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12.16 Derivative of the inverse

Derivative of the inverse

Theorem

Let \(f:[a,b]\rightarrow[c,d]\) be bijective and continuous, let \(y_0\in(c,d)\) and \(x_0=f^{-1}(y_0)\) where \(x_0\in(a,b),f(x_0)=y_0\)

If \(f\) is differentiable at \(x_0\) and \(f^{-1}(x_0)\neq 0\), then \(f^{-1}\) is differentiable at \(y_0\) **** and \(\left(f^{-1}\right)^{\prime}\left(y_0\right)=\frac{1}{f^{\prime}\left(x_0\right)}=\frac{1}{f^{\prime}\left(f^{-1}\left(y_0\right)\right)}\)

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Proof

NTP \(\lim_{y\to y_0}\frac{f^{-1}\left(y\right)-f^{-1}\left(y_0\right)}{y-y_0}=\frac{1}{f^{\prime}\left(x_0\right)}\;\iff\;\lim_{h\to0}\frac{f^{-1}\left(y_0+h\right)-f^{-1}\left(y_0\right)}{h}=\frac{1}{f^{\prime}\left(x_0\right)}\)

Let \(k(h)=f^{-1}\left(y_0+h\right)-f^{-1}\left(y_0\right)=f^{-1}\left(y_0+h\right)-x_0\)

We know that \(k(h)\) is continuous because \(f^{-1}\) is continuous and \(x_0+h\) is continuous

In particular \(\lim_{h\to0}k\left(h\right)=k\left(0\right)=0\)​

Note that \(x_0+k(h)=f^{-1}\left(y_0+h\right)\Rightarrow f\left(x_0+k\left(h\right)\right)=f\left(f^{-1}\left(y_0+h\right)\right)=y_0+h\Rightarrow h=f\left(x_0+k\left(h\right)\right)-y_0=f\left(x_0+k\left(h\right)\right)-f\left(x_0\right)\)​

Then back to original part, we have \(\frac{f^{-1}\left(y_0+h\right)-x_0}{h}=\frac{k\left(h\right)}{f\left(x_0+k\left(h\right)\right)-f\left(x_0\right)}=\cfrac{1}{\frac{f\left(x_0+k\left(h\right)\right)-f\left(x_0\right)}{k\left(h\right)}}\)

Since \(f\) is injective and \(f^{-1}(y_0)=x_0\), then \(k(h)=0\) only when \(h=0\).

Also \(f(x_0+k(h))-f(x_0)=0\) only if \(k(h)=0\) only if \(h=0\)

Thus \(\lim_{h\to0}\frac{f^{-1}\left(y_0+h\right)-f^{-1}\left(y_0\right)}{h}=\lim_{h\to0}\cfrac{1}{\frac{f\left(x_0+k\left(h\right)\right)-f\left(x_0\right)}{k\left(h\right)}}\)

Then combine \(\lim_{h\to0}\frac{f\left(x_0+h\right)-f\left(x_0\right)}{h}=f^{\prime}\left(x_0\right)\) with \(\lim_{h\to0}k\left(h\right)=0\), we have \(\lim_{h\to0}\frac{f\left(x_0+k\left(h\right)\right)-f\left(x_0\right)}{k\left(h\right)}=f^{\prime}\left(x_0\right)\)​

Therefore \(\lim_{h\to0}\frac{f^{-1}\left(y_0+h\right)-f^{-1}\left(y_0\right)}{h}=\frac{1}{f^{\prime}\left(x_0\right)}\)

Remark

Note that if you already know that \(f^{-1}\) is differentiable at \(y_0\), then ..... then ...... then we can apply chain rule if we prove \(f\) is differentiable at \(x_0\) and \(f^{-1}\) is differentiable at \(f(x_0)\), then since \(f(f^{-1}(y))=y\), we get \((f^{\prime}(f^{-1}(y))\cdot(f^{-1})^{\prime}(y)=1\), then \(\left(f^{-1}\right)^{\prime}\left(y_0\right)=\frac{1}{f^{\prime}\left(f^{-1}\left(y_0\right)\right)}\)

Example

\(f(x)=\arcsin\left(x\right)\), \(f'(x)=?\)

Since \(\sin(\arcsin(x))=x\), then \(\cos(\arcsin(x))\cdot\arcsin^{\prime}x=1\), then \(\arcsin^{\prime}\left(x\right)=\frac{1}{\cos\left(\arcsin\left(x\right)\right)}\)

Since \(\cos(x)=\sqrt{1-\sin^2(x)}\), then \(\cos(\arcsin(x))=\sqrt{1-\sin^2(\arcsin(x))}=\sqrt{1-x^2}\)

Thus \(\arcsin^{\prime}\left(x\right)=\frac{1}{\cos\left(\arcsin\left(x\right)\right)}=\frac{1}{\sqrt{1-x^2}}\)

Fundamental theorems of the derivative

Fundamental theorems of continuous functions

  1. Intermediate Value Theorem

  2. MAX Min Theorem

Main tool: Lemma: If f: X \to \mathbb{R} is continuous in x_0 and f(x_0) > 0, f(x_0) \neq 0, f(x_0) <...

How about for differentiable functions?

Main tool is a lemma

Pre-Lemma

If \(\lim_{x \to x_0} \phi(x) = \ell\) and \(\ell > 0\), then \(\exists \delta_0 > 0\) such that \(\phi(x) > 0 \, \forall x \in (x_0 - \delta_0, x_0 + \delta_0)\).

Also, \(\lim_{n \to \infty} a_n = \ell \text{ and } \ell \geq 0 \Rightarrow \exists n_0 \text{ such that } a_n > 0 \text{ for } n > n_0.\)

Proof:

Take \(\epsilon = \frac{\ell}{2} = \frac{|\ell|}{2}\).

Lemma

If \(f: (a, b) \to \mathbb{R}\) is differentiable at \(x_0 \in (a, b)\) and \(f'(x_0) > 0\), then \(\exists \delta_0 > 0\) such that if \(x_0 - \delta_0 < x_1 < x_0 < x_2 < x_0 + \delta_0\) \(\Rightarrow\,f(x_1)<f(x_0)<f(x_2).\)

Proof

We have \(\lim_{x\to x_0}\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}=f^{\prime}\left(x_0\right)>0\), apply the pre-lemma, \(\exists\delta_0:\forall x\in\left(x_0-\delta,x_0+\delta\right),\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}>0\)

Then if \(x_1\) is in \((x_0-\delta,x_0)\Rightarrow x_1-x_0<0\Rightarrow f\left(x_1\right)-f\left(x_0\right)<0\Rightarrow f\left(x_1\right)<f\left(x_0\right)\)

If \(x_2\) is in \((x_0-\delta,x_0)\Rightarrow x_2-x_0>0\Rightarrow f\left(x_2\right)-f\left(x_0\right)>0\Rightarrow f\left(x_2\right)>f\left(x_0\right)\)

Definition

Let \(f:X\to \R\) a function and \(x_0\in X\). We say that

  • \(x_0\) is a global or absolute maximal point of \(f\) if \(f(x_0)\geq f(x),\forall x\in X\)​

\(x_0\) is a global or absolute minimal point of \(f\) if \(f(x_0)\leq f(x),\forall x\in X\) * \(X\subseteq \R\), \(x_0\) is a local or relative maximal point of \(f\) if \(\delta_0>0:f\left(x_0\right)\geq f\left(x\right).\forall x\in X\cap\left(x_0-\delta,x_0+\delta\right)\)​

\(X\subseteq \R\), \(x_0\) is a local or relative minimal point of \(f\) if \(\delta_0>0:f\left(x_0\right)\leq f\left(x\right).\forall x\in X\cap\left(x_0-\delta,x_0+\delta\right)\) * If \(X=[a,b]\) and \(x_0\in(a,b)\), we say that \(x_0\) is a singular point of if \(f'(x_0)=0\) (we are assuming \(f\) is differentiable at \(x_0\)​)

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Theorem (Fermat)

If \(f:(a,b)\rightarrow\R\) is differentiable then every local maximal or minimal is a singular point.

That is: if \(f(x_0)\geq f(x),\forall x\in(x_0-\delta,x_0+\delta)\Rightarrow f^{\prime}\left(x_0\right)=0\)​

Proof

Three possibilities \(\begin{cases}f^{\prime}\left(x_0\right)>0\Rightarrow\left(\text{use lemma}\right)\exists x_0-\delta_0<x_1<x_0<x_2<x_0+\delta_0\text{ s.t. }\,f(x_1)<f(x_0)<f(x_2)\\ f^{\prime}\left(x_0\right)<0\Rightarrow\left(\text{use lemma}\right)\exists x_0-\delta_0<x_1<x_0<x_2<x_0+\delta_0\text{ s.t. }\,f(x_1)>f(x_0)>f(x_2)\\ f^{\prime}\left(x_0\right)=0\end{cases}\)

Since the first two are contradiction to \(f(x_0)\geq f(x),\forall x\in(x_0-\delta,x_0+\delta)\)

Thus the only possibility is \(f'(x_0)=0\)

Theorem (Rolle)

Let \(f:[a,b]\rightarrow \R\) be continuous and differentiable on \((a,b)\). If \(f(a)=f(b)\), then \(\exists c\in(a,b)\) such that \(f'(c)=0\)

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Proof

Since \(f:[a,b]\rightarrow \R\) is continuous, \(\exists x_{\max},x_{\min}\in\left\lbrack a,b\right\rbrack\) such that \(x_{\max}\) is a global maximal, then \(x_{\max}\) is local maximal and \(x_{\min}\) is a global minimal, then \(x_{\min}\) is local minimal

By Fermat theorem, \(x_{\max}\) and \(x_{\min}\) are singular if they are in \((a,b)\) that is \(f^{\prime}(x_{\max})=f^{\prime}(x_{\min})=0\)

Take \(c=x_{\max}\text{ or }x_{\min}\) the one that is in \((a,b)\).

But it could happen that both \(x_{\text{max}},x_{\text{min}}\in\left\lbrace a,b\right\rbrace\).

But, in this case, since \(f(a)=f(b)\) \(\Rightarrow f(x_{\text{max}})=f(x_{\min})=f(a)=f(b)\\\Rightarrow f\left(x)=f\left(a\right)=f\left(b\right)\;\forall x\in[a,b]\Rightarrow f^{\prime}(x\right)=0\;\forall x\in[a,b]\)

So we can take \(c\) any point in \([a,b]\)

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