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If $y=\sec \left( {{\tan }^{-1}}x \right)$, then find the value of $\dfrac{dy}{dx}$ .
(a) $\dfrac{x}{\sqrt{1+{{x}^{2}}}}$
(b) $x\sqrt{1+{{x}^{2}}}$
(c) $\sqrt{1+{{x}^{2}}}$
(d) $\dfrac{1}{\sqrt{1+{{x}^{2}}}}$
(e) $\dfrac{x}{1+{{x}^{2}}}$

Answer
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Hint: Consider ‘y’ as composite function of f(g(x)) where f(x) is $\sec x$ and g(x) is ${{\tan }^{-1}}x$ and then use the identity
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$
Where ${{f}^{'}}\left( g\left( x \right) \right)$ is differentiation of f(x) keeping g(x) as it is and $g'(x)$ means differentiating g(x) irrespective of f(x).

“Complete step-by-step answer:”

We are given with the function
$y=\sec \left( {{\tan }^{-1}}x \right)$
Now we are asked to find $\left( \dfrac{dy}{dx} \right)$ which means we have to differentiate ‘y’ with respect to ‘x’.
Let us consider two functions f(x) and g(x) where f(x) be $\sec x$ and g(x) be ${{\tan }^{-1}}x$.
So we can write,
$y=\sec \left( {{\tan }^{-1}}x \right)$ as $y=f\left( g\left( x \right) \right)$
Now we have to differentiate ‘y’ with respect to ‘x’ using the identity,
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right)\times g'\left( x \right)$
Here \[f'\left( g\left( x \right) \right)\] means differentiating f(x) keeping g(x) constant and here $g'\left( x \right)$ means differentiating g(x) independently irrespective of f(x).
So by using the formula which are,
$\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x$, $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$
We get,
\[~\dfrac{dy}{dx}=\sec \left( {{\tan }^{-1}}x \right).\tan \left( {{\tan }^{-1}}x \right).\dfrac{1}{\left( 1+{{x}^{2}} \right)}..........(i)\]
Now here we can use the identity,
$\begin{align}
  & \tan \left( {{\tan }^{-1}}x \right)=x \\
 & \sec \left( {{\tan }^{-1}}x \right)=\sqrt{1+{{x}^{2}}} \\
\end{align}$
By using these identities, equation (i) can be written as
$\dfrac{dy}{dx}=\sqrt{1+{{x}^{2}}}.x.\dfrac{1}{1+{{x}^{2}}}$
By rationalizing the above equation, we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{\sqrt{1+{{x}^{2}}}}$
Therefore this is the required differentiation.
Hence the correct answer is option (a).

Note: There is alternative way of solving the problem is by converting $y=\sec \left( {{\tan }^{-1}}x \right)$ as $y=\sqrt{1+{{x}^{2}}}$ and using $f\left( x \right)=\sqrt{x}$ and $g\left( x \right)=\left( 1+{{x}^{2}} \right)$ . Thus solving same by the identity
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$ to get desired result.