Question

Find the derivative of tan$\sqrt{x}$ w.r.t x, using first principle.

Answer 1

First principle method,

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

$\Rightarrow f^{\prime }\left(x\right)={lim}_{t\to x}\frac{\tan \sqrt{t}-\tan \sqrt{x}}{t-x}$

$={lim}_{t\to x}\frac{\frac{\tan \sqrt{t}-\tan \sqrt{x}}{1+\tan \sqrt{t}\tan \sqrt{x}}}{t-x}\times (1+\tan \sqrt{t}\tan \sqrt{x})$

$={lim}_{t\to x}\left(\frac{\tan (\sqrt{t}-\sqrt{x})}{t-x}\right)\times {lim}_{t\to x}(1+\tan \sqrt{t}\tan \sqrt{x})$

$=(1+\tan \sqrt{t}\tan \sqrt{x}){lim}_{t\to x}\left(\frac{\tan (\sqrt{t}-\sqrt{x})}{\sqrt{t}-\sqrt{x}}\right)\frac{1}{\sqrt{t}+\sqrt{x}}$

$=(1+{\tan }^2\sqrt{x}){lim}_{(\sqrt{t}-\sqrt{x})\to 0}\left(\frac{\tan (\sqrt{t}-\sqrt{x})}{\sqrt{t}-\sqrt{x}}\right)\frac{1}{\sqrt{t}+\sqrt{x}}$

$=\left({\sec }^2\sqrt{x}\right)\left(1\right)\left(\frac{1}{\sqrt{x}+\sqrt{x}}\right)$

$f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}{\sec }^2\sqrt{x}$