Exercises: Derivatives 1 – Serlo
Computing derivatives with differential quotients
BearbeitenExrecises: derivative and differentiability
BearbeitenExercise (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
So does not exist.
Exercise (Determining limits with the differential quotient)
Let be differentiable at . Show that the following limits exist
- for
How to get to the proof? (Determining limits with the differential quotient)
Since in is differentiable, there is
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
BearbeitenExercise (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
BearbeitenApplying the computation rules
BearbeitenExercise (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
- with
Proof of computational laws
BearbeitenExercise (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
- for and
- for and