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)<f(c) \ \ \ \ \ (30)

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)<f(c)

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<c \Rightarrow \frac{f(x)-f(c)}{x-c}\geq 0

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

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