Criterion for constancy[Bearbeiten]
Exercise (Proof of identities)
Show that

for all 
Proof (Proof of identities)
Part 1: The function
is differentiable by the chain- and difference rule with
So
is constant. Since further there is
we have
.
Part 2:
is differentiable according to the sum rule, since the arcus-functions are differentiable. Further there is
Hence
is constant. Further there is
since
and
. So
establishing the assertion.
Proof (Logarithm representations of
and
)
Part 1:
The function
is differentiable, see examples for derivatives, with
By the chain- and sum rule also
is differentiable with
So we get
. But now,
since
, and there is
So
, and hence
.
Part 2:
is differentiable, as well, with
By the factor-, chain- and quotient rule, also
is differentiable with
So we have
. Since
by
, as well as
there is again
, and hence
.
Proof (General solution of a differential equation)
Part 1: There is
and
So
and
satisfy the differential equations.
Part 2:We define (as given in the hint) the auxiliary functions
These are differentiable by the product-, sum- and difference rule with
and
by the criterion for constancy, there is now
with
. Further there is
So
and
.
Part 3: If further
and
then
Exercise (Monotony of the exponential function)
Show using the monotony criterion that:
- For all
there is 
is strictly monotonously increasing.
Hint: use 1. in order to prove 2.
Exercise (Condition for monotonicity of a cubic function)
Let
. Provide a condition for
such that
is strictly monotonously increasing on all of
.
Solution (Condition for monotonicity of a cubic function)
For
being strictly monotonously increasing on all of
, we need by the monotony criterion that
holds for all
.
Fall 1: 
Fall 2: 
Fall 3: 
Here, we have
However, this is never fulfilled for all
. So in this case
is never strictly monotonous increasing.
Hint: Consider the auxiliary function
.
Proof (Applying the monotony criterion)
As stated in the hint we consider
is differentiable according to the product rule with
But now
and by assumption
. So there is

for all
by the monotony criterion,
is monotonously decreasing. Since further there is
we have

for all
Therefore, we also have

for all
Derivative and extrema[Bearbeiten]
Exercise (Extrema of functions 1)
Investigate whether the following functions have local/global extrema. Determine and characterise these if they exist.


Solution (Extrema of functions 1)
Part 1:
Part 1: Local extrema of 
is differentiable on
according to the quotient rule with
According to the sufficient criterion for the existence of an extremum
, there must be
. Now
So
and
are the candidates for local extrema in
. Now there is
The case
and
is not possible. Thus
holds on
.
Further there is
So
holds on
and on
.
By the sufficient criterion,
is a (strict) local minimum and
is a (strict) local maximum of
.
Part 2: Global extrema of 
Part 2:
Part 1: Local extrema of 
is differentiable on
by the chain rule with
Graph of the function

Since now
and
there is,
. By the necessary criterion for extrema,
has no local extrema on
.
Since
is continuous on
, it follows from
for all
that
is strictly monotonously decreasing on
. Therefore
has a local maximum at
.
Part 2: Global extrema of 
Using the same argument as in part 1, it follows that
is even a global maximum of
.
Exercise (Extrema of functions 2)
Investigate whether the following functions are continuous, differentiability and/or have local/global extrema:
Solution (Extrema of functions 2)
Part 1: continuity and differentiability
Continuity:
On
the function
is continuous as a polynomial. The function
is continuous on
as a composition of the continuous functions
,
and
. At
there is
In addition, since
and by continuity of the exponential function
So
is continuous at zero and hence on all of
.
Differentiability:
On
the function
is differentiable as a polynomial, with
On
the function
is differentiable by the chain- and product rule as it is a composition of differentiable functions
,
and
. There is
At
there is, according to L'Hospital's rule,
So
is not differentiable at zero.
Part 2: Local and global extrema
Local extrema:
On
there is
. So
cannot have local extrema there.
On
however
So
is a candidate for a possible extremum. Further
So
has a strict local minimum in
.
Now we still have to examine
. Since
is not differentiable there, our necessary and sufficient criteria are not applicable. However, there is

for all
and

for all
Thus
is strictly monotonously increasing on
, and strictly monotonously decreasing on
. Since
is continuous at zero, it follows that

for all
So
has a strict local maximum in
.
Global extrema:
There is
Graph of the function

and
Therefore
is unbounded from above and below, and has no global extrema.
Exercise (Extrema of functions 3)
Show that the function
has exactly two local extrema, and determine their type.
Solution (Extrema of functions 3)
Candidates for the extreme values are obtained from our necessary condition
Graph of the auxiliary function

Since the zeros of
cannot be calculated explicitly, we need to examine this function more closely. There is

,
and 
Because of continuity and 2.
with the intermediate value theorem has (at least) two zeros
and
.
Because of 1., the function
is strictly monotonously increasing on
and strictly monotonously decreasing on
. Thus
is respectively injective on
and
and thus has exactly the two zeros
and
.
For the derivative of
we now have
According to our first sufficient criterion,
has a strict local maximum at
and a strict local minimum at
.
Computing limits via L'Hospital[Bearbeiten]
Solution (L'Hospital 1)
Part 1:
Part 2:
L'Hospital's rule is not applicable here. However, the function
is continuous at the point
, and therefore there is
Part 3:
Part 4:
This limit value does not exist. First it can be decomposed into
For the left-hand limit there is now with Part 1:
Analogously, however, for the right-hand limit:
So
, and hence
does not exist.
Part 5:
Part 6:
L'Hospital can be applied here, but it is useless:
Instead, it makes sense to use the definitions of
and
, and then transform the quotient:
Part 7:
L'Hospital cannot be applied here because the enumerator
for
diverges (improperly). Instead, the fraction can be estimated as follows:
Using the squeeze theorem, it follows
.
Part 8:
Part 9:
Part 10:
Solution (L'Hospital 2)
Part 1:
and since the limit exists, the application of L'Hospital is justified.
Part 2:
and since the limit exists, the application of L'Hospital is justified.
Part 3:
and since the limit exists, the application of L'Hospital is justified.
Part 4:
Part 5:
For the expression in the exponent there is now
and since the limit exists, the application of L'Hospital is justified.
By continuity of
at
we now get
Part 6: First we have: If the limit
exists, then the sequence limit
also exists.
Further:
For the expression in the exponent there is now
and since the limit exists, the application of L'Hospital is justified.
By continuity of
at
we now get
And now, we also have
.
Part 7:
For the expression in the exponent there is now
and since the limit exists, the application of L'Hospital is justified.
By continuity of
at
we now get
Part 8:
For the expression in the exponent there is now
and since the limit exists, the application of L'Hospital is justified.
By continuity of
at
we now get
Part 9:
and since the limit exists, the application of L'Hospital is justified.
Part 10:
and since the limit exists, the application of L'Hospital is justified.