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Question: If \[y = {\cos ^{ - 1}}(4{x^3} - 3x)\] then \[\dfrac{{dy}}{{dx}} = ?\] A.\[\dfrac{3}{{\sqrt {1 - {...

If y=cos⁡−1(4x3−3x)y = {\cos ^{ - 1}}(4{x^3} - 3x) then 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. −43x2−1\dfrac{{ - 4}}{{3{x^2} - 1}}

Explanation

Solution

We substitute the value of x as cos⁡θ\cos \theta and then use the trigonometric formula of cos⁡3θ\cos 3\theta to convert the equation in the bracket in simpler form. Cancel the inverse of the given function from the function itself and we get an equation in y and x. Use a method of differentiation to differentiate both sides of the equation.

  • cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta
  • Differentiation ofxn{x^n}with respect to x is given by ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}

Complete step-by-step solution:
We are given the equation y=cos⁡−1(4x3−3x)y = {\cos ^{ - 1}}(4{x^3} - 3x)
Substitute the value of x=cos⁡θx = \cos \theta in the given equation
(When x=cos⁡θx = \cos \theta , then θ=cos⁡−1x\theta = {\cos ^{ - 1}}x)
⇒y=cos⁡−1(4cos⁡3θ−3cos⁡θ)\Rightarrow y = {\cos ^{ - 1}}(4{\cos ^3}\theta - 3\cos \theta )
Since we know the trigonometric formula cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta
Substitute the value of cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta in RHS of the equation
⇒y=cos⁡−1(cos⁡3θ)\Rightarrow y = {\cos ^{ - 1}}(\cos 3\theta )
Cancel inverse cosine function by cosine function in right hand side of the equation
⇒y=3θ\Rightarrow y = 3\theta
Substitute the value of θ=cos⁡−1x\theta = {\cos ^{ - 1}}xin right hand side of the equation
⇒y=3cos⁡−1x\Rightarrow y = 3{\cos ^{ - 1}}x
Now differentiate both sides of equation with respect to x
⇒dydx=ddx(3cos⁡−1x)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {3{{\cos }^{ - 1}}x} \right)
We can take out constant value from differentiation in right hand side of the equation
⇒dydx=3ddx(cos⁡−1x)\Rightarrow \dfrac{{dy}}{{dx}} = 3\dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right)
We know ddx(cos⁡−1x)=−11−x2\dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right) = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}, substitute this value of differentiation in right hand side of the equation
⇒dydx=3×−11−x2\Rightarrow \dfrac{{dy}}{{dx}} = 3 \times \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}
Multiply the values in numerator in right hand side of the equation
⇒dydx=−31−x2\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }}
∴\therefore The value of dydx\dfrac{{dy}}{{dx}} is −31−x2\dfrac{{ - 3}}{{\sqrt {1 - {x^2}} }}

∴\therefore The correct option is B.

Note: Many students make the mistake of differentiating the given equation directly, which creates a long and confusing answer. Keep in mind when differentiating the inverse trigonometric function the angle inside the function should be as simple as possible so we can substitute the value of angle directly in the differentiation. Also, when we clearly see some part of the equation equal to one of the trigonometric identities, we substitute that value of the equation as a trigonometric term so as to convert it into an easier formula.