# 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

${\displaystyle {}g(x):=\sum _{n=0}^{\infty }a_{n}x^{n}\,}$

denote a power series which converges on the open interval ${\displaystyle {}]-r,r[}$, and represents there a function ${\displaystyle {}f\colon ]-r,r[\rightarrow \mathbb {R} }$. Then the formally differentiated power series

${\displaystyle {}{\tilde {g}}(x):=\sum _{n=1}^{\infty }na_{n}x^{n-1}\,}$

is convergent on ${\displaystyle {}]-r,r[}$. The function ${\displaystyle {}f}$ is differentiable in every point of the interval, and

${\displaystyle {}f'(x)={\tilde {g}}(x)\,}$

holds.

### Proof

The proof requires a detailed study of power series.
${\displaystyle \Box }$

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

## Corollary

A function given by a power series is infinitely often differentiable on its interval of convergence.

### Proof

This follows immediately from fact.

${\displaystyle \Box }$

## Theorem

${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \exp x,}$
is

differentiable with

${\displaystyle {}\exp \!'(x)=\exp x\,.}$

### Proof

Due to fact, we have

{\displaystyle {}{\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}}}
${\displaystyle \Box }$

## Theorem

${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto a^{x},}$

with base ${\displaystyle {}a>0}$, is differentiable with

${\displaystyle {}{\left(a^{x}\right)}'={\left(\ln a\right)}a^{x}\,.}$

### Proof

By definition, we have

${\displaystyle {}a^{x}=\exp {\left(x\,\ln a\right)}\,.}$

The derivative with respect to ${\displaystyle {}x}$ equals

${\displaystyle {}{\left(a^{x}\right)}'={\left(\exp {\left(x\,\ln a\right)}\right)}'={\left(\ln a\right)}\exp '(x\,\ln a)={\left(\ln a\right)}\exp {\left(x\,\ln a\right)}={\left(\ln a\right)}a^{x}\,,}$

due to fact and the chain rule.

${\displaystyle \Box }$

## Remark

${\displaystyle {}y(x)=a^{x}\,,}$

the relation

${\displaystyle {}y'={\left(\ln a\right)}y\,}$

holds, due to fact. Hence, there is a proportional relationship between the function ${\displaystyle {}y}$ and its derivative ${\displaystyle {}y'}$, and ${\displaystyle {}\ln a}$ is the factor. This is still true if ${\displaystyle {}a^{x}}$ is multiplied with a constant. If we consider ${\displaystyle {}y}$ as a function depending on time ${\displaystyle {}x}$, then ${\displaystyle {}y'(x)}$ describes the growing behavior at that point of time. The equation ${\displaystyle {}y'={\left(\ln a\right)}y}$ means that the instantaneous growing rate is always proportional with the magnitude of the function. Such an increasing behavior (or decreasing behavior, if ${\displaystyle {}a<1}$) 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

${\displaystyle {}y'=by\,}$

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

${\displaystyle {}y(x)=ce^{bx}\,.}$

## Corollary

The derivative of the natural logarithm

${\displaystyle \ln \colon \mathbb {R} _{+}\longrightarrow \mathbb {R} ,x\longmapsto \ln x,}$
is
${\displaystyle \ln \!'\colon \mathbb {R} _{+}\longrightarrow \mathbb {R} ,x\longmapsto {\frac {1}{x}}.}$

### Proof

As the logarithm is the inverse function of the exponential function, we can apply fact and get

${\displaystyle {}\ln '(x)={\frac {1}{\exp '(\ln x)}}={\frac {1}{\exp(\ln x)}}={\frac {1}{x}}\,,}$

using fact.

${\displaystyle \Box }$

## Corollary

Let ${\displaystyle {}\alpha \in \mathbb {R} }$. Then the function

${\displaystyle f\colon \mathbb {R} _{+}\longrightarrow \mathbb {R} _{+},x\longmapsto x^{\alpha },}$

is differentiable, and its derivative is

${\displaystyle {}f'(x)=\alpha x^{\alpha -1}\,.}$

### Proof

By definition, we have

${\displaystyle {}x^{\alpha }=\exp {\left(\alpha \,\ln x\right)}\,.}$

The derivative with respect to ${\displaystyle {}x}$ equals

${\displaystyle {}{\left(x^{\alpha }\right)}'={\left(\exp {\left(\alpha \,\ln x\right)}\right)}'={\frac {\alpha }{x}}\cdot \exp {\left(\alpha \,\ln x\right)}={\frac {\alpha }{x}}x^{\alpha }=\alpha x^{\alpha -1}\,}$

using fact, fact and the chain rule.

${\displaystyle \Box }$

## Theorem

${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \sin x,}$

is differentiable, with

${\displaystyle {}\sin \!'(x)=\cos x\,,}$

and the cosine function

${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \cos x,}$

is differentiable, with

${\displaystyle {}\cos \!'(x)=-\sin x\,.}$

### Proof

${\displaystyle \Box }$

## Theorem

${\displaystyle \mathbb {R} \setminus {\left({\frac {\pi }{2}}+\mathbb {Z} \pi \right)}\longrightarrow \mathbb {R} ,x\longmapsto \tan x,}$

is differentiable, with

${\displaystyle {}\tan \!'(x)={\frac {1}{\cos ^{2}x}}\,,}$

and the cotangent function

${\displaystyle \mathbb {R} \setminus \mathbb {Z} \pi \longrightarrow \mathbb {R} ,x\longmapsto \cot x,}$

is differentiable, with

${\displaystyle {}\cot \!'(x)=-{\frac {1}{\sin ^{2}x}}\,.}$

### Proof

Using the quotient rule, fact, and the circle equation, we get

{\displaystyle {}{\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.

${\displaystyle \Box }$