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MHTCET 10 May 2024 Shift 2 Paper

Question : 136 of 150

Marks: +1, -0

Derivative of

\tan ^{-1} \sqrt{\frac{1-x}{1+x}}

\cos ^{-1}\left(4 x^{3}-3 x\right)

Solution:

Concept:

Simplify both inverse trigonometric functions using standard trigonometric substitutions, then use the chain rule.

Explanation:

Let

y = \tan^{-1}\sqrt{\frac{1-x}{1+x}}

and

z = \cos^{-1}(4x^3 - 3x)

We need

\frac{dy}{dz}

, which can be written as

\frac{dy/dx}{dz/dx}

Put

x = \cos 2\theta_1

y

Then

\frac{1-\cos 2\theta_1}{1+\cos 2\theta_1} = \frac{2\sin^2\theta_1}{2\cos^2\theta_1} = \tan^2\theta_1

\sqrt{\frac{1-\cos 2\theta_1}{1+\cos 2\theta_1}} = \tan\theta_1

Thus

y = \tan^{-1}(\tan\theta_1) = \theta_1 = \frac{\cos^{-1}x}{2}

Differentiating, we get

\frac{dy}{dx} = -\frac{1}{2\sqrt{1-x^2}}

Now put

x = \cos\theta

z

Then

4x^3 - 3x = 4\cos^3\theta - 3\cos\theta = \cos 3\theta

z = \cos^{-1}(\cos 3\theta) = 3\theta = 3\cos^{-1}x

Differentiating, we get

\frac{dz}{dx} = -\frac{3}{\sqrt{1-x^2}}

Therefore,

\frac{dy}{dz} = \frac{-\frac{1}{2\sqrt{1-x^2}}}{-\frac{3}{\sqrt{1-x^2}}} = \frac{1}{6}

Answer:

\frac{1}{6}

, which is option D.

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