Now we will calculate some examples of derivatives from the table above. Often it comes down to determining the differential quotient of the function, i.e. a limit value. But sometimes it is also useful to use the calculation rules from the chapter before.
We start with some simple derivatives:
Theorem (Derivative of a constant function)
Every constant function
f
≡
c
{\displaystyle f\equiv c}
is differentiable on all of
R
{\displaystyle \mathbb {R} }
with derivative
0
{\displaystyle 0}
.
Proof (Derivative of a constant function)
Let
x
~
∈
R
{\displaystyle {\tilde {x}}\in \mathbb {R} }
. Then there is
f
′
(
x
~
)
=
lim
x
→
x
~
f
(
x
)
−
f
(
x
~
)
x
−
x
~
=
lim
x
→
x
~
c
−
c
x
−
x
~
=
lim
x
→
x
~
0
x
−
x
~
=
lim
x
→
x
~
0
=
0
{\displaystyle f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {f(x)-f({\tilde {x}})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {c-c}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {0}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}0=0}
Power functions with natural numbers as powers
Bearbeiten
Now we turn to the derivative of power functions with natural powers. First we will deal with a few special cases:
Example (Derivative of the identity and the square function)
The functions
f
:
R
→
R
,
f
(
x
)
=
x
{\displaystyle f:\mathbb {R} \to \mathbb {R} ,\ f(x)=x}
and
g
:
R
→
R
,
g
(
x
)
=
x
2
{\displaystyle g:\mathbb {R} \to \mathbb {R} ,\ g(x)=x^{2}}
are differentiable on all of
R
{\displaystyle \mathbb {R} }
. Further there is for
x
~
∈
R
{\displaystyle {\tilde {x}}\in \mathbb {R} }
:
f
′
(
x
~
)
=
lim
x
→
x
~
x
−
x
~
x
−
x
~
=
lim
x
→
x
~
1
=
1
{\displaystyle f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {x-{\tilde {x}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}1=1}
as well as
g
′
(
x
~
)
=
lim
x
→
x
~
x
2
−
x
~
2
x
−
x
~
=
lim
x
→
x
~
(
x
+
x
~
)
(
x
−
x
~
)
x
−
x
~
=
lim
x
→
x
~
(
x
+
x
~
)
=
2
x
~
{\displaystyle g'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {x^{2}-{\tilde {x}}^{2}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {(x+{\tilde {x}})(x-{\tilde {x}})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}(x+{\tilde {x}})=2{\tilde {x}}}
For the derivative of
g
{\displaystyle g}
we used the 3rd binomial formula
(
x
+
y
)
(
x
−
y
)
=
x
2
−
y
2
{\displaystyle (x+y)(x-y)=x^{2}-y^{2}}
.
Exercise (Derivative of a power function)
Compute die derivative von
h
:
R
→
R
,
h
(
x
)
=
x
3
{\displaystyle h:\mathbb {R} \to \mathbb {R} ,\ h(x)=x^{3}}
Solution (Derivative of a power function)
For
x
~
∈
R
{\displaystyle {\tilde {x}}\in \mathbb {R} }
there is
h
′
(
x
~
)
=
lim
x
→
x
~
x
3
−
x
~
3
x
−
x
~
=
lim
x
→
x
~
(
x
2
+
x
x
~
+
x
~
2
)
(
x
−
x
~
)
x
−
x
~
=
lim
x
→
x
~
(
x
2
+
x
x
~
+
x
~
2
)
=
3
x
~
2
{\displaystyle h'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {x^{3}-{\tilde {x}}^{3}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {(x^{2}+x{\tilde {x}}+{\tilde {x}}^{2})(x-{\tilde {x}})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}(x^{2}+x{\tilde {x}}+{\tilde {x}}^{2})=3{\tilde {x}}^{2}}
Instead of using the identity
x
3
−
x
~
3
=
(
x
2
+
x
x
~
+
x
~
2
)
(
x
−
x
~
)
{\displaystyle x^{3}-{\tilde {x}}^{3}=(x^{2}+x{\tilde {x}}+{\tilde {x}}^{2})(x-{\tilde {x}})}
, we could also have calculated
x
3
−
x
~
3
x
−
x
~
=
x
2
+
x
x
~
+
x
~
2
{\displaystyle {\tfrac {x^{3}-{\tilde {x}}^{3}}{x-{\tilde {x}}}}=x^{2}+x{\tilde {x}}+{\tilde {x}}^{2}}
using polynomial division.
Now we turn to the general case, i.e. the derivative of
x
↦
x
n
{\displaystyle x\mapsto x^{n}}
for
n
∈
N
{\displaystyle n\in \mathbb {N} }
:
Proof (Derivative of power functions)
For
x
~
∈
R
{\displaystyle {\tilde {x}}\in \mathbb {R} }
, so there is
f
′
(
x
~
)
=
lim
x
→
x
~
x
n
−
x
~
n
x
−
x
~
=
lim
x
→
x
~
∑
k
=
0
n
−
1
x
k
x
~
n
−
1
−
k
=
∑
k
=
0
n
−
1
x
~
k
x
~
n
−
1
−
k
=
∑
k
=
0
n
−
1
x
~
n
−
1
=
n
x
~
n
−
1
{\displaystyle f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {x^{n}-{\tilde {x}}^{n}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}\sum _{k=0}^{n-1}x^{k}{\tilde {x}}^{n-1-k}=\sum _{k=0}^{n-1}{\tilde {x}}^{k}{\tilde {x}}^{n-1-k}=\sum _{k=0}^{n-1}{\tilde {x}}^{n-1}=n{\tilde {x}}^{n-1}}
We used the geometric sum formula
∑
k
=
0
n
−
1
x
k
x
~
n
−
1
−
k
=
x
n
−
x
~
n
x
−
x
~
{\displaystyle \sum _{k=0}^{n-1}x^{k}{\tilde {x}}^{n-1-k}={\tfrac {x^{n}-{\tilde {x}}^{n}}{x-{\tilde {x}}}}}
and the continuity of the polynomial function
x
↦
∑
k
=
0
n
−
1
x
k
x
~
n
−
1
−
k
{\displaystyle x\mapsto \sum _{k=0}^{n-1}x^{k}{\tilde {x}}^{n-1-k}}
.
Using the calculation rules for derivatives we can now calculate the derivatives of polynomial functions and rational functions:
We can already differentiate power functions with natural powers. Now we investigate those with negative integer exponents.
Example (Derivative of the hyperbolic function)
The power function
f
:
R
∖
{
0
}
,
f
(
x
)
=
x
−
1
=
1
x
{\displaystyle f:\mathbb {R} \setminus \{0\},\ f(x)=x^{-1}={\frac {1}{x}}}
is differentiable on
R
∖
{
0
}
{\displaystyle \mathbb {R} \setminus \{0\}}
and there is
f
′
(
x
~
)
=
lim
x
→
x
~
1
x
−
1
x
~
x
−
x
~
=
lim
x
→
x
~
x
~
−
x
x
x
~
x
−
x
~
=
lim
x
→
x
~
x
~
−
x
x
x
~
(
x
−
x
~
)
=
lim
x
→
x
~
−
(
x
−
x
~
)
x
x
~
(
x
−
x
~
)
=
lim
x
→
x
~
−
1
x
x
~
=
−
1
x
~
2
{\displaystyle f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {{\frac {1}{x}}-{\frac {1}{\tilde {x}}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {\frac {{\tilde {x}}-x}{x{\tilde {x}}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {{\tilde {x}}-x}{x{\tilde {x}}(x-{\tilde {x}})}}=\lim _{x\to {\tilde {x}}}{\frac {-(x-{\tilde {x}})}{x{\tilde {x}}(x-{\tilde {x}})}}=\lim _{x\to {\tilde {x}}}{\frac {-1}{x{\tilde {x}}}}=-{\frac {1}{{\tilde {x}}^{2}}}}
for
x
~
∈
R
∖
{
0
}
{\displaystyle {\tilde {x}}\in \mathbb {R} \setminus \{0\}}
.
Exercise (Derivative of
x
↦
1
x
2
{\displaystyle x\mapsto {\tfrac {1}{x^{2}}}}
)
Prove that the power function
f
:
R
∖
{
0
}
,
f
(
x
)
=
x
−
2
=
1
x
2
{\displaystyle f:\mathbb {R} \setminus \{0\},\ f(x)=x^{-2}={\frac {1}{x^{2}}}}
is differentiable on
R
∖
{
0
}
{\displaystyle \mathbb {R} \setminus \{0\}}
and compute its derivative.
Solution (Derivative of
x
↦
1
x
2
{\displaystyle x\mapsto {\tfrac {1}{x^{2}}}}
)
For
x
~
∈
R
∖
{
0
}
{\displaystyle {\tilde {x}}\in \mathbb {R} \setminus \{0\}}
there is
f
′
(
x
~
)
=
lim
x
→
x
~
1
x
2
−
1
x
~
2
x
−
x
~
=
lim
x
→
x
~
x
~
2
−
x
2
x
2
x
~
2
x
−
x
~
=
lim
x
→
x
~
(
x
~
−
x
)
(
x
~
+
x
)
x
2
x
~
2
(
x
−
x
~
)
=
lim
x
→
x
~
−
(
x
−
x
~
)
(
x
+
x
~
)
x
2
x
~
2
(
x
−
x
~
)
=
lim
x
→
x
~
−
(
x
+
x
~
)
x
2
x
~
2
=
−
2
x
~
x
~
4
=
−
2
x
~
3
{\displaystyle f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {{\frac {1}{x^{2}}}-{\frac {1}{{\tilde {x}}^{2}}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {\frac {{\tilde {x}}^{2}-x^{2}}{x^{2}{\tilde {x}}^{2}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {({\tilde {x}}-x)({\tilde {x}}+x)}{x^{2}{\tilde {x}}^{2}(x-{\tilde {x}})}}=\lim _{x\to {\tilde {x}}}{\frac {-(x-{\tilde {x}})(x+{\tilde {x}})}{x^{2}{\tilde {x}}^{2}(x-{\tilde {x}})}}=\lim _{x\to {\tilde {x}}}{\frac {-(x+{\tilde {x}})}{x^{2}{\tilde {x}}^{2}}}=-{\frac {2{\tilde {x}}}{{\tilde {x}}^{4}}}=-{\frac {2}{{\tilde {x}}^{3}}}}
In the general case
x
−
n
=
1
x
n
{\displaystyle x^{-n}={\tfrac {1}{x^{n}}}}
with
n
∈
N
{\displaystyle n\in \mathbb {N} }
there is
Theorem (Derivative of the power function with negative integer powers)
The power function
f
:
R
∖
{
0
}
,
f
(
x
)
=
x
−
n
=
1
x
n
{\displaystyle f:\mathbb {R} \setminus \{0\},\ f(x)=x^{-n}={\frac {1}{x^{n}}}}
is differentiable on
R
∖
{
0
}
{\displaystyle \mathbb {R} \setminus \{0\}}
, and for
x
~
∈
R
∖
{
0
}
{\displaystyle {\tilde {x}}\in \mathbb {R} \setminus \{0\}}
there is
f
′
(
x
~
)
=
−
n
x
n
−
1
{\displaystyle f'({\tilde {x}})=-{\frac {n}{x^{n-1}}}}
Proof (Derivative of the power function with negative integer powers)
For
x
~
∈
R
∖
{
0
}
{\displaystyle {\tilde {x}}\in \mathbb {R} \setminus \{0\}}
there is
f
′
(
x
~
)
=
lim
x
→
x
~
1
x
n
−
1
x
~
n
x
−
x
~
=
lim
x
→
x
~
x
~
n
−
x
n
x
n
x
~
n
x
−
x
~
=
lim
x
→
x
~
1
x
n
x
~
n
x
~
n
−
x
n
x
−
x
~
=
lim
x
→
x
~
1
x
n
x
~
n
−
(
x
~
n
−
x
n
)
x
−
x
~
=
lim
x
→
x
~
1
x
n
x
~
n
⋅
(
−
∑
k
=
0
n
−
1
x
k
x
~
n
−
1
−
k
)
=
1
x
~
2
n
⋅
(
−
∑
k
=
0
n
−
1
x
~
n
−
1
)
=
−
1
x
~
2
n
⋅
n
x
~
n
−
1
=
−
n
x
~
n
−
1
{\displaystyle {\begin{aligned}f'({\tilde {x}})&=\lim _{x\to {\tilde {x}}}{\frac {{\frac {1}{x^{n}}}-{\frac {1}{{\tilde {x}}^{n}}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {\frac {{\tilde {x}}^{n}-x^{n}}{x^{n}{\tilde {x}}^{n}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {1}{x^{n}{\tilde {x}}^{n}}}{\frac {{\tilde {x}}^{n}-x^{n}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {1}{x^{n}{\tilde {x}}^{n}}}{\frac {-({\tilde {x}}^{n}-x^{n})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {1}{x^{n}{\tilde {x}}^{n}}}\cdot \left(-\sum _{k=0}^{n-1}x^{k}{\tilde {x}}^{n-1-k}\right)\\&={\frac {1}{{\tilde {x}}^{2n}}}\cdot \left(-\sum _{k=0}^{n-1}{\tilde {x}}^{n-1}\right)=-{\frac {1}{{\tilde {x}}^{2n}}}\cdot n{\tilde {x}}^{n-1}=-{\frac {n}{{\tilde {x}}^{n-1}}}\end{aligned}}}
Exercise (Derivative of the power function)
Prove
(
1
x
n
)
′
=
−
n
x
n
+
1
{\displaystyle ({\tfrac {1}{x^{n}}})'={\tfrac {-n}{x^{n+1}}}}
using the quotient rule
Solution (Derivative of the power function)
For
x
~
∈
R
∖
{
0
}
{\displaystyle {\tilde {x}}\in \mathbb {R} \setminus \{0\}}
there is by the quotient rule
(
1
x
~
n
)
′
=
0
⋅
x
~
n
−
1
⋅
n
x
~
n
−
1
(
x
~
n
)
2
=
−
n
x
~
n
−
1
x
~
2
n
=
−
n
x
~
2
n
−
(
n
−
1
)
=
−
n
x
~
n
+
1
{\displaystyle \left({\frac {1}{{\tilde {x}}^{n}}}\right)'={\frac {0\cdot {\tilde {x}}^{n}-1\cdot n{\tilde {x}}^{n-1}}{({\tilde {x}}^{n})^{2}}}={\frac {-n{\tilde {x}}^{n-1}}{{\tilde {x}}^{2n}}}={\frac {-n}{{\tilde {x}}^{2n-(n-1)}}}={\frac {-n}{{\tilde {x}}^{n+1}}}}
Remark: Of course we can also apply the inverse rule directly, and thus get the same result
(
1
x
~
n
)
′
=
−
n
x
~
n
−
1
x
~
2
n
=
−
n
x
~
n
+
1
{\displaystyle \left({\frac {1}{{\tilde {x}}^{n}}}\right)'=-{\frac {n{\tilde {x}}^{n-1}}{{\tilde {x}}^{2n}}}={\frac {-n}{{\tilde {x}}^{n+1}}}}
Let us look again at the derivatives rule in the last case, i.e.
f
′
(
x
~
)
=
−
n
x
~
n
−
1
=
−
n
x
~
−
n
−
1
{\displaystyle f'({\tilde {x}})=-{\tfrac {n}{{\tilde {x}}^{n-1}}}=-n{\tilde {x}}^{-n-1}}
for
n
∈
N
{\displaystyle n\in \mathbb {N} }
. If we put
k
=
−
n
∈
Z
−
{\displaystyle k=-n\in \mathbb {Z} ^{-}}
, we get
f
′
(
x
~
)
=
k
x
~
k
−
1
{\displaystyle f'({\tilde {x}})=k{\tilde {x}}^{k-1}}
. The derivative rule is hence the same as for
x
n
{\displaystyle x^{n}}
with
n
∈
N
{\displaystyle n\in \mathbb {N} }
. So we can summarize the two cases and get
Now we investigate the derivative of root functions. We start again with the simplest case:
Example (Derivative of the square root function)
The square root function
f
:
R
+
→
R
,
f
(
x
)
=
x
{\displaystyle f:\mathbb {R} ^{+}\to \mathbb {R} ,\ f(x)={\sqrt {x}}}
is differentiable on
R
+
{\displaystyle \mathbb {R} ^{+}}
and for
x
~
∈
R
+
{\displaystyle {\tilde {x}}\in \mathbb {R} ^{+}}
there is
f
′
(
x
~
)
=
lim
x
→
x
~
x
−
x
~
x
−
x
~
=
lim
x
→
x
~
x
−
x
~
(
x
)
2
−
(
x
~
)
2
=
lim
x
→
x
~
x
−
x
~
(
x
−
x
~
)
(
x
+
x
~
)
=
lim
x
→
x
~
1
x
+
x
~
=
1
2
x
~
{\displaystyle {\begin{aligned}f'({\tilde {x}})&=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt {x}}-{\sqrt {\tilde {x}}}}{x-{\tilde {x}}}}\\[1em]&=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt {x}}-{\sqrt {\tilde {x}}}}{({\sqrt {x}})^{2}-({\sqrt {\tilde {x}}})^{2}}}\\[1em]&=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt {x}}-{\sqrt {\tilde {x}}}}{({\sqrt {x}}-{\sqrt {\tilde {x}}})({\sqrt {x}}+{\sqrt {\tilde {x}}})}}\\[1em]&=\lim _{x\to {\tilde {x}}}{\frac {1}{{\sqrt {x}}+{\sqrt {\tilde {x}}}}}\\[1em]&={\frac {1}{2{\sqrt {\tilde {x}}}}}\end{aligned}}}
Question: Why is the square root function in
x
~
=
0
{\displaystyle {\tilde {x}}=0}
not differentiable, although it is defined and continuous there?
For the differential quotient there is
lim
x
→
0
+
x
−
0
x
−
0
=
lim
x
→
0
+
x
x
=
lim
x
→
0
+
1
x
=
∞
{\displaystyle \lim _{x\to 0+}{\frac {{\sqrt {x}}-{\sqrt {0}}}{x-0}}=\lim _{x\to 0+}{\frac {\sqrt {x}}{x}}=\lim _{x\to 0+}{\frac {1}{\sqrt {x}}}=\infty }
So it does not exist. Hence, we have non-differentiability.
Exercise (Derivative of the cubic root function)
Compute the derivative of the cubic root function
f
:
R
+
→
R
,
f
(
x
)
=
x
3
{\displaystyle f:\mathbb {R} ^{+}\to \mathbb {R} ,\ f(x)={\sqrt[{3}]{x}}}
Solution (Derivative of the cubic root function)
For
x
~
∈
R
+
{\displaystyle {\tilde {x}}\in \mathbb {R} ^{+}}
there is
f
′
(
x
~
)
=
lim
x
→
x
~
x
3
−
x
~
3
x
−
x
~
=
lim
x
→
x
~
x
3
−
x
~
3
(
x
3
)
3
−
(
x
~
3
)
3
=
lim
x
→
x
~
x
3
−
x
~
3
(
x
3
−
x
~
3
)
(
(
x
3
)
2
+
x
3
⋅
x
~
3
+
(
x
~
3
)
2
)
=
lim
x
→
x
~
1
(
x
3
)
2
+
x
3
⋅
x
~
3
+
(
x
~
3
)
2
=
1
3
(
x
~
3
)
2
{\displaystyle {\begin{aligned}f'({\tilde {x}})&=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt[{3}]{x}}-{\sqrt[{3}]{\tilde {x}}}}{x-{\tilde {x}}}}\\[1em]&=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt[{3}]{x}}-{\sqrt[{3}]{\tilde {x}}}}{({\sqrt[{3}]{x}})^{3}-({\sqrt[{3}]{\tilde {x}}})^{3}}}\\[1em]&=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt[{3}]{x}}-{\sqrt[{3}]{\tilde {x}}}}{({\sqrt[{3}]{x}}-{\sqrt[{3}]{\tilde {x}}})(({\sqrt[{3}]{x}})^{2}+{\sqrt[{3}]{x}}\cdot {\sqrt[{3}]{\tilde {x}}}+({\sqrt[{3}]{\tilde {x}}})^{2})}}\\[1em]&=\lim _{x\to {\tilde {x}}}{\frac {1}{({\sqrt[{3}]{x}})^{2}+{\sqrt[{3}]{x}}\cdot {\sqrt[{3}]{\tilde {x}}}+({\sqrt[{3}]{\tilde {x}}})^{2}}}\\[1em]&={\frac {1}{3({\sqrt[{3}]{\tilde {x}}})^{2}}}\end{aligned}}}
Now let us consider the general case of the
k
{\displaystyle k}
-th root function. Here there is
Proof (Derivative of the
k
{\displaystyle k}
-th root function)
For
x
~
∈
R
+
{\displaystyle {\tilde {x}}\in \mathbb {R} ^{+}}
there is
f
′
(
x
~
)
=
lim
x
→
x
~
x
k
−
x
~
k
x
−
x
~
=
lim
x
→
x
~
x
k
−
x
~
k
(
x
k
)
k
−
(
x
~
k
)
k
=
lim
x
→
x
~
1
(
x
k
)
k
−
(
x
~
k
)
k
x
k
−
x
~
k
=
lim
x
→
x
~
1
∑
l
=
0
k
−
1
(
x
k
)
l
(
x
~
k
)
k
−
1
−
l
=
1
∑
l
=
0
k
−
1
(
x
~
k
)
k
−
1
=
1
k
(
x
~
k
)
k
−
1
=
1
k
x
~
k
−
1
k
{\displaystyle f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt[{k}]{x}}-{\sqrt[{k}]{\tilde {x}}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt[{k}]{x}}-{\sqrt[{k}]{\tilde {x}}}}{({\sqrt[{k}]{x}})^{k}-({\sqrt[{k}]{\tilde {x}}})^{k}}}=\lim _{x\to {\tilde {x}}}{\frac {1}{\frac {({\sqrt[{k}]{x}})^{k}-({\sqrt[{k}]{\tilde {x}}})^{k}}{{\sqrt[{k}]{x}}-{\sqrt[{k}]{\tilde {x}}}}}}=\lim _{x\to {\tilde {x}}}{\frac {1}{\sum _{l=0}^{k-1}({\sqrt[{k}]{x}})^{l}({\sqrt[{k}]{\tilde {x}}})^{k-1-l}}}={\frac {1}{\sum _{l=0}^{k-1}({\sqrt[{k}]{\tilde {x}}})^{k-1}}}={\frac {1}{k({\sqrt[{k}]{\tilde {x}}})^{k-1}}}={\frac {1}{k{\sqrt[{k}]{{\tilde {x}}^{k-1}}}}}}
This can now be generalised
The (generalized) exponential function and generalized power functions
Bearbeiten
In this section we prove that the derivative of the exponential function is again the exponential function. So we can determine the derivative of the generalized exponential and power function.
Theorem (Derivative of the exponential function)
The exponential function
f
:
R
→
R
,
f
(
x
)
=
exp
(
x
)
{\displaystyle f:\mathbb {R} \to \mathbb {R} ,\ f(x)=\exp(x)}
is differentiable on
R
{\displaystyle \mathbb {R} }
, and for
x
~
∈
R
{\displaystyle {\tilde {x}}\in \mathbb {R} }
there is
f
′
(
x
~
)
=
exp
(
x
~
)
{\displaystyle f'({\tilde {x}})=\exp({\tilde {x}})}
How to get to the proof? (Derivative of the exponential function)
For this derivative it is more useful to use the
h
{\displaystyle h}
method
f
′
(
x
~
)
=
lim
h
→
0
f
(
x
~
+
h
)
−
f
(
x
~
)
h
{\displaystyle f'({\tilde {x}})=\lim _{h\to 0}{\frac {f({\tilde {x}}+h)-f({\tilde {x}})}{h}}}
Because in this case we know the limit value
lim
h
→
0
exp
(
h
)
−
1
h
=
1
{\displaystyle \lim _{h\to 0}{\frac {\exp(h)-1}{h}}=1}
Furthermore we need the functional equation of the exponential function
exp
(
x
+
y
)
=
exp
(
x
)
exp
(
y
)
{\displaystyle \exp(x+y)=\exp(x)\exp(y)}
Proof (Derivative of the exponential function)
For
x
~
∈
R
{\displaystyle {\tilde {x}}\in \mathbb {R} }
there is
f
′
(
x
~
)
=
lim
h
→
0
f
(
x
~
+
h
)
−
f
(
x
~
)
h
=
lim
h
→
0
exp
(
x
~
+
h
)
−
exp
(
x
~
)
h
=
lim
h
→
0
exp
(
x
~
)
exp
(
h
)
−
exp
(
x
~
)
h
=
lim
h
→
0
exp
(
x
~
)
exp
(
h
)
−
1
h
=
exp
(
x
~
)
⋅
lim
h
→
0
exp
(
h
)
−
1
h
=
exp
(
x
~
)
⋅
1
=
exp
(
x
~
)
{\displaystyle {\begin{aligned}f'({\tilde {x}})&=\lim _{h\to 0}{\frac {f({\tilde {x}}+h)-f({\tilde {x}})}{h}}=\lim _{h\to 0}{\frac {\exp({\tilde {x}}+h)-\exp({\tilde {x}})}{h}}=\lim _{h\to 0}{\frac {\exp({\tilde {x}})\exp(h)-\exp({\tilde {x}})}{h}}\\&=\lim _{h\to 0}\exp({\tilde {x}}){\frac {\exp(h)-1}{h}}=\exp({\tilde {x}})\cdot \lim _{h\to 0}{\frac {\exp(h)-1}{h}}=\exp({\tilde {x}})\cdot 1=\exp({\tilde {x}})\end{aligned}}}
Using the chain rule, the derivatives of the generalized exponential function
x
↦
a
x
{\displaystyle x\mapsto a^{x}}
for
a
∈
R
+
{\displaystyle a\in \mathbb {R} ^{+}}
and the generalized power function
x
↦
x
r
{\displaystyle x\mapsto x^{r}}
for
r
∈
R
{\displaystyle r\in \mathbb {R} }
can be calculated:
Proof (Derivative of the generalized exponential function)
For
x
~
∈
R
{\displaystyle {\tilde {x}}\in \mathbb {R} }
there is
f
′
(
x
~
)
=
exp
(
x
~
ln
(
a
)
)
⋅
ln
(
a
)
=
ln
(
a
)
a
x
~
{\displaystyle f'({\tilde {x}})=\exp({\tilde {x}}\ln(a))\cdot \ln(a)=\ln(a)a^{\tilde {x}}}
Exercise (Derivative of the generalized exponential function)
Prove that the derivative of the generalized power function at
x
~
∈
R
+
{\displaystyle {\tilde {x}}\in \mathbb {R} ^{+}}
is
r
x
~
r
−
1
{\displaystyle r{\tilde {x}}^{r-1}}
.
Proof (Derivative of the generalized exponential function)
For
x
~
∈
R
{\displaystyle {\tilde {x}}\in \mathbb {R} }
the chain rule yields
f
′
(
x
~
)
=
exp
(
ln
(
x
~
)
r
)
⋅
r
x
~
=
r
x
~
r
1
x
~
=
r
x
~
r
−
1
{\displaystyle f'({\tilde {x}})=\exp(\ln({\tilde {x}})r)\cdot {\frac {r}{\tilde {x}}}=r{\tilde {x}}^{r}{\frac {1}{\tilde {x}}}=r{\tilde {x}}^{r-1}}
Now we turn to the derivative of the natural and generalised logarithm function. Since the natural logarithm is the inverse of the exponential function, we can deduce its derivative directly from rule for derivatives of inverse function :
Theorem (Derivative of the natural logarithm function)
The natural logarithm function
g
:
R
+
→
R
,
g
(
x
)
=
ln
x
{\displaystyle g:\mathbb {R} ^{+}\to \mathbb {R} ,\ g(x)=\ln x}
is differentiable on
R
+
{\displaystyle \mathbb {R} ^{+}}
. For
x
~
∈
R
+
{\displaystyle {\tilde {x}}\in \mathbb {R} ^{+}}
there is
g
′
(
y
)
=
1
y
{\displaystyle g'(y)={\frac {1}{y}}}
The derivative can also be calculated directly using the differential quotient. If you want to try this, we recommend the corresponding exercise (missing) .
Using the derivative of the natural logarithm function we can now immediately conclude
Proof (Derivative of the generalized logarithm function)
From the derivative rule for the multiple of a function, we get that for all
y
∈
R
+
{\displaystyle y\in \mathbb {R} ^{+}}
:
g
′
(
y
)
=
1
ln
a
1
y
=
1
y
ln
a
{\displaystyle g'(y)={\frac {1}{\ln a}}{\frac {1}{y}}={\frac {1}{y\ln a}}}
If the derivative of the natural logarithm is not available, we can calculate it using the theorem of the derivative of the inverse function.
Theorem (Derivative of the sine function)
The sine function is differentiable. For all
x
~
∈
R
{\displaystyle {\tilde {x}}\in \mathbb {R} }
there is:
sin
′
(
x
~
)
=
cos
(
x
~
)
{\displaystyle \sin '({\tilde {x}})=\cos({\tilde {x}})}
Proof (Derivative of the sine function)
For
x
~
∈
R
{\displaystyle {\tilde {x}}\in \mathbb {R} }
there is
sin
′
(
x
~
)
=
lim
h
→
0
sin
(
x
~
+
h
)
−
sin
(
x
~
)
h
↓
sin
(
x
+
y
)
=
sin
(
x
)
cos
(
y
)
+
cos
(
x
)
sin
(
y
)
=
lim
h
→
0
sin
(
x
~
)
cos
(
h
)
+
cos
(
x
~
)
sin
(
h
)
−
sin
(
x
~
)
h
=
sin
(
x
~
)
⋅
lim
h
→
0
cos
(
h
)
−
1
h
⏟
=
0
+
cos
(
x
~
)
⋅
lim
h
→
0
sin
(
h
)
h
⏟
=
1
=
cos
(
x
~
)
{\displaystyle {\begin{aligned}\sin '({\tilde {x}})&=\lim _{h\to 0}{\frac {\sin({\tilde {x}}+h)-\sin({\tilde {x}})}{h}}\\[0.3em]&\color {Gray}\left\downarrow \ \sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\right.\\[0.3em]&=\lim _{h\to 0}{\frac {\sin({\tilde {x}})\cos(h)+\cos({\tilde {x}})\sin(h)-\sin({\tilde {x}})}{h}}\\[0.3em]&=\sin({\tilde {x}})\cdot \underbrace {\lim _{h\to 0}{\frac {\cos(h)-1}{h}}} _{=0}+\cos({\tilde {x}})\cdot \underbrace {\lim _{h\to 0}{\frac {\sin(h)}{h}}} _{=1}\\[0.3em]&=\cos({\tilde {x}})\end{aligned}}}
Theorem (Derivative of the cosine function)
The cosine function is differentiable with
cos
′
(
x
~
)
=
−
sin
(
x
~
)
{\displaystyle \cos '({\tilde {x}})=-\sin({\tilde {x}})}
Proof (Derivative of the cosine function)
cos
′
(
x
~
)
=
lim
h
→
0
cos
(
x
~
+
h
)
−
cos
(
x
~
)
h
↓
cos
(
x
+
y
)
=
cos
(
x
)
cos
(
y
)
−
sin
(
x
)
sin
(
y
)
=
lim
h
→
0
cos
(
x
~
)
cos
(
h
)
−
sin
(
x
~
)
sin
(
h
)
−
cos
(
x
~
)
h
=
cos
(
x
~
)
⋅
lim
h
→
0
cos
(
h
)
−
1
h
⏟
=
0
−
sin
(
x
~
)
⋅
lim
h
→
0
sin
(
h
)
h
⏟
=
1
=
−
sin
(
x
~
)
{\displaystyle {\begin{aligned}\cos '({\tilde {x}})&=\lim _{h\to 0}{\frac {\cos({\tilde {x}}+h)-\cos({\tilde {x}})}{h}}\\[0.5em]&\color {Gray}\left\downarrow \ \cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)\right.\\[0.5em]&=\lim _{h\to 0}{\frac {\cos({\tilde {x}})\cos(h)-\sin({\tilde {x}})\sin(h)-\cos({\tilde {x}})}{h}}\\[0.5em]&=\cos({\tilde {x}})\cdot \underbrace {\lim _{h\to 0}{\frac {\cos(h)-1}{h}}} _{=0}-\sin({\tilde {x}})\cdot \underbrace {\lim _{h\to 0}{\frac {\sin(h)}{h}}} _{=1}\\[0.5em]&=-\sin({\tilde {x}})\end{aligned}}}
Theorem (Derivative of the tangent function)
The tangent function
tan
:
D
=
R
∖
{
π
2
+
k
π
∣
k
∈
Z
}
→
R
,
tan
(
x
)
=
sin
(
x
)
cos
(
x
)
{\displaystyle \tan :D=\mathbb {R} \setminus \{{\tfrac {\pi }{2}}+k\pi \mid k\in \mathbb {Z} \}\to \mathbb {R} ,\ \tan(x)={\frac {\sin(x)}{\cos(x)}}}
is differentiable on
D
{\displaystyle D}
, and for
x
~
∈
D
{\displaystyle {\tilde {x}}\in D}
there is
tan
′
(
x
~
)
=
1
cos
2
(
x
~
)
=
1
+
tan
2
(
x
~
)
{\displaystyle \tan '({\tilde {x}})={\frac {1}{\cos ^{2}({\tilde {x}})}}=1+\tan ^{2}({\tilde {x}})}
Exercise (Derivative of the cotangent function)
The cotangent function
cot
:
D
=
R
∖
{
k
π
∣
k
∈
Z
}
→
R
,
cot
(
x
)
=
1
tan
(
x
)
=
cos
(
x
)
sin
(
x
)
{\displaystyle \cot :D=\mathbb {R} \setminus \{k\pi \mid k\in \mathbb {Z} \}\to \mathbb {R} ,\ \cot(x)={\frac {1}{\tan(x)}}={\frac {\cos(x)}{\sin(x)}}}
is differentiable on
D
{\displaystyle D}
, and for
x
~
∈
D
{\displaystyle {\tilde {x}}\in D}
there is
cot
′
(
x
~
)
=
−
1
sin
2
(
x
~
)
=
−
1
−
cot
2
(
x
~
)
{\displaystyle \cot '({\tilde {x}})=-{\frac {1}{\sin ^{2}({\tilde {x}})}}=-1-\cot ^{2}({\tilde {x}})}
The derivatives of secant and cosecant can be found in the corresponding exercise.
Using the rule for derivatives of the inverse function we can differentiate the arc-functions (which are inverses of sine, cosine, etc.)
Theorem (Derivative of the arcsin/arccos function)
The inverse functions of the trigonometric functions
arcsin
{\displaystyle \arcsin }
,
arccos
{\displaystyle \arccos }
are differentiable with
arcsin
′
(
x
~
)
=
1
1
−
x
~
2
for all
x
~
∈
(
−
1
,
1
)
,
arccos
′
(
x
~
)
=
−
1
1
−
x
~
2
for all
x
~
∈
(
−
1
,
1
)
,
{\displaystyle {\begin{aligned}\arcsin '({\tilde {x}})&={\frac {1}{\sqrt {1-{\tilde {x}}^{2}}}}\quad {\text{ for all }}{\tilde {x}}\in (-1,1),\\\arccos '({\tilde {x}})&=-{\frac {1}{\sqrt {1-{\tilde {x}}^{2}}}}\quad {\text{ for all }}{\tilde {x}}\in (-1,1),\\\end{aligned}}}
Proof (Derivative of the arcsin/arccos function)
Derivative of
arcsin
{\displaystyle \arcsin }
:
For the sine function
sin
:
(
−
π
2
,
π
2
)
→
R
{\displaystyle \sin :(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})\to \mathbb {R} }
there is:
sin
′
=
cos
{\displaystyle \sin '=\cos }
. So the function is differentiable, and since
cos
(
x
)
>
0
{\displaystyle \cos(x)>0}
for all
x
∈
(
−
π
2
,
π
2
)
{\displaystyle x\in (-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})}
, it is strictly monotonously increasing on this interval. Further,
sin
(
(
−
π
2
,
π
2
)
)
=
(
−
1
,
1
)
{\displaystyle \sin((-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}))=(-1,1)}
. So
sin
:
(
−
π
2
,
π
2
)
→
(
−
1
,
1
)
{\displaystyle \sin :(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})\to (-1,1)}
is surjective. The inverse function is the arc sine function
arcsin
:
(
−
1
,
1
)
→
(
−
π
2
,
π
2
)
→
R
{\displaystyle \arcsin :(-1,1)\to (-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})\to \mathbb {R} }
From the theorem about the derivative of the inverse we now have for every
x
~
∈
(
−
1
,
1
)
{\displaystyle {\tilde {x}}\in (-1,1)}
:
arcsin
′
(
x
~
)
=
1
sin
′
(
arcsin
(
x
~
)
)
=
1
cos
(
arcsin
(
x
~
)
)
=
trigonometric
Pythagoras
1
1
−
sin
2
(
arcsin
(
x
~
)
)
=
1
1
−
x
~
2
{\displaystyle \arcsin '({\tilde {x}})={\frac {1}{\sin '(\arcsin({\tilde {x}}))}}={\frac {1}{\cos(\arcsin({\tilde {x}}))}}{\underset {\text{Pythagoras}}{\overset {\text{trigonometric}}{=}}}{\frac {1}{\sqrt {1-\sin ^{2}(\arcsin({\tilde {x}}))}}}={\frac {1}{\sqrt {1-{\tilde {x}}^{2}}}}}
Derivative of
arccos
{\displaystyle \arccos }
:
For the cosine function
cos
:
(
0.
π
)
→
R
{\displaystyle \cos :(0.\pi )\to \mathbb {R} }
there is:
cos
′
=
−
sin
{\displaystyle \cos '=-\sin }
. So the function is differentiable, and because of
−
sin
(
x
)
|
(
0
,
π
)
<
0
{\displaystyle -\sin(x)|_{(0,\pi )}<0}
, strictly monotonously decreasing. Further,
cos
(
(
0
,
π
)
)
=
(
−
1
,
1
)
{\displaystyle \cos((0,\pi ))=(-1,1)}
. So
cos
:
(
0
,
π
)
→
(
−
1
,
1
)
{\displaystyle \cos :(0,\pi )\to (-1,1)}
is surjective. The inverse function
arccos
:
(
−
1
,
1
)
→
(
0
,
π
)
→
R
{\displaystyle \arccos :(-1,1)\to (0,\pi )\to \mathbb {R} }
is differentiable according to the theorem about the derivative of the inverse function, and for every
x
~
∈
(
−
1
,
1
)
{\displaystyle {\tilde {x}}\in (-1,1)}
there is:
arccos
′
(
x
~
)
=
1
cos
′
(
arccos
(
x
~
)
)
=
1
−
sin
(
arcsin
(
x
~
)
)
=
trigonometric
Pythagoras
−
1
1
−
cos
2
(
arccos
(
x
~
)
)
=
−
1
1
−
x
~
2
{\displaystyle \arccos '({\tilde {x}})={\frac {1}{\cos '(\arccos({\tilde {x}}))}}={\frac {1}{-\sin(\arcsin({\tilde {x}}))}}{\underset {\text{Pythagoras}}{\overset {\text{trigonometric}}{=}}}-{\frac {1}{\sqrt {1-\cos ^{2}(\arccos({\tilde {x}}))}}}=-{\frac {1}{\sqrt {1-{\tilde {x}}^{2}}}}}
Theorem (Derivative of the arctan/ arccot function)
The inverse functions of the trigonometric functions
arctan
{\displaystyle \arctan }
,
arccot
{\displaystyle \operatorname {arccot} }
are differentiable, and there is
arctan
′
(
x
~
)
=
1
1
+
x
~
2
for all
x
~
∈
R
arccot
′
(
x
~
)
=
−
1
1
+
x
~
2
for all
x
~
∈
R
{\displaystyle {\begin{aligned}\arctan '({\tilde {x}})&={\frac {1}{1+{\tilde {x}}^{2}}}\quad {\text{ for all }}{\tilde {x}}\in \mathbb {R} \\\operatorname {arccot} '({\tilde {x}})&=-{\frac {1}{1+{\tilde {x}}^{2}}}\quad {\text{ for all }}{\tilde {x}}\in \mathbb {R} \end{aligned}}}
Proof (Derivative of the arctan/ arccot function)
For the tangent function
tan
|
(
−
π
2
,
π
2
)
{\displaystyle \tan |_{(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})}}
there is:
tan
′
=
1
+
tan
2
>
0
{\displaystyle \tan '=1+\tan ^{2}>0}
. So the function is differentiable and strictly monotonically increasing. Further,
tan
(
(
−
π
2
,
π
2
)
)
=
R
{\displaystyle \tan((-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}))=\mathbb {R} }
. So
tan
:
(
−
π
2
,
π
2
)
→
R
{\displaystyle \tan :(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})\to \mathbb {R} }
is surjective. The inverse function
arccos
:
(
−
1
,
1
)
→
(
0
,
π
)
→
R
{\displaystyle \arccos :(-1,1)\to (0,\pi )\to \mathbb {R} }
is hence differentiable, and now for
x
~
∈
(
−
1
,
1
)
{\displaystyle {\tilde {x}}\in (-1,1)}
there is:
arctan
′
(
x
~
)
=
1
tan
′
(
arctan
(
x
~
)
)
=
1
1
+
tan
2
(
arctan
(
x
~
)
)
=
1
1
+
x
~
2
{\displaystyle \arctan '({\tilde {x}})={\frac {1}{\tan '(\arctan({\tilde {x}}))}}={\frac {1}{1+\tan ^{2}(\arctan({\tilde {x}}))}}={\frac {1}{1+{\tilde {x}}^{2}}}}
And finally, we determine the derivatives of the hyperbolic functions
sinh
{\displaystyle \sinh }
,
cosh
{\displaystyle \cosh }
and
tanh
{\displaystyle \tanh }
:
Theorem (Derivative of hyperbolic functions)
The functions
sinh
:
R
→
R
,
sinh
(
x
)
=
e
x
−
e
−
x
2
cosh
:
R
→
R
,
cosh
(
x
)
=
e
x
+
e
−
x
2
tanh
:
R
→
R
,
tanh
(
x
)
=
sinh
(
x
)
cosh
(
x
)
{\displaystyle {\begin{aligned}\sinh :\mathbb {R} \to \mathbb {R} ,\ \sinh(x)={\frac {e^{x}-e^{-x}}{2}}\\\cosh :\mathbb {R} \to \mathbb {R} ,\ \cosh(x)={\frac {e^{x}+e^{-x}}{2}}\\\tanh :\mathbb {R} \to \mathbb {R} ,\ \tanh(x)={\frac {\sinh(x)}{\cosh(x)}}\end{aligned}}}
are differentiable, and there is
sinh
′
(
x
)
=
cosh
(
x
)
cosh
′
(
x
~
)
=
sinh
(
x
~
)
tanh
′
(
x
~
)
=
1
cosh
2
(
x
~
)
=
tanh
2
(
x
~
)
−
1
{\displaystyle {\begin{aligned}\sinh '(x)&=\cosh(x)\\\cosh '({\tilde {x}})&=\sinh({\tilde {x}})\\\tanh '({\tilde {x}})&={\frac {1}{\cosh ^{2}({\tilde {x}})}}=\tanh ^{2}({\tilde {x}})-1\\\end{aligned}}}
Proof (Derivative of hyperbolic functions)
The derivatives follow directly from the calculation rules. We show only the derivative of
sinh
{\displaystyle \sinh }
. The other two are left to you for practice.
According to the factor and difference rule
sinh
(
x
)
=
1
2
e
x
−
1
2
e
−
x
{\displaystyle \sinh(x)={\tfrac {1}{2}}e^{x}-{\tfrac {1}{2}}e^{-x}}
for all
x
∈
R
{\displaystyle x\in \mathbb {R} }
is differentiable, and there is
sinh
′
(
x
)
=
1
2
e
x
−
1
2
e
−
x
(
−
1
)
=
1
2
e
x
+
1
2
e
−
x
=
cosh
(
x
)
{\displaystyle \sinh '(x)={\tfrac {1}{2}}e^{x}-{\tfrac {1}{2}}e^{-x}(-1)={\tfrac {1}{2}}e^{x}+{\tfrac {1}{2}}e^{-x}=\cosh(x)}