Question

If $y = {\tan ^{ - 1}}\dfrac{x}{{\sqrt {1 - {x^2}} }}$ then find $\dfrac{{dy}}{{dx}}$
(A) $\dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}$
(B) $\dfrac{x}{{\sqrt {1 - {x^2}} }}$
(C) $\dfrac{1}{{\sqrt {1 - {x^2}} }}$
(D) $\dfrac{{\sqrt {1 - {x^2}} }}{x}$

Answer 1

Hint: We have to find the derivative of $y = {\tan ^{ - 1}}\dfrac{x}{{\sqrt {1 - {x^2}} }}$ with respect to $x$. The differentiation of a function shows the rate of a function at a given point. Firstly, we will assume $x$ as $\sin \alpha $. Then after putting the value of $x$ in $y$ and simplifying we will differentiate it with respect to $x$ for getting the final answer.

Formula Used:
$\sqrt {1 - {{\sin }^2}\alpha } = \cos \alpha $
$\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$

Complete step by step Solution:
Given, $y = {\tan ^{ - 1}}\dfrac{x}{{\sqrt {1 - {x^2}} }}$ (1)
We will assume \[x = \sin \alpha \]
Putting in (1)
$y = {\tan ^{ - 1}}\dfrac{{\sin \alpha }}{{\sqrt {1 - {{\sin }^2}\alpha } }}$
We know
$\sqrt {1 - {{\sin }^2}\alpha } = \cos \alpha $
Using this identity
$y = {\tan ^{ - 1}}\dfrac{{\sin \alpha }}{{\cos \alpha }}$
We know
$\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$
Using this identity
$y = {\tan ^{ - 1}}(\tan \alpha )$
$ \Rightarrow y = \alpha $ (2)
\[x = \sin \alpha \]
${\sin ^{ - 1}}x = \alpha $
Putting in (2)
$y = {\sin ^{ - 1}}x$
Differentiating with respect to $x$
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}$

Hence, the correct option is C.

Note: Students can make mistakes while differentiating with respect to $x$. So, they have to pay attention. We also have to take care of what we assume the function is. We have to assume functions correctly so that we can solve problems easily and get the correct answer without any errors.