Question

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.$\dfrac{a}{{\sqrt {1 - {a^2}{x^2}} }}$

Answer 1

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 $\sin x$