Question

The derivative of \[{\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - x{}^2}}} \right)\] with respect to ${\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)$.
A. $\dfrac{{\sqrt {1 - {x^2}} }}{{1 + {x^2}}}$
B. $\dfrac{1}{{\sqrt {1 - {x^2}} }}$
C. $\dfrac{2}{{1 + {x^2}}}$
D. $\dfrac{{2\sqrt {1 - {x^2}} }}{{1 + {x^2}}}$

Answer 1

Hint: We have to find the derivative of \[{\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - x{}^2}}} \right)\] with respect to ${\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)$. The differentiation of a function shows the rate of a function at a given point. Firstly, we will assume \[{\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - x{}^2}}} \right)\] as $u$then differentiate it with respect to $x$. And assume ${\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)$ as $v$ then differentiate with respect to $x$ . Then, for the final answer we will divide $du$ by $dv$.

Complete step by step solution: 
We assume \[{\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - x{}^2}}} \right)\] as $u$
$ \Rightarrow u = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$ (1)
Let $x = \tan \alpha $
Putting in (1)
$ \Rightarrow u = {\tan ^{ - 1}}\left( {\dfrac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }}} \right)$
We know that $\tan 2\alpha = \dfrac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }}$
$ \Rightarrow u = {\tan ^{ - 1}}\left( {\tan 2\alpha } \right)$
$ \Rightarrow u = 2\alpha $ (2)
$x = \tan \alpha $
$ \Rightarrow {\tan ^{ - 1}}x = \alpha $
Putting in (2)
$u = 2{\tan ^{ - 1}}x$
Differentiating with respect to $x$
$\dfrac{{du}}{{dx}} = \dfrac{2}{{1 + {x^2}}}$ (3)
Let $v = {\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)$ (4)
Let $x = \sin \alpha $
Putting in (4)
$v = {\cos ^{ - 1}}\left( {\sqrt {1 - {{\sin }^2}\alpha } } \right)$
$ \Rightarrow v = {\cos ^{ - 1}}(\cos \alpha )$
$v = \alpha $ (5)
${\sin ^{ - 1}}x = \alpha $
Putting in (5)
$v = {\sin ^{ - 1}}x$
Differentiating with respect to $x$
$\dfrac{{dv}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}$ (6)
Dividing (6) by (3)
$\dfrac{{\dfrac{{du}}{{dx}}}}{{\dfrac{{dv}}{{dx}}}} = \dfrac{{\dfrac{2}{{1 + {x^2}}}}}{{\dfrac{1}{{\sqrt {1 - {x^2}} }}}}$
$ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{2\sqrt {1 - {x^2}} }}{{1 + {x^2}}}$
Hence, the correct answer is (D).

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