# 4. Differentiate the functions with respect to x in $\sec (\tan (\sqrt x) )$

Given function is
$f(x)=\sec (\tan (\sqrt x) )$
when we differentiate it w.r.t. x.
Lets take $t = \sqrt x$ .  then,
$f(t) = \sec (\tan t)$
take $\tan t = k$. then,
$f(k) = \sec k$
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$                                          (By  chain rule)
$\frac{df(k)}{dk} = \frac{d(\sec k )}{dk} = \sec k \tan k = \sec(\tan\sqrt x)\tan(\tan\sqrt x)$
$(\because k = \tan t \ and \ t = \sqrt x)$
$\frac{df(t)}{dt} = \frac{d(\tan t )}{dt} = \sec^2 t =\sec^2 (\sqrt x) \ \ \ \ \ \ (\because t = \sqrt x)$
$\frac{dt}{dx} = \frac{d(\sqrt x)}{dx} = \frac{1}{2\sqrt x}$
Now,
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx} =\sec(\tan \sqrt x)\tan(\tan \sqrt x).\sec^2 (\sqrt x) . \frac{1}{2\sqrt x}$
Therefore, the answer is   $\frac{\sec(\tan \sqrt x).\tan(\tan \sqrt x).\sec^2 (\sqrt x)}{2\sqrt x}$

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