Solve \[\dfrac{d}{{dx}}{\sin ^{ - 1}}\left( {2ax\sqrt {1 - {a^2}{x^2}} } \right) = \]
A.\[\dfrac{{2a}}{{\sqrt {{a^2} - {x^2}} }}\]
B.\[\dfrac{a}{{\sqrt {{a^2} - {x^2}} }}\]
C.\[\dfrac{{2a}}{{\sqrt {1 - {a^2}{x^2}} }}\]
D.\[\dfrac{a}{{\sqrt {1 - {a^2}{x^2}} }}\]

Hint: These types of questions are of specified form where you have to substitute the value \[x\] and \[a\] together then find the derivation , sometimes the variable \[a\] is not given , so just substitute the \[x\] . The substitution will take place in such a way that it will form an identity or formula to simplify the expression .

Complete step-by-step answer:
Given : \[y = {\sin ^{ - 1}}\left( {2ax\sqrt {1 - {a^2}{x^2}} } \right)\]
Now substituting the value of \[ax = \sin \theta \] , we get
\[ = {\sin ^{ - 1}}\left( {2\sin \theta \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} } \right)\] , on solving we get
\[ = {\sin ^{ - 1}}\left( {2\sin \theta \sqrt {1 - {{\sin }^2}\theta } } \right)\] ,
now using the basic trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\] , we get
\[ = {\sin ^{ - 1}}\left( {2\sin \theta \sqrt {{{\cos }^2}\theta } } \right)\] , on simplifying we get
\[ = {\sin ^{ - 1}}\left( {2\sin \theta \cos \theta } \right)\] ,
Now using the identity of \[\sin 2x = 2\sin x\cos x\] , we get
\[ = {\sin ^{ - 1}}\left( {\sin 2\theta } \right)\] , on solving we get
\[ = 2\theta \] .
Now putting the values of \[\theta \] we get ,
\[y = 2{\sin ^{ - 1}}\left( {ax} \right)\]
Now differentiating the expression with respect to \[x\] we get ,
\[\dfrac{{dy}}{{dx}} = 2{\sin ^{ - 1}}\left( {ax} \right)\] ,
Using the formula for differentiation of \[{\sin ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}\] we get
\[\dfrac{{dy}}{{dx}} = 2\left[ {\dfrac{1}{{\sqrt {1 - {{\left( {ax} \right)}^2}} }} \times a} \right]\]
We have multiplied by \[a\] , using the chain rule of derivatives .
\[\dfrac{{dy}}{{dx}} = \dfrac{{2a}}{{\sqrt {1 - {{\left( {ax} \right)}^2}} }}\] , on simplifying we get
\[\dfrac{{dy}}{{dx}} = \dfrac{{2a}}{{\sqrt {1 - {a^2}{x^2}} }}\] .
Therefore , option ( C ) is the correct answer for the given question .
So, the correct answer is “Option C”.

Note: You have to be cautious about the chain rule of the derivatives and the specified form of the questions of the derivatives to check whether we have to have a substitute or not . If you directly apply the formula for the inverse of \[{\sin ^{ - 1}}x\] it will become more complex and take more time to solve . The inverse of function is not cancelled out with same trigonometric function like \[{\sin ^{ - 1}}(\sin x)\] , it is just like equating the angles as the angles in range of \[\sin x\] and domain of \[\sin x\]