitenLet 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
BearbeitenLet 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:
BearbeitenLet 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"
BearbeitenConsider 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:
BearbeitenLet be and by the chain rule. Then, is and by the chain rule as well and for all there is
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:
- Why is the domain of only ?
- What of domain of ?
Solutions:
- There is is well-defined
- 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
BearbeitenThe 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:
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
BearbeitenWe 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
BearbeitenThe 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:
Induction step: .
Example (Generalized product rule)
The function
is differentiable, since , and are differentiable for all . In addition
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:
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
BearbeitenThe 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
- 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
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
- for
- 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
BearbeitenJust 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:
Induction step: . For there is