We saw that (i) $\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}}$

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

We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$

$\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$

$\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}$

$\therefore \quad \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x$

$\Rightarrow \quad \frac{d}{d x}\left(\cos ^{-1} x\right)=0-\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{-1}{\sqrt{1-x^2}}$

$\text { Il y, } \quad \frac{d}{d x}\left(\cot ^{-1} x\right)=-\frac{d}{d x}\left(\tan ^{-1} x\right)=-\frac{1}{1+x^2}$

Let's calculate $\frac{d}{d x}\left(\sec ^{-1} x\right)$.

Let $y=\sec ^{-1} x$ Then $x=\sec y$

Diff. w.r.t. $x$, we get

$1=\frac{d}{d x}(\sec y) =\frac{d}{d y}(\sec y) \cdot \frac{d y}{d x}$

(by chain rule)

$=\sec y \tan y \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x} =\frac{1}{\sec y \tan y}$

Since, $\sec y=x, \quad \tan ^2 y=\sec ^2 y-1=x^2-1$

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

Recall : $\sec ^{-1} x \in\left(0, \frac{\pi}{2}\right)$ if $x \in(1, \infty)$

$2 \sec ^{-1} x \in\left(\frac{\pi}{2}, \pi\right) \text { if } x \in(-\infty,-1)$

( $\sec ^{-1} x$ is not defined if $-1<x<1$ )

$\therefore y=\sec ^{-1} x \in\left(0, \frac{\pi}{2}\right) \text { if } x \in(1, \infty)$

$\in\left(\frac{\pi}{2}, \pi\right] \text { if } x \in(-\infty,-1)$

$\Rightarrow \quad \tan y \geqslant 0 \quad \text { if } x \in(1, \infty)$

$~ \qquad\leqslant 0 \quad \text { if } x \in(-\infty,-1)$

$\tan y= \begin{cases}\sqrt{x^2-1} \text { if } x \geqslant 1 \\ \sqrt{x^2-1} \text { if } x \leqslant-1\end{cases}$

$\tan \left(\sec ^{-1} x\right)$

$\therefore \quad \frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{\sec y \tan y}= \begin{cases}\frac{1}{x \sqrt{x^2-1}} & \text { if } x \geqslant 1, \ \qquad \frac{1}{-x \sqrt{x^2-1}} & \text { if } x \leqslant 1\end{cases}$

$\text { Since }|x|= x \text { if } x \geqslant 0 \quad or \quad -x\text { if } x<0$

Sometimes we have am equation relating $y$ to $x$ , but it is difficult to write explicitly $y$ as a fr. of $x$'

For Example: $~ ~ y + \sin = x$

Find $=\frac{dy}{dx}$

Here instead of trying to write $y$ as a function of $x$, we will do implicit differentiation as follows.

y + sin y $=x$

$\Rightarrow \frac{d}{dx} =(y + \sin y) = \frac{d}{dx}(x)$$\quad$

$\Rightarrow\frac{dy}{dx} + \frac{d}{dx}(\sin y)=1$

$\Rightarrow\frac{dy}{dx} + \frac{d}{dy} \sin y \frac{dy}{dx}=1$$\quad$

$\Rightarrow\frac{dy}{dx}[1 + \cos y] = 1$

$\Rightarrow\frac{dy}{dx} = \frac{1}{1 + \cos y}$ , provided $\cos y \neq-1$

Note that by doing have, we need not get $\frac{dy}{dx}$ as a function of $x$ only.

$y=exp(x)$

Putting $x=1$, we get

$\exp (1) =1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{n !}+\cdots$

$=\sum_{k=0}^{\infty} \frac{1}{k !}$

We know that $\exp (1)$ is a real, and we denote this by $e$ (Euler's constant).

So, $e=\exp (1)=\sum_{k=0}^{\infty} \frac{1}{k !}$

We can show that $2<e<3$ (In fact, $e \approx 2.718 \ldots$

$e=1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots>1+\frac{1}{1 !}=2$ .

Also, $e=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots$

$<1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots$

(because $\frac{1}{(n+1) !}<\frac{1}{2^n}$ for $n \geqslant 2$ )

Recall: Geometric series

$a+a r+a r^2+\cdots=\frac{a}{1-r} \text { if }|r|<1$

Putting $a=1, r = \frac{1}{2}$ gives

$1+\frac{1}{2}+\frac{1}{2^2}+\cdots=\frac{1}{1-1 / 2}=2$

$\therefore \quad e<1+2=3$

Fact: It can be proved that $e$ is an irrational number.

Also,

$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n=e .$

$\lim_{x \rightarrow 0}(1+x)^{1 / x}=e .$

Let’s calculate $\lim _{h \rightarrow 0} \frac{\exp (h)-1}{h}$

Notation: We write $\exp (x)$ as $e^x$.

$e^h=\exp (h)=1+\frac{h}{1 !}+\frac{h^2}{2 !}+\cdots+\frac{h^h}{n !}+\cdots$

$\Rightarrow \quad e^h-1=h+\frac{h^2}{2 !}+\frac{h^3}{3 !}+\cdots+\frac{h^h}{n !}+\cdots$

$\Rightarrow \quad \frac{e^h-1}{h}=1+\frac{h}{2 !}+\frac{h^2}{3 !}+\cdots+\frac{h^{h-1}}{n !}+\cdots \text { for } h \neq 0 .$

$\lim _{h \rightarrow 0} \frac{e^h-1}{h}=1$

Derivative of $e^x=\exp (x)$

$f(x) =e^x$

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

$=\lim _{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}=\lim _{h \rightarrow 0} \frac{e^x e^h-e^x}{h}$

$=e^x \lim _{h \rightarrow 0}\left(\frac{e^h-1}{h}\right)=e^x .$

(1) $\frac{d}{d x}\left(e^{5 x}\right)=\frac{d}{d(5 x)}\left(e^{5 x}\right) \cdot \frac{d}{d x}(5 x)=e^{5 x} \cdot 5$

(2) $\quad \frac{d}{d x}\left(e^{x^2}\right)=e^{x^2} \cdot \frac{d}{d x}\left(x^2\right)=2 x e^{x^2}$.

(3) $\quad \frac{d}{d x} \tan ^{-1}\left(e^x\right)=\frac{1}{1+\left(e^x\right)^2} \cdot \frac{d}{d x}\left(e^{x}\right)=\frac{e^x}{1+e^{2 x}}$