Question

If $y = \sin^{ - 1}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + \sec^{ - 1}\left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$. Then what is the value of $\dfrac{{dy}}{{dx}}$?
A. $\dfrac{4}{{\left( {1 - {x^2}} \right)}}$
B. $\dfrac{4}{{\left( {1 + {x^2}} \right)}}$
C. $\dfrac{1}{{\left( {1 + {x^2}} \right)}}$
D. $\dfrac{{ - 4}}{{\left( {1 + {x^2}} \right)}}$

Answer 1

Hint: Solve the given trigonometric equation by substituting $x = \tan\theta $. Further simplify the equation using the identities of $\sin 2A$ and $\cos 2A$. Then substitute the value of $\theta $ and simplify the equation. In the end, differentiate the equation with respect to $x$ and get the required answer.

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

Complete step by step solution:
The given trigonometric equation is $y = \sin^{ - 1}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + \sec^{ - 1}\left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$.
Let’s simplify the above equation.
Substitute $x = \tan\theta $ in the above equation.
$y = \sin^{ - 1}\left( {\dfrac{{2\tan\theta }}{{1 + \tan^{2}\theta }}} \right) + \sec^{ - 1}\left( {\dfrac{{1 + \tan^{2}\theta }}{{1 - \tan^{2}\theta }}} \right)$
$ \Rightarrow y = \sin^{ - 1}\left( {\dfrac{{2\tan\theta }}{{1 + \tan^{2}\theta }}} \right) + \cos^{ - 1}\left( {\dfrac{{1 - \tan^{2}\theta }}{{1 + \tan^{2}\theta }}} \right)$ $\because \sec^{ - 1}A = \cos^{ - 1}\left( {\dfrac{1}{A}} \right)$
Now apply the formulas of $\sin 2A$, and $\cos 2A$.
We get,
$y = \sin^{ - 1}\left( {\sin2\theta } \right) + \cos^{ - 1}\left( {\cos2\theta } \right)$
Now apply the inverse trigonometric rules $\sin^{ - 1}\left( {\sin A} \right) = A$, and $\cos^{ - 1}\left( {\cos A} \right) = A$.
$y = 2\theta + 2\theta $
$ \Rightarrow y = 4\theta $
Resubstitute the value of $\theta $.
$y = 4\tan^{ - 1}x$

Differentiate the above equation with respect to $x$.
$\dfrac{{dy}}{{dx}} = 4\dfrac{d}{{dx}}\left( {\tan^{ - 1}x} \right)$
Apply the formula of a derivative $\dfrac{d}{{dx}}\left( {\tan^{ - 1}x} \right) = \dfrac{1}{{1 + {x^2}}}$.
$\dfrac{{dy}}{{dx}} = 4\left( {\dfrac{1}{{1 + {x^2}}}} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{{\left( {1 + {x^2}} \right)}}$

Option ‘B’ is correct

Note: The first principle of differentiation can also be used to help solve the problem. Trigonometric identities and formulas are used to simplify the function in the questions, and basic differentiation results are then used to determine the right answer. When changing a variable's value, we must be careful.