Question

If $y = {\cos ^{ - 1}}\left( {\dfrac{{x - {x^{ - 1}}}}{{x + {x^{ - 1}}}}} \right)$, then $\dfrac{{dy}}{{dx}} =$
A. $\dfrac{{ - 2}}{{1 + {x^2}}}$
B. $\dfrac{2}{{1 + {x^2}}}$
C. $\dfrac{1}{{1 + {x^2}}}$
D. $\dfrac{{ - 1}}{{1 + {x^2}}}$

Answer 1

Here, we have to find the derivative of $y = {\cos ^{ - 1}}\left( {\dfrac{{x - {x^{ - 1}}}}{{x + {x^{ - 1}}}}} \right)$. For that, first of all substitute ${x^{ - 1}} = \dfrac{1}{x}$ and then take the LCM and simplify. Then take out minus (-) common from numerator and use the formula
$\Rightarrow {\cos ^{ - 1}}\left( { - \theta } \right) = \pi - \cos \theta$ and
$\Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right) = 2{\tan ^{ - 1}}x$
Now, we can easily find the derivative of the given function.

Complete step-by-step answer:
In this question, we are given a trigonometric equation and we need to find its derivative.
The given equation is : $y = {\cos ^{ - 1}}\left( {\dfrac{{x - {x^{ - 1}}}}{{x + {x^{ - 1}}}}} \right)$ - - - - - - - - - - - - - (1)
Now, there is no direct formula for differentiating equation (1), so we need to use some trigonometric formulas and relations to find its derivative.
First of all, x inverse means reciprocal of x. Therefore, substitute $\dfrac{1}{x}$ instead of ${x^{ - 1}}$ in equation (1), we get
$\Rightarrow y = {\cos ^{ - 1}}\left( {\dfrac{{x - \dfrac{1}{x}}}{{x + \dfrac{1}{x}}}} \right)$
Now, taking LCM, we get
$\Rightarrow y = {\cos ^{ - 1}}\left( {\dfrac{{\dfrac{{{x^2} - 1}}{x}}}{{\dfrac{{{x^2} + 1}}{x}}}} \right)$
Here, x gets cancelled. Therefore, we get
$\Rightarrow y = {\cos ^{ - 1}}\left( {\dfrac{{{x^2} - 1}}{{{x^2} + 1}}} \right)$- - - - - - - - - (2)
Now, take out minus (-) common from the numerator in equation (2), we get
$\Rightarrow y = {\cos ^{ - 1}}\left[ { - \left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)} \right]$ - - - - - - - - - (3)
Now, we know that ${\cos ^{ - 1}}\left( { - \theta } \right) = \pi - \cos \theta$. Therefore, equation (3) becomes
$\Rightarrow y = \pi - {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ - - - - - - (4)
Now, we know that ${\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right) = 2{\tan ^{ - 1}}x$.
Therefore, putting this value in equation (4), we get
$\Rightarrow y = \pi - 2{\tan ^{ - 1}}x$
Now, differentiating the above equation, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = 0 - 2\dfrac{d}{{dx}}{\tan ^{ - 1}}x$
Now, we know that the derivative of ${\tan ^{ - 1}}x$ is equal to $\dfrac{1}{{1 + {x^2}}}$.
Therefore,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 2}}{{1 + {x^2}}}$

So, the correct answer is “Option A”.

Note: Here is the proof for ${\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right) = 2{\tan ^{ - 1}}x$.
Let us take LHS.
$\Rightarrow LHS = {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$
Now, let $x = \tan \theta$
$\Rightarrow \theta = {\tan ^{ - 1}}x$
$\Rightarrow LHS = {\cos ^{ - 1}}\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) \\\ \Rightarrow LHS = {\cos ^{ - 1}}\left( {\cos 2\theta } \right) \\\ \Rightarrow LHS = 2\theta \\\ \Rightarrow LHS = 2{\tan ^{ - 1}}x \\\$