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

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

Complete step-by-step answer:
Given : \[y = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)\]
The expression will have different values in different quadrants .
Let \[x = \sin \theta \] , we get
 \[y = {\sin ^{ - 1}}\left( {3\sin \theta - 4{{\sin }^3}\theta } \right)\] ,
 on simplifying we get
 \[y = {\sin ^{ - 1}}\left( {\sin 3\theta } \right)\] , as \[\left( {\sin 3\theta = 3\sin \theta - {{\sin }^3}\theta } \right)\]
 \[y = \pm 3\theta \]
Now , on differentiating with respect to \[\theta \] we get ,
 \[\dfrac{{dy}}{{d\theta }} = \pm 3\] ……….. equation (a)
Now , for the second function we have
 \[z = {\sin ^{ - 1}}x\]
Let \[x = \sin \theta \] we get ,
 \[z = {\sin ^{ - 1}}\left( {\sin \theta } \right)\] ,
 on simplifying we get
 \[z = \theta \]
Now , on differentiating with respect \[\theta \] to we get ,
 \[\dfrac{{dz}}{{d\theta }} = 1\] ………….. equation (b)
Now , from equation (a) and equation (b) we have ,
 \[\dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dz}}{{d\theta }}}} = \dfrac{3}{1}\] , on solving we get
On simplifying we get ,
 \[\dfrac{{dy}}{{dz}} = 3\]

Note: Implicit differentiation is the procedure of differentiating an implicit equation with respect to the desired variable \[x\] while treating the other variables as unspecified functions of \[x\] . The value of \[{\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)\] can vary with change in \[x\] as it moves from one quadrant to another . We have to consider both the possible answers for \[\dfrac{{dy}}{{d\theta }} = \pm 3\] 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 .