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Question: If \(y = {\cos ^{ - 1}}(4{x^3} - 3x)\), then what is the value of \(\dfrac{{dy}}{{dx}}\)? A. \(\df...

If y=cos⁡−1(4x3−3x)y = {\cos ^{ - 1}}(4{x^3} - 3x), then what is the value of dydx\dfrac{{dy}}{{dx}}?
A. 31−x2\dfrac{3}{{\sqrt {1 - {x^2}} }}
B. −31−x2\dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }}
C. 41−x2\dfrac{4}{{\sqrt {1 - {x^2}} }}
D. −4(3x2−1)\dfrac{{ - 4}}{{\left( {3{x^2} - 1} \right)}}

Explanation

Solution

In the question, we are given an equation in terms of y and x. We are asked to differentiate this equation. Start by making this equation simpler. This can be done by using certain trigonometric formulas. Substitute a different value for xx using those formulas. And then simplify the equation. When the equation is completely simplified, it can be differentiated.

Formula used: 1) cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta
2) cos⁡−1(cos⁡θ)=θ{\cos ^{ - 1}}(\cos \theta ) = \theta
3)d(cos⁡−1θ)dx=−11−x2\dfrac{{d({{\cos }^{ - 1}}\theta )}}{{dx}} = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}

Complete step-by-step answer:
We are given that y=cos⁡−1(4x3−3x)y = {\cos ^{ - 1}}(4{x^3} - 3x).
We will start by substituting x=cos⁡θx = \cos \theta ............…. (1)
This is done because the equation (4x3−3x)(4{x^3} - 3x) is similar to the formula- cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta . We intend to simplify the equation by substituting x=cos⁡θx = \cos \theta .
Substituting (1) in the given equation,
⇒y=cos⁡−1(4cos⁡3θ−3cos⁡θ)\Rightarrow y = {\cos ^{ - 1}}(4{\cos ^3}\theta - 3\cos \theta )
Putting 4cos⁡3θ−3cos⁡θ=cos⁡3θ4{\cos ^3}\theta - 3\cos \theta = \cos 3\theta
⇒y=cos⁡−1(cos⁡3θ)\Rightarrow y = {\cos ^{ - 1}}(\cos 3\theta )
Using the formula- cos⁡−1(cos⁡θ)=θ{\cos ^{ - 1}}(\cos \theta ) = \theta
⇒y=3θ\Rightarrow y = 3\theta
Now, as per the question we need to find dydx\dfrac{{dy}}{{dx}}. But it can only be found if our final equation is in the terms of y and x. But here, our equation is in y and θ\theta . So, we will use equation (1) to bring our equation in the terms of y and x.
Using (1),
⇒x=cos⁡θ\Rightarrow x = \cos \theta
Shifting cos to the other side (it will result in inverse trigonometric ratio),
⇒θ=cos⁡−1x\Rightarrow \theta = {\cos ^{ - 1}}x
Putting this in y=3θy = 3\theta ,
⇒y=3cos⁡−1x\Rightarrow y = 3{\cos ^{ - 1}}x
Now, we will differentiate the final equation in terms of x,
⇒dydx=−31−x2\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }}

Therefore −31−x2\dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }} is the required answer.

Note: 1) The student can differentiate the given equation without simplifying as well, but that will be lengthy and a little confusing. Due to this reason, the given method is preferred.
2) The students must be aware of all the trigonometric identities as they are helpful in solving differential equations also.