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Question: Derivative of \[{\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)\] with respect to \[{\sin ^{ - 1}}x\] is...

Derivative of sin⁡−1(3x−4x3){\sin ^{ - 1}}\left( {3x - 4{x^3}} \right) with respect to sin⁡−1x{\sin ^{ - 1}}x is

Explanation

Solution

Hint : The given question is an example of implicit derivative and the range of the xx is to be observed carefully . From the option we can say that the range of xx is less than or equal to 12\dfrac{1}{2} . We have to make a substitution for xx so as to make it the identity of sin⁡3θ\sin 3\theta .

Complete step-by-step answer :
Given : y=sin⁡−1(3x−4x3)y = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)
The expression will have different values in different quadrants .
Let x=sin⁡θx = \sin \theta , we get
y=sin⁡−1(3sin⁡θ−4sin⁡3θ)y = {\sin ^{ - 1}}\left( {3\sin \theta - 4{{\sin }^3}\theta } \right) ,
on simplifying we get
y=sin⁡−1(sin⁡3θ)y = {\sin ^{ - 1}}\left( {\sin 3\theta } \right) , as (sin⁡3θ=3sin⁡θ−sin⁡3θ)\left( {\sin 3\theta = 3\sin \theta - {{\sin }^3}\theta } \right)
y=±3θy = \pm 3\theta
Now , on differentiating with respect to θ\theta we get ,
dydθ=±3\dfrac{{dy}}{{d\theta }} = \pm 3 ……….. equation (a)
Now , for the second function we have
z=sin⁡−1xz = {\sin ^{ - 1}}x
Let x=sin⁡θx = \sin \theta we get ,
z=sin⁡−1(sin⁡θ)z = {\sin ^{ - 1}}\left( {\sin \theta } \right) ,
on simplifying we get
z=θz = \theta
Now , on differentiating with respect θ\theta to we get ,
dzdθ=1\dfrac{{dz}}{{d\theta }} = 1 ………….. equation (b)
Now , from equation (a) and equation (b) we have ,
dydθdzdθ=31\dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dz}}{{d\theta }}}} = \dfrac{3}{1} , on solving we get
On simplifying we get ,
dydz=3\dfrac{{dy}}{{dz}} = 3

Note : Implicit differentiation is the procedure of differentiating an implicit equation with respect to the desired variable xx while treating the other variables as unspecified functions of xx . The value of sin⁡−1(3x−4x3){\sin ^{ - 1}}\left( {3x - 4{x^3}} \right) can vary with change in xx as it moves from one quadrant to another . We have to consider both the possible answers for dydθ=±3\dfrac{{dy}}{{d\theta }} = \pm 3 as it is given in the options . In an implicit function we have to differentiate the two functions separately with respect to the corresponding function , then divide the equations obtained from differentiation to get the desired result .