Solution (Derivative of the inverse function)
Part 1:
Proof step: Existence and computation of
Proof step: Existence and computation of
Part 2:
Proof step: Computation of
There is
So
Proof step: Computation of
Derivative of a general logarithm function
[Bearbeiten]
Exercise (Derivative of a general logarithm function)
Show that the function
with base is bijective and differentiable on . Show further that the inverse function
is differentiable on all of with derivative .
Solution (Derivative of a general logarithm function)
Proof step: is bijective and differentiable
Fall 1:
Fall 2:
Proof step: exists and is differentiable
Proof (derivatives on and area-functions)
Differentiability of :
For the cotangent function there is:. Thus the function is differentiable and strictly monotonically decreasing, and thus injective. Further, . Thus is bijective. The inverse function
is differentiable according to the theorem on the derivative of the inverse function, and for there is:
Differentiability of :
The hyperbolic sine function is differentiable with . Thus it is strictly monotonously increasing, and hence injective. Further, . So it is also surjective. The inverse function
is differentiable by the theorem on the derivative of the inverse function , and for there is:
Differentiability of :
The hyperbolic cosine function is differentiable with on . Thus it is strictly monotonically increasing, and hence injective. Further, . So it is also surjective. The inverse function
is differentiable by to the theorem on the derivative of the inverse function, and for there is:
Differentiability of :
For the cotangent function there is: . Thus the function is strictly monotonically increasing, and thus injective. Further, . Thus is bijective. The inverse function
is differentiable by the theorem on the derivative of the inverse function, and for there is:
Exercise (Non-differentiable functions at zero)
Let . Show that:
- Let with and for all . Then and are not simultaneously differentiable at zero.
- Let , and let be differentiable at zero. Further let and for all . Then is not differentiable at zero.
Exercise (Arbitrarily often differentiable function)
Show that the function is arbitrarily often differentiable and for all there is:
Proof (Arbitrarily often differentiable function)
The proof goes by induction over :
Theorem whose validity shall be proven for the :
1. Base case:
1. inductive step:
2a. inductive hypothesis:
2b. induction theorem:
2b. proof of induction step:
Exercise (Exactly one/two/three times differentiable functions)
Provide an example of a
- function
- function that is differentiable, but not continuously differentiable on
- function
Solution (Exactly one/two/three times differentiable functions)
Solution sub-exercise 1:
or
or
Solution sub-exercise 2:
or
Solution sub-exercise 3:
or
or
Solution sub-exercise 4:
or
or
Hint
We can successively proceed with the construction of functions and obtain some for all , with
or
or
Exercise (Application of the Leibniz rule)
Determine the following derivatives using the Leibniz rule
- for
Solution (Application of the Leibniz rule)