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Question: Differentiate \({\cos ^{ - 1}}(4{x^3} - 3x)\) with respect to \(x\)....

Differentiate cos⁡−1(4x3−3x){\cos ^{ - 1}}(4{x^3} - 3x) with respect to xx.

Explanation

Solution

First consider the variable xx to be cos⁡θ\cos \theta and then see the new argument of the given inverse function and simplify it with help of multiple angle formula of cosine function, then further simplify the inverse function and finally find the differentiation.Multiple angle formula of cos,cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x = 4{\cos ^3}x - 3\cos x.

Formula used:
Trigonometric identity for multiple angle of cosine: cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x = 4{\cos ^3}x - 3\cos x
Derivative of cos⁡−1x:  −11−x2{\cos ^{ - 1}}x:\;\dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}

Complete step by step answer:
To differentiate cos⁡−1(4x3−3x){\cos ^{ - 1}}(4{x^3} - 3x) with respect to xx, we will first try to simplify given trigonometric expression, in order to simplify it let us consider xx to be cos⁡θ\cos \theta , that is then the expression can be written as
cos⁡−1(4cos⁡3θ−3cos⁡θ){\cos ^{ - 1}}(4{\cos ^3}\theta - 3\cos \theta )
See the argument of the above inverse trigonometric expression, we are familiar to it as it is a trigonometric identity that can be written as follows
4cos⁡3θ−3cos⁡θ=cos⁡3θ4{\cos ^3}\theta - 3\cos \theta = \cos 3\theta
Replacing 4cos⁡3θ−3cos⁡θ  with  cos⁡3θ4{\cos ^3}\theta - 3\cos \theta \;{\text{with}}\;\cos 3\theta in the above expression, we will get
cos⁡−1(cos⁡3θ){\cos ^{ - 1}}(\cos 3\theta )
Simplifying it further we will get
3θ3\theta
Now as we have considered x=cos⁡θ⇒cos⁡−1x=θx = \cos \theta \Rightarrow {\cos ^{ - 1}}x = \theta , we will get
3cos⁡−1x3{\cos ^{ - 1}}x
So we have get cos⁡−1(4x3−3x)=3cos⁡−1x{\cos ^{ - 1}}(4{x^3} - 3x) = 3{\cos ^{ - 1}}x
Now differentiating it with respect to xx
d(cos⁡−1(4x3−3x))dx=d(3cos⁡−1x)dx\dfrac{{d\left( {{{\cos }^{ - 1}}(4{x^3} - 3x)} \right)}}{{dx}} = \dfrac{{d\left( {{{3\cos }^{ - 1}}x} \right)}}{{dx}}
We know that d(cos⁡−1x)dx=−31−x2\dfrac{{d\left( {{{\cos }^{ - 1}}x} \right)}}{{dx}} = \dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }}

Therefore d(cos⁡−1(4x3−3x))dx=d(cos⁡−1x)dx=−31−x2\dfrac{{d\left( {{{\cos }^{ - 1}}(4{x^3} - 3x)} \right)}}{{dx}} = \dfrac{{d\left( {{{\cos }^{ - 1}}x} \right)}}{{dx}} = \dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }} is the required derivative.

Note: When differentiating either inverse trigonometric function or trigonometric function, always try to simplify their argument first, then further use the derivative formulae to directly differentiate them. Simplifying the argument makes the process easier as we have seen in this problem.This problem could be solved by one more way that is with the help of the first principle of differentiation in which we can differentiate a function by computing its limit. Try to solve this question with this method and match the answer.