## Real Analysis – Differential Calculus II

 Theorem 60 (Differentiability of the composite function) Let ${D, E \in C}$, ${g:D\rightarrow E}$, ${f:E\rightarrow\mathbb{R}}$ and ${c\in D\cap D'}$. If ${g}$ is differentiable in ${c}$ and ${f}$ is differentiable in ${g(c)}$, then ${f\circ g}$ in ${c}$ and it is $\displaystyle (f\circ g)'(c)=f'(g(c))g'(c) \quad\mathrm{if}\quad \varphi=f(t) \quad\mathrm{with}\quad t=g(x) \ \ \ \ \ (24)$ $\displaystyle (f\circ g)'(x)=f'(g(x))g'(x) \quad\mathrm{if}\quad \varphi=f(g(x)) \ \ \ \ \ (25)$ Using Leibniz’s notation we can also write the previous theorem as $\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx} \ \ \ \ \ (26)$ A notation that formally suggests that we can cancel out the ${dt}$. Proof: Let ${a=g(c)}$. Since ${f}$ is differentiable in ${a}$ by Theorem 57 it is $\displaystyle f(t)=f(a)+(f'(a)+\varphi (t)(t-a)\quad \forall t \in E$ with ${\varphi}$ continuous in ${a}$. Taking ${g(x)=t}$ and ${g(c)=a}$ it is $\displaystyle f(g(x))=f(g(c))+f'(g(c))+\varphi(g(x))(g(x)-g(c))\quad\forall x \in D$ Hence $\displaystyle \frac{f(g(x))-f(g(c))}{x-c}=(f'(g(c))+\varphi(g(x)))\frac{g(x)-g(c)}{x-c } \ \ \ \ \ (27)$ Since ${g}$ is differentiable in ${c}$ it also is continuous in ${c}$ by Corollary 59. Then ${\varphi (g(x))}$ also is continuous in ${c}$ (by Theorem 43). Hence $\displaystyle \lim_{x\rightarrow c}\varphi(g(x))=\varphi (g(c))=\varphi(a)=0$ Taking the limit ${x\rightarrow c}$ in 27 it is $\displaystyle \lim_{x\rightarrow c}\frac{f(g(x))-f(g(c))}{x-c}=f'(g(c))$ Which is to say $\displaystyle (f \circ g)'(c)=f'(g(c)g'(c)$ $\Box$

As an application of Theorem 60 let us look into some simple examples.

1. ${\left( e^{g(x)} \right)'=?}$

Now ${e^{g(x)}=f(g(x))}$ and let ${t=g(x)}$. Hence

{\begin{aligned} \left( e^{g(x)} \right)' &= \left(e^t\right)'g'(x)\\ &= e^t g'(x)\\ &= e^{g(x)}g'(x) \end{aligned}} Hence

$\displaystyle \left( e^{g(x)} \right)'=g'(x) e^{g(x)})$

2. Let ${\alpha\in\mathbb{R}}$ and ${x>0}$ and calculate ${\left( x^\alpha \right)'}$.

{\begin{aligned} \left( x^\alpha\right)'&=\left( e^{\alpha\log x}\right)'\\ &=(\alpha\log x)'e^{\alpha\log x}\\ &=\dfrac{\alpha}{x}e^{\alpha\log x}\\ &=\dfrac{\alpha}{x}x^\alpha\\ &=\alpha x^{\alpha -1} \end{aligned}}

which generalizes the know rule for integer exponents.

Hence

$\displaystyle \left( x^\alpha\right)'= \alpha x^{\alpha -1}\quad \forall\alpha\in\mathbb{R},\forall x>0$

3. ${(\log g(x))'=?}$

Like in the first example the construction of interest is ${\log g(x)=f(g(x))}$ where ${f(t)=\log t}$ and ${t=g(x)}$.

Hence

{\begin{aligned} (\log g(x))'&=(\log t)'g'(x)\\ &= \dfrac{1}{t}g'(x)\\ &=\dfrac{g'(x)}{g(x)} \end{aligned}}

Hence for ${g(x)>0}$

$\displaystyle (\log g(x))'=\frac{g'(x)}{g(x)}$

In particular one can calculate ${(\log |x|)'}$

$\displaystyle (\log |x|)'=\frac{|x|'}{|x|}=\begin{cases} \dfrac{1}{|x|}\quad x>0\\-\dfrac{1}{|x|}\quad x<0 \end{cases}$

Since ${-|x|=x}$ for ${x<0}$ it always is

$\displaystyle (\log |x|)'=\frac{1}{x}\quad\forall x\neq 0$

 Theorem 61 (Differentiability of the inverse function) Let ${D\subset\mathbb{R}}$, ${f:D\rightarrow\mathbb{R}}$ an injective function and ${c\in D\cap D'}$. If ${f}$ is differentiable in ${c}$ ${f'(c)\neq 0}$ ${f^{-1}}$ is continuous then ${f^{-1}}$ is differentiable and it is $\displaystyle \left( f^{-1} \right)'(f(c))=\frac{1}{f(c)} \ \ \ \ \ (28)$ In Leibniz’s notation one introduces ${y=f(x)}$, then ${x=f^{-1}(y)}$ and the differentiability of the inverse function equation is $\displaystyle \frac{dx}{dx}=\frac{1}{\frac{dy}{dx}} \ \ \ \ \ (29)$ Proof: Omitted. $\Box$

Just like in Theorem 60 we will state an application of the previous theorem.

Let ${y=\sin x}$ and ${x\in [-\pi /2,\pi /2]}$, then ${x=\arcsin y}$.

Now

• ${f(x)}$ is differentiable in all points of the interval
• ${f'(x)=\cos x\neq 0}$ for all points contained in the interval ${[-\pi /2,\pi /2]}$
• ${\arcsin y}$ is continuous ${[-1,1]}$

Then

{\begin{aligned} (\arcsin y)' &= \left( f^{-1}(y) \right)'\\ &=\dfrac{1}{f'(x)}\\ &=\dfrac{1}{\cos x}\\ &=\dfrac{1}{\sqrt{1-\sin^2x}}\\ &=\dfrac{1}{\sqrt{1-y^2}} \end{aligned}}

Finally

$\displaystyle (\arcsin y)'=\frac{1}{\sqrt{1-y^2}} \quad y \in [-1,1]$

Or, writing in a notation that is more usual

$\displaystyle (\arcsin x)'=\frac{1}{\sqrt{1-x^2}} \quad x \in [-1,1]$

In general one can define superior derivatives by using recursion.

Let us denote the nth derivative of ${f}$ by ${f^{(n)}}$ (since for the 50th derivative it isn’t very practical to use ${'}$ fifty times). One first define ${f^{(0)}=f}$. Now for ${f^{(n+1)}}$ it is

$\displaystyle f^{(n+1)}=\left( f^{(n)} \right)'$

That is to say that

1. ${f'=\dfrac{df}{dx}}$
2. ${f''=\dfrac{d}{dx}\dfrac{df}{dx}=\left( \dfrac{d}{dx} \right)^2 f=\dfrac{d^2}{dx^2}f}$
3. ${f'''=\dfrac{d}{dx}\dfrac{d^2}{dx^2}f=\dfrac{d^3}{dx^3}f}$
4. ${f^{(n)}=\left( \dfrac{d}{dx} \right)^n f=\dfrac{d^n f}{dx^n}}$

Given the previous discussion it makes sense to introduce the following definition

 Definition 39 A function ${f}$ is said to be ${n}$ times differentiable in ${c}$ if all ${f^{(n)}}$ exist and are finite.

We already know that a differentiable function is continuous (via Corollary 59 in Real Analysis – Differential Calculus I) , but is it that the derivative of a differentiable function also is a continuous function?

As an (counter-)example let us look into the following function:

$\displaystyle f(x)=\begin{cases} x^2\sin (1/x) \quad &x\neq 0\\ 0 & x=0 \end{cases}$

It is easy to see that ${f}$ is differentiable in ${\mathbb{R}}$

$\displaystyle f'(x)=\begin{cases} 2x\sin (1/x)-\cos (1/x) & x\neq 0\\ 0 & x=0 \end{cases}$

but ${f'}$ isn’t continuous for ${x=0}$. The reader is invited to calculate ${\displaystyle\lim_{x\rightarrow 0^+}f'(x)}$ and ${\displaystyle\lim_{x\rightarrow 0^-}f'(x)}$.

Apparently the derivative of a function either is continuous or it is strongly discontinuous. That being said it is obvious that it makes sense to introduce differentiability classes, which classifies a function according to its derivatives properties.

 Definition 40 A function ${f}$ is said to be of class ${C^n}$ if it is ${n}$ times differentiable and ${f^{(n)}}$ is continuous.

It is easy to see that a function that is of class ${C^{n+1}}$ also is of classs ${C^n}$.

A function is said to be of class ${c^\infty}$ if it has finite derivatives in all orders (which are necessarily continuous).

If ${f,g}$ are ${n}$ times differentible in ${c}$ then ${f+g}$, ${fg}$, ${f/g\quad g(c)\neq 0}$ are also ${n}$ times differentiable in ${c}$.

 Definition 41 Let ${D\subset\mathbb{R}}$, ${f:D\rightarrow\mathbb{R}}$ and ${c\in D}$. ${c}$ is said to be a relative maximum of ${f}$ if $\displaystyle \exists r>0:x\in V (c,r)\cap D \Rightarrow f(x)

 Definition 42 Let ${D\subset\mathbb{R}}$, ${f:D\rightarrow\mathbb{R}}$ and ${c\in D}$. ${c}$ is said to be a relative minimum of ${f}$ if $\displaystyle \exists r>0:x\in V (c,r)\cap D \Rightarrow f(x)>f(c) \ \ \ \ \ (31)$

 Theorem 62 (Interior Extremum Theorem) Let ${I\in\mathbb{R}}$ and ${c}$ is an interior point of ${I}$. If ${f}$ has a relative extremum in ${c}$ and ${f'(c)}$ exists then ${f'(c)=0}$ Proof: Let us suppose without loss of generality that ${f}$ has a relative maximum in ${c}$. Since ${c}$ is an interior point of ${I}$ and ${f'(c)}$ exists, ${f_+(c)}$ and ${f_-(c)}$ exist and are equal. It is ${f_+'(c)=\displaystyle\lim_{x\rightarrow c^+}\dfrac{f(x)-f(c)}{x-c}}$ From our hypothesis $\displaystyle \exists r>0:x\in V (c,r)\cap D \Rightarrow f(x) Hence $\displaystyle x\in V(c,r)\cap I\quad\mathrm{and}\quad x>c \Rightarrow \frac{f(x)-f(c)}{x-c}\leq 0$ Then by corollary 31 (Real Analysis – Limits and Continuity II) it is $\displaystyle f_+'(c)=\lim_{x\rightarrow c^+}\dfrac{f(x)-f(c)}{x-c}\leq 0$ Likewise $\displaystyle x\in V(c,r)\cap I\quad\mathrm{and}\quad x Hence $\displaystyle f_-'(c)=\lim_{x\rightarrow c^-}\dfrac{f(x)-f(c)}{x-c}\geq 0$ Since ${f_+'(c)=f_-'(c)=f'(c)}$ it has to be ${f_+'(c)=f_-'(c)=0}$ and consequently ${f'(c)=0}$. $\Box$

One can visualize the previous theorem in the following way. Imagine that you have a relative maximum in a given interval for a continuous function. For some vicinity of that point we must have function values that are inferior to ${c}$. Since we are assuming that ${c}$ is a maximum values to its left are increasing as we approach ${c}$ and values to its right are decreasing as we mover further away from ${c}$.

Hence for its left side the derivative of ${f}$ has positive values, while to its left the derivative of ${f}$ has negative values, since we also assume that the derivative exists in ${c}$ we can reason by continuity that its value is ${0}$.

 Theorem 63 (Rolle’s Theorem) Let ${[a,b]\subset\mathbb{R}}$ and ${f}$ continuous such as ${f:[a,b]\rightarrow \mathbb{R}}$. If ${f}$ is differentiable ${]a,b[}$ and ${f(a)=f(b)}$ there exists a point ${c\in ]a,b[}$ such as ${f'(c)=0}$. Proof: Since ${f}$ is continuous in the compact interval ${[a,b]}$ it has a maximum and a minimum in ${[a,b]}$ (see Extreme Value Theorem which is theorem 55 in Real Analysis – Limits and Continuity VII). If for ${c\in ]a,b[}$ ${f(c)}$ is either a maximum or a minimum then by Theorem 62 ${f'(c)=0}$. Let ${m}$ denote the minimum and ${M}$ denote the maximum. and let us analyze the case were the extrema values occur at the extremities of the interval. Since by hypothesis ${f(a)=f(b)}$ then ${m=M}$. In this case ${f}$ is constant and it trivially is ${f'(c)=0\quad\forall c\in [a,b]}$ $\Box$

 Corollary 64 Let ${I\in\mathbb{R}}$, ${f}$ continuous such as ${f:I\rightarrow\mathbb{R}}$. If ${f}$ is differentiable in the interior of ${I}$ and ${f'}$ doesn’t vanish in the interior of ${I}$, then ${f}$ doesn’t have more than one ${0}$ in ${I}$. Proof: Let us use the method of reductio ad absurdum that ${f}$ vanishes for two points of ${I}$ (${a}$ and ${b}$). applying Theorem 63 in ${[a,b]}$ (since ${f(a)=f(b)}$) there exists ${c}$ in ${]a,b[}$ such as ${f'(c)=0}$. Hence ${f'}$ vanishes in the interior of ${I}$ which contradicts our hypothesis. $\Box$