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  • Differentiate the functions with respect to x in sec tan x ^ 1 /2

4. Differentiate the functions with respect to x in 

\sec (\tan (\sqrt x) )

Answers (1)

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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}

Posted by

Gautam harsolia

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