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

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

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

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

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

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

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

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

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

$1=\frac{d}{d x}(\sec y) =\frac{d}{d y}(\sec y) \cdot \frac{d y}{d x} \text { (by Chain rule) }$

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

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

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

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

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

Since $|x|={\begin{cases}{}x \text { if } x \geqslant 0 \\ -x \text { if } x<0\end{cases}}$,

$\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{|x| \sqrt{x^2-1}} \quad \text { if }|x| \geqslant 1$

Hence, $\frac{d}{d x}\left(\operatorname{cosec}^{-1} x\right)=\frac{-1}{|x| \sqrt{x^2-1}}$ if $|x| \geqslant 1$

Sometimes we have an equation relating

$y$ to $x$, but it is difficult to write explicitly $y$ as a f(x). of $x$.

For example: $y+\sin y=x$

Find $\frac{d y}{d x}$.

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

$y+\sin y=x$

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

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

$\Rightarrow \frac{d y}{d x}+\frac{d}{d y}(\sin y) \frac{d y}{d x}=1 \quad \text { (By Chain rule) }$

$\Rightarrow \quad \frac{d y}{d x}[1+\cos y]=1$

$\Rightarrow \frac{d y}{d x}=\frac{1}{1+\cos y} \text {, provided cosy } \neq-1 .$

Note that by doing implicit differentiation, we need not get $\frac{d y}{d x}$ as a function of $x$ only.

We define

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

$=\sum_{n=0}^{\infty} \frac{x^n}{n !}$

= $\lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{x^k}{k !}$

Fact: The above series converges (to a finite real number) for every $x \in \mathbb{R}$.

Thus $\exp (x)$ is defined for every real no $x$.

(7) $\exp (x+y)=\exp (x) \exp (y)$

(Compare with $a^{m+n}=a^m \cdot a^n$ if $m, n \in \mathbb{N}$ ) Using (7) and (1), we get

$\exp (-x) =\frac{1}{\exp (x)}$

$(\because 1=\exp (0) =\exp (x+(-x))=\exp (x) \exp (-x))$

So, (6) follows from (5).

Putting $x=1$, we get

$\begin{aligned}\exp (1) & =1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{n !}+\cdots \& =\sum_{k=0}^{\infty} \frac{1}{k !}\end{aligned}$

We know that $\exp (1)$ in 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, $e=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots$

$<1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots \cdot \quad \quad$ (because $\frac{1}{(n+1) !}<\frac{1}{2^n}$ for $n \geqslant 2$ )

Recall: Geometric series $\quad\quad a+a r+a r^2+\cdots=\frac{a}{1-r} \text { if }|\gamma|<1$

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

$1+\frac{1}{2}+\frac{1}{2^2}+\cdots=\frac{1}{1-x_2}=2$

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

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

$\begin{aligned}& \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n=e . \& \lim _{x \rightarrow 0}(1+x)^{1 / x}=e .\end{aligned}$

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

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

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

$f(x)=e^x$

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

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

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

So, $\frac{d}{d x}\left(e^x\right)=e^x$