# 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.

$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$

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.

Last updated date: 23rd Sep 2023

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