Recall: Composition of two functions.

Notation: $(f \circ g)(x)=f(g(x))$

Suppose $g(x)$ in differentiable at $x=a$ and $f(x)$ is differentiable at $x=g(a)$.

Then is $(f \circ g)(x)$ differentiable at $x=a$ ?

If so, $(f \circ g)^{\prime}(a)=$ ?

Let $k(x)=(f \circ g)(x)=f(g(x))$

To check whether $k(x)$ is differentiable at $x=a$, we need to $100 k$ at the limit $\lim _{h \rightarrow 0} \frac{k(a+h)-k(a)}{h}$

i.e. $\lim _{h \rightarrow 0} \frac{f(g(a+h))-f(g(a))}{h}$

If $g(a+h)-g(a) \neq 0$ for all small $h \neq 0$ then

$\begin{gathered}\frac{f(g(a+h))-f(g(a))}{h}=\frac{f(g(a+h))-f(g(a))}{g(a+h)-g(a)} \cdot \frac{g(a+h)-g(h)}{h} \ \end{gathered}$

$\text { if } g(a+h)-g(a) \neq 0$

we are given that $g$ is diff. at $a$, so

$\lim _{h \rightarrow 0} \frac{g(a+h)-g(a)}{h}=g^{\prime}(a)$

Also, $f$ in diff. at $g(a)$, so

$\lim _{h \rightarrow 0} \frac{f(g(a+h))-f(g(a))}{g(a+h)-g(a)}=f^{\prime}(g(a))$

because as $h \rightarrow 0, g(a+h) \rightarrow g(a)(\because g$ is continuous at a)

$\therefore \lim _{b \rightarrow g(a)} \frac{f(y)-f(g(a)}{y-g(a)}=f^{\prime}(g(a))$.

So, if $g(a+h) \neq g(a)$ for small $h$, then

$(f \circ g)^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a) \text {. }$

In fact, the above formula is always true.

(Chain Rule): Let $f$ & $g$ be functions such that $f$ is diff. at $g(a)$ and $g$ is diff. at $a$. Then $f \circ g$ is diff.

at $a$ and the derivative is given by

$(f \circ g)^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)$

We need to show that

$\lim _{h \rightarrow 0} \frac{f(g(a+h))-f(g(a))}{h}=f^{\prime}(g(a)) g^{\prime}(a)$

Define $\phi(y)= \begin{cases}\frac{f(y)-f(g(a))}{y-g(a)} & \text { if } y \neq g(a) \\ f^{\prime}(g(a)) & \text { if } y=g(a)\end{cases}$

$\begin{aligned}\lim _{y \rightarrow g(a)} \phi(y) & =\lim _{y \rightarrow g(a)} \frac{f(y)-f(g(a))}{y-g(a)} \\& =f^{\prime}(g(a))[\because f \text { is ff at } g(a)] \\& =\phi(g(a))\end{aligned}$

$\Rightarrow \phi$ is continuous at $g(a)$.

(if $y \neq g(a)$ then $\phi(y)=\frac{f(y)-f(g(a))}{y-g(a)}$

$\Rightarrow f(y)-f(g(a))= \phi(y)[y-g(a)]$

if $y=g(a)$, the L.H.S $=0=$ R.H.S. )

Put $y=g(a+h)$ :

$f(g(a+h))-f(g(a))=\phi(g(a+h))[g(a+h)-g(a)]$

In R.H.S., $\lim _{h \rightarrow 0} \frac{g(a+h)-g(a)}{h}=g^{\prime}(a)$

and $\quad \lim _{h \rightarrow 0} \phi(g(a+h))=\phi(g(a))[\because \phi$ is cont. at g(a)]

$=f^{\prime}(g(a))$

$\therefore \lim _{h \rightarrow 0} \frac{(f \circ g)(a+h)-(f \circ g)(a)}{h}=f^{\prime}(g(a)) g^{\prime}(a)$.

ie. $(f \circ g)^{\prime}(a)=f^{\prime}(g(a)) \cdot g^{\prime}(a)$

This proves the chain rule.

$\quad g(a+h)-g(a) \neq 0$ might not hold true for small enough $h$.

e.g. $\quad g(x)= x^2 sin(\frac{1}{x}), \quad x \neq 0$, otherwise $g(x)=0$

Claim: $g$ is diff. at $0 ~ \& ~ g^{\prime}(a)=0$

$\lim _{h \rightarrow 0} (\frac{g(0+h)-g(0)}{h}) =\lim _{h \rightarrow 0} \frac{h^2 \sin \left(\frac{1}{h}\right)-0}{h}$

$=\lim _{h \rightarrow 0} h \sin \left(\frac{1}{h}\right)=0$

$\left|\sin \frac{1}{h}\right| \leq 1 \Rightarrow 0 \leq h \sin \left(\frac{1}{h}\right)|\leq| h \mid$

$\text { By Sandwich thin., } \left.\lim _{h \rightarrow \infty} h \sin \left(\frac{h}{h}\right)=0\right)$

But, $g\left(\frac{1}{m \pi}\right)=0$ for all m $\in \mathbb{Z}${0}

because $\sin (m \pi)=0 \quad \forall m \in \mathbb{Z}$

For any $\delta>0, \frac{1}{m \pi} \in(-\delta, \delta)$ for some m $\in \mathbb{Z}$.

Write $y=g(x)$ & $u=f(y)$

ie. $u=f(g(x))=(f \circ g)(x)$

The chain rule $(f \circ g)^{\prime}(x)=f^{\prime}(g(x)) g^{\prime}(x)$

ie. $\frac{d u}{d x}=\frac{d u}{d y} \cdot \frac{d y}{d x} \quad\left(\begin{array}{l}y=g(x) \Rightarrow \frac{d y}{d x}=g^{\prime}(x) \ \left.u=f(y) \Rightarrow \frac{d u}{d y}=f^{\prime}(y)\right)\end{array}\right.$

$u=f(y) ; \quad y=g(x)$

$\frac{d u}{d x}=\frac{d u}{d y} \cdot \frac{d y}{d x}$

Let $f(x)=\left(x^2+1\right)^3$. Find $f^{\prime}(x)$.

$\rightarrow$ Expand $\left(x^2+1\right)^3=x^6+3 x^4+3 x^2+1$

$\therefore f^{\prime}(x)=6 x^5+12 x^3+6 x$

$\rightarrow$ Use product rule: $f(x)=\left(x^2+1\right)\left(x^2+1\right)^2$

$f^{\prime}(x) =2 x\left(x^2+1\right)^2+\left(x^2+1\right) \cdot \frac{d}{d x}\left(x^2+1\right)^2$

$\frac{d}{d x}\left(x^2+1\right)^2 =2 x\left(x^2+1\right)+\left(x^2+1\right) \cdot 2 x$

$=4 x\left(x^2+1\right)$

$\therefore \quad f^{\prime}(x) =2 x\left(x^2+1\right)^2+4 x\left(x^2+1\right)^2$

$=6 x\left(x^2+1\right)^2$

$=6 x\left(x^4+2 x^2+1\right)$

Using chain rule:

$f(x)=\left(x^2+1\right)^3=(g(x))^3 ; g(x)=x^2+1$

By chain rule, $f^{\prime}(x)=3(g(x))^2 \cdot g^{\prime}(x)$

$\begin{aligned}& =3\left(x^2+1\right)^2 \cdot 2 x \& =6 x\left(x^2+1\right)^2\end{aligned}$

or, $\quad y=\left(x^2+1\right)^3=u^3 ; u=x^2+1$

$\frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x} =3 u^2 \cdot 2 x$

$=3\left(x^2+1\right)^2 \cdot 2 x$

$=6 x\left(x^2+1\right)^2$

$\rightarrow$ Expand $\left(x^2+1\right)^3=x^6+3 x^4+3 x^2+1$

$\therefore f^{\prime}(x)=6 x^5+12 x^3+6 x$

$\rightarrow$ Use product rule: $f(x)=\left(x^2+1\right)\left(x^2+1\right)^2$

$f^{\prime}(x) =2 x\left(x^2+1\right)^2+\left(x^2+1\right) \cdot \frac{d}{d x}\left(x^2+1\right)^2$

$\frac{d}{d x}\left(x^2+1\right)^2 =2 x\left(x^2+1\right)+\left(x^2+1\right) \cdot 2 x$

$=4 x\left(x^2+1\right)$

$\therefore f^{\prime}(x) =2 x\left(x^2+1\right)^2+4 x\left(x^2+1\right)^2$

$=6 x\left(x^2+1\right)^2$

$=6 x\left(x^4+2 x^2+1\right)$

Find $\frac{d}{d x}\left[\sin \left(x^2\right)\right]$

$y =\sin u, u=x^2$

$\frac{d y}{d u} =\cos u, \frac{d u}{d x}=2 x$

$\frac{d y}{d x} =\frac{d y}{d u} \cdot \frac{d u}{d x}=\cos \left(x^2\right) \cdot 2 x$

Find $\frac{d}{d x}\left[\sin ^2\left(x^3\right)\right]$

$y=\sin ^2\left(x^3\right)=\left[\sin \left(x^3\right)\right]^2$

$\frac{d y}{d x}=2 \sin \left(x^3\right) \cdot \frac{d}{d x}\left[\sin \left(x^3\right)\right]=2 \sin \left(x^3\right) \cdot \cos \left(x^3\right) \cdot 3 x^2$

$y= \sin \left(\sin \left(\cos \left(x^3+x\right)\right)\right)$

$\frac{d y}{d x}= \cos \left(\sin \left(\cos \left(x^3+x\right)\right)\right) \cdot \frac{d}{d x}\left[\sin \left(\cos \left(x^3+x\right)\right)\right]$

= $\cos \left(\sin \left(\cos \left(x^3+x\right)\right)\right) \cdot \cos \left[\cos \left(x^3+x\right)\right].{(-\sin (x^2+x))} \cdot(3 x^2+1)$

$\frac{d}{d x}\left(\sin ^{-1} x\right), \frac{d}{d x}\left(\cos ^{-1} x\right), \frac{d}{d x}\left(\tan ^{-1} x\right), \cdots$

Let $y=f(x)$

Suppose $f(x)$ has an inverse $g(x)$.

ie. $\quad f(g(x))=g(f(x))=x$.

$y=f(x) \Leftrightarrow x =f^{-1}(y)=g(y) .$

$\therefore \quad g^{\prime}(y) =\frac{d x}{d y}$

$x=f(g(x)) \Rightarrow 1 =\frac{d}{d x}[f(g(x))]=f^{\prime}(g(x)) \cdot g^{\prime}(x)$

So, if $f^{\prime}(g(x)) \neq 0$ then $g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))}$

If $f^{-1}$ denotes the inverse of $f$, then

$\left.\frac{d}{d x}\left[f^{-1}(x)\right]=\frac{1}{f^{\prime}\left(f^{-1}(x)\right.}\right] \cdot f^{\prime}\left(f^{\prime}(x)\right) \neq 0 .$

eg. $\frac{d}{d x}\left[\sin ^{-1} x\right]=\frac{1}{f^{\prime}\left(\sin ^{-1} x\right)}$, where $f(x)=\sin x$

$=\frac{1}{\cos \left(\sin ^{-1} x\right)}$

If $y=\sin ^{-1} x ; x=\sin y$

$\Rightarrow \cos ^2 y=1-\sin ^2 y=1-x^2$

$\Rightarrow \cos y= \pm \sqrt{1-x^2} .$

$\sin ^{-1} x$ is defined for $x \in[-1,1]$

$\sin ^{-1} x \in\left(0, \frac{\pi}{2}\right)$ if $x \in(0,1)$

$\sin ^{-1} x \in\left(-\frac{\pi}{2}, 0\right)$ if $x \in(-1,0)$

But $\cos \theta$ is positive if $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

$\therefore \quad \cos \left(\sin ^{-1} x\right)>0 \quad \forall x \in(-1,1) .$

$\therefore \cos \left(\sin ^{-1} x\right)=\sqrt{1-x^2} \quad \forall x \in(-1,1)$

$\therefore \frac{d}{d x}\left[\sin ^{-1} x\right]=\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{d x}\left[\tan ^{-1} x\right]=?$

$y=\tan ^{-1} x \Leftrightarrow x=\tan y$

$\quad \therefore \frac{d x}{d y}=\sec ^2 y$

$\therefore \frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}=\frac{1}{\sec ^2 y}=\frac{1}{1+\tan ^2 y}=\frac{1}{1+x^2}$

$\therefore \frac{d}{d x}\left[\tan ^{-1} x\right]=\frac{1}{1+x^2}$