Question

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}}}} $

Answer 1

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.

Complete step-by-step 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}}} $ .
So, the correct answer is “Option A”.

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.