Question

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

Answer 1

$\tan \sqrt{x}$

Using first principle,

$f\left(x\right)=\tan \sqrt{x}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\left[f\right(x+h)-f(x\left)\right]/h$

$=\underset{h\to 0}{lim}[\tan \sqrt{(x+h)}-\tan \sqrt{x}]/h$

$=\underset{h\to 0}{lim}\left[\right\{\sin \sqrt{(x+h)}/\cos \sqrt{(x+h)}\}-\{\sin \sqrt{2}\cos \sqrt{2}\}/h$

$=\underset{h\to 0}{lim}[\sin \sqrt{x+h}-\sqrt{x}]/\cos \sqrt{x+h}.\cos x/h$

$=\underset{h\to 0}{lim}\left\{\right[\sin \sqrt{x}(1+1/2.h/x+(1/2\left)\right(1/2-1\left)\right(h/x){]}_{+}^2.....-1]\}/h\cos \sqrt{x+h}.\cos \sqrt{x}$.

$=\underset{h\to 0}{lim}\left[\sin \right\{h/2\sqrt{x}+$ terms containing higher powers of $h/x\left]/\right(h/2\sqrt{x}).2\sqrt{x}.\cos \sqrt{x+h})\cos \sqrt{x}$

$=\underset{h\to 0}{lim}\left[\sin \right(h/2\sqrt{x})+$ terms containing higher powers of $h/x)/h\sqrt{x}.2\sqrt{2}.(\cos \sqrt{x+h}).\cos \sqrt{x}]$

$=1/(2\sqrt{x}.{\cos }^{2\sqrt{x}})+C$

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