# Power series/R/Properties of derivations/No proof/Section

Many important functions, like the exponential function or the trigonometric functions, are represented by a power series. The following theorem shows that these functions are differentiable, and that the derivative of a power series is itself a power series, given by differentiating the individual terms of the series.

## Theorem

Let
    $g(x):=\sum _{n=0}^{\infty }a_{n}x^{n}\,$
denote a power series which converges on the open interval $]-r,r[$, and represents there a function $f\colon ]-r,r[\rightarrow \mathbb {R}$. Then the formally differentiated power series
    ${\tilde {g}}(x):=\sum _{n=1}^{\infty }na_{n}x^{n-1}\,$
is convergent on $]-r,r[$. The function $f$ is differentiable in every point of the interval, and
    $f'(x)={\tilde {g}}(x)\,$

### Proof

$\Box$
In the formulation of the theorem, we have distinguished between $g$ for the power series and $f$ for the function, defined by the series, in order to stress the roles they play. This distinction is now not necessary anymore.

## Corollary
### Proof

This follows immediately from fact.
$\Box$

## Theorem

The exponential function
    $\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \exp x,$
differentiable with
    $\exp \!'(x)=\exp x\,.$

### Proof

Due to fact, we have
    ${\begin{aligned}\exp \!'(x)&={\left(\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\right)}'\\&=\sum _{n=1}^{\infty }{\left({\frac {x^{n}}{n!}}\right)}'\\&=\sum _{n=1}^{\infty }{\frac {n}{n!}}x^{n-1}\\&=\sum _{n=1}^{\infty }{\frac {1}{(n-1)!}}x^{n-1}\\&=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\\&=\exp x.\end{aligned}}$
$\Box$

## Theorem

The exponential function
    $\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto a^{x},$
with base $a>0$, is differentiable with
    ${\left(a^{x}\right)}'={\left(\ln a\right)}a^{x}\,.$

### Proof

By definition, we have
    $a^{x}=\exp {\left(x\,\ln a\right)}\,.$
The derivative with respect to $x$ equals
    ${\left(a^{x}\right)}'={\left(\exp {\left(x\,\ln a\right)}\right)}'={\left(\ln a\right)}\exp '(x\,\ln a)={\left(\ln a\right)}\exp {\left(x\,\ln a\right)}={\left(\ln a\right)}a^{x}\,,$
due to fact and the chain rule.
$\Box$

## Remark

For a real exponential function
    $y(x)=a^{x}\,,$
the relation
    $y'={\left(\ln a\right)}y\,$
holds, due to fact. Hence, there is a proportional relationship between the function $y$ and its derivative $y'$, and $\ln a$ is the factor. This is still true if $a^{x}$ is multiplied with a constant. If we consider $y$ as a function depending on time $x$, then $y'(x)$ describes the growing behavior at that point of time. The equation $y'={\left(\ln a\right)}y$ means that the instantaneous growing rate is always proportional with the magnitude of the function. Such an increasing behavior (or decreasing behavior, if $a<1$) occurs in nature for a population, if there is no competition for resources, and if the dying rate is neglectable (the number of mice is then proportional with the number of mice born). A condition of the form
    $y'=by\,$
is an example of a differential equation. This is an equation for a function, which expresses a condition for the derivative. A solution for such a differential equation is a differentiable function which fulfills the condition on its derivative. The differential equation just mentioned are fulfilled by the functions
    $y(x)=ce^{bx}\,.$

## Corollary

The derivative of the natural logarithm
    $\ln \colon \mathbb {R} _{+}\longrightarrow \mathbb {R} ,x\longmapsto \ln x,$
    $\ln \!'\colon \mathbb {R} _{+}\longrightarrow \mathbb {R} ,x\longmapsto {\frac {1}{x}}.$

### Proof

As the logarithm is the inverse function of the exponential function, we can apply fact and get
    $\ln '(x)={\frac {1}{\exp '(\ln x)}}={\frac {1}{\exp(\ln x)}}={\frac {1}{x}}\,,$
using fact.
$\Box$

## Corollary

Let $\alpha \in \mathbb {R}$. Then the function
    $f\colon \mathbb {R} _{+}\longrightarrow \mathbb {R} _{+},x\longmapsto x^{\alpha },$
is differentiable, and its derivative is
    $f'(x)=\alpha x^{\alpha -1}\,.$

### Proof

By definition, we have
    $x^{\alpha }=\exp {\left(\alpha \,\ln x\right)}\,.$
The derivative with respect to $x$ equals
    ${\left(x^{\alpha }\right)}'={\left(\exp {\left(\alpha \,\ln x\right)}\right)}'={\frac {\alpha }{x}}\cdot \exp {\left(\alpha \,\ln x\right)}={\frac {\alpha }{x}}x^{\alpha }=\alpha x^{\alpha -1}\,$
using fact, fact and the chain rule.
$\Box$

## Theorem

The sine function
    $\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \sin x,$
is differentiable, with
    $\sin \!'(x)=\cos x\,,$
and the cosine function
    $\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \cos x,$
is differentiable, with
    $\cos \!'(x)=-\sin x\,.$

### Proof

$\Box$

## Theorem

The tangent function
    $\mathbb {R} \setminus {\left({\frac {\pi }{2}}+\mathbb {Z} \pi \right)}\longrightarrow \mathbb {R} ,x\longmapsto \tan x,$
is differentiable, with
    $\tan \!'(x)={\frac {1}{\cos ^{2}x}}\,,$
and the cotangent function
    $\mathbb {R} \setminus \mathbb {Z} \pi \longrightarrow \mathbb {R} ,x\longmapsto \cot x,$
is differentiable, with
    $\cot \!'(x)=-{\frac {1}{\sin ^{2}x}}\,.$

### Proof

Using the quotient rule, fact, and the circle equation, we get
    ${\begin{aligned}(\tan x)^{\prime }&={\left({\frac {\sin x}{\cos x}}\right)}^{\prime }\\&={\frac {(\cos x)(\cos x)-(\sin x)(-\sin x)}{\cos ^{2}x}}\\&={\frac {1}{\cos ^{2}x}}.\end{aligned}}$
The derivative of the cotangent function follows in the same way.
$\Box$