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The differential coefficient of $\sec \left( {{\tan }^{-1}}x \right)$ is:(a) $\dfrac{x}{1+{{x}^{2}}}$ (b) $x\sqrt{1+{{x}^{2}}}$ (c) $\dfrac{1}{\sqrt{1+{{x}^{2}}}}$ (d) $\dfrac{x}{\sqrt{1+{{x}^{2}}}}$

Last updated date: 23rd Jun 2024
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Hint: Start by finding the derivative of $\sec \left( {{\tan }^{-1}}x \right)$ using the chain rule. We know that $\dfrac{d\left( \sec x \right)}{dx}=\sec x\tan x$ and $\dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}=\dfrac{1}{1+{{x}^{2}}}$ . Once you get the derivative report the coefficient of $\sec \left( {{\tan }^{-1}}x \right)$ as your answer. You may have to use the identity $\tan \left( {{\tan }^{-1}}x \right)=x$ for simplification of your answer.

Let us start with finding the derivative of $\sec \left( {{\tan }^{-1}}x \right)$ . For finding the derivative, we will use the chain rule of differentiation. According to the chain rule of differentiation, we know $\dfrac{d\left( f\left( g(x) \right) \right)}{dx}=f'\left( g(x) \right)\times g'(x)$ . For $\sec \left( {{\tan }^{-1}}x \right)$ , $f\left( x \right)=\sec x$ and $g(x)={{\tan }^{-1}}x$ .
We also know that $\dfrac{d\left( \sec x \right)}{dx}=\sec x\tan x$ . .
$\therefore \dfrac{d\left( \sec \left( {{\tan }^{-1}}x \right) \right)}{dx}=\sec \left( {{\tan }^{-1}}x \right)\tan \left( {{\tan }^{-1}}x \right)\times \dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}$
Now, we know that ${{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}$ . So, if we use this in our equation, we get
$\dfrac{d\left( \sec \left( {{\tan }^{-1}}x \right) \right)}{dx}=\sec \left( {{\tan }^{-1}}x \right)\tan \left( {{\tan }^{-1}}x \right)\times \dfrac{1}{1+{{x}^{2}}}$
Now, according to the rules of inverse trigonometric function, we know $\tan \left( {{\tan }^{-1}}x \right)=x$ . If we use this in our equation, we get
$\dfrac{d\left( \sec \left( {{\tan }^{-1}}x \right) \right)}{dx}=\sec \left( {{\tan }^{-1}}x \right)\times x\times \dfrac{1}{1+{{x}^{2}}}$
$\Rightarrow \dfrac{d\left( \sec \left( {{\tan }^{-1}}x \right) \right)}{dx}=\sec \left( {{\tan }^{-1}}x \right)\times \dfrac{x}{1+{{x}^{2}}}$
Now, we are asked the differential coefficient of $\sec \left( {{\tan }^{-1}}x \right)$ . So, from the above result, we can clearly see that the coefficient of $\sec \left( {{\tan }^{-1}}x \right)$ in the derivative of $\sec \left( {{\tan }^{-1}}x \right)$ is $\dfrac{x}{1+{{x}^{2}}}$ .
Note: The first thing to keep in mind is that according to the rule of trigonometric inverse functions, $\tan \left( {{\tan }^{-1}}x \right)=x$ , but ${{\tan }^{-1}}\left( \tan x \right)=x$ if and only if it is mentioned that $x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ . The other thing to keep in mind is while you report the answers, report the exact matched option, as the options given are very similar and misleading.