Question

Differentiate $se{c}^{-1}x$ w.r.to x by first principle.

Answer 1

DIfferentiate $se{c}^{-1}x$ wrt x by first principle

From the First Principle we have

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

Here , we have

$f(x+h)=se{c}^{-1}(x+h)$

$f\left(x\right)=se{c}^{-1}\left(x\right)$

$\therefore$ Using the First Principle we have

$f{}^{\prime }\left(x\right)={lim}_{h\to 0}\frac{se{c}^{-1}(x+h)-se{c}^{-1}x}{h}$

Let $sec\theta =x$

$\theta =se{c}^{-1}\left(x\right)$

$\theta =ta{n}^{-1}\sqrt{x^2-1}$

$se{c}^{-1}\left(x\right)=ta{n}^{-1}\sqrt{x^2-1}$

$se{c}^{-1}\left(x\right)=se{c}^{-1}\left(x\right)$ , When $x\ge 1$

$se{c}^{-1}\left(x\right)=\Pi -ta{n}^{-1}\sqrt{x^2-1},x\le 1$

$\therefore$ $se{c}^2\theta -ta{n}^2\theta =1$

$se{c}^2\theta =1+ta{n}^2\theta$

$sec\theta =\sqrt{1+ta{n}^2\theta }$

Therefore, On Putting this value

$f{}^{\prime }\left(x\right)={lim}_{h\to 0}\frac{se{c}^{-1}(x+h)-se{c}^{-1}x}{h}$

$f{}^{\prime }\left(x\right)={lim}_{h\to 0}\frac{ta{n}^{-1}\sqrt{1-(x+h{)}^2}-ta{n}^{-1}\sqrt{1-x^2}}{h}$

Using $ta{n}^{-1}a-ta{n}^{-1}b=ta{n}^{-1}\left[\frac{a-b}{1+ab}\right]$

$={lim}_{h\to 0}ta{n}^{-1}\left[\frac{\sqrt{(x+h{)}^2-1}-\sqrt{x^2-1}}{1+\left(\sqrt{(x+h{)}^2-1}\right)\left(\sqrt{x^2-1}\right)}\right]$