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

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

holds.

Proof

The proof requires a detailed study of power series.

$\Box$

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

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

Proof

This follows immediately from fact.

$\Box$

Theorem

The exponential function

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

is

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

$\ln \colon \mathbb {R} _{+}\longrightarrow \mathbb {R} ,x\longmapsto \ln x,$

is

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

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

See exercise.

$\Box$

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.

$\Box$