Question

If $y = {\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ),\dfrac{{ - 1}}{{\sqrt 2 }} \leqslant x \leqslant \dfrac{1}{{\sqrt 2 }}$, then $\dfrac{{dy}}{{dx}}$ is equal to
A. $\dfrac{x}{{\sqrt {(1 - {x^2})} }}$
B. $\dfrac{1}{{\sqrt {(1 - {x^2})} }}$
C. $\dfrac{2}{{\sqrt {(1 - {x^2})} }}$
D. $\dfrac{{2x}}{{\sqrt {(1 - {x^2})} }}$

Answer 1

Hint: In such questions, always let $x$ be a trigonometric function where maximum times by taking $x$ as the trigonometric function same as the function outside the angle helps to get the answer easily i.e., Let, $x$ be $\sin \theta $and put in given condition then solve further.

Formula Used:
Pythagorean Identity –
${\sin ^2}\theta + {\cos ^2}\theta = 1$
$ \Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta $
Trigonometry formulas involving double angle identity –
$\sin 2\theta = 2\sin \theta \cdot \cos \theta $
Inverse trigonometry formula –
$\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }},x \ne \pm 1$

Complete step by step solution:
Given that,
$y = {\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) - - - - - (1)$
where, $\dfrac{{ - 1}}{{\sqrt 2 }} \leqslant x \leqslant \dfrac{1}{{\sqrt 2 }}$
Let, $x = \sin \theta $
Which implies that, $\theta = {\sin ^{ - 1}}x - - - - - (2)$
Put the value of $x$ in equation (1)
$y = {\sin ^{ - 1}}\left( {2\left( {\sin \theta } \right)\sqrt {1 - {{\left( {\sin \theta } \right)}^2}} } \right)$
$y = {\sin ^{ - 1}}\left( {2\sin \theta \sqrt {{{\cos }^2}\theta } } \right)$
$y = {\sin ^{ - 1}}\left( {2\sin \theta \cdot \cos \theta } \right)$
$y = {\sin ^{ - 1}}\left( {\sin 2\theta } \right)$
$y = 2\theta $
$y = 2{\sin ^{ - 1}}x$
Differentiate the above equation with respect to $x$,
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {2{{\sin }^{ - 1}}x} \right)$
$\dfrac{{dy}}{{dx}} = 2\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right)$
$\dfrac{{dy}}{{dx}} = 2\left( {\dfrac{1}{{\sqrt {1 - {x^2}} }}} \right)$
$\dfrac{{dy}}{{dx}} = \dfrac{2}{{\sqrt {1 - {x^2}} }}$

Option ‘C’ is correct

Note: To differentiate any function, first try to make the function easy or same as any trigonometric formula. Main point for differentiation is that the function should be a trigonometric function not a variable and the last answer will be in the form of given variables. Use more and more formulas to solve such types of questions and apply them carefully.