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MHTCET 15 May 2024 Shift 2 Paper

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Question : 116 of 150

Marks: +1, -0

The derivative of

\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)

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

is

$\frac{2}{3}$
$\frac{1}{2}$
$\frac{3}{2}$
1

Solution:

Concept:

Use substitution

x = \sin \theta

to simplify the inverse trigonometric expressions, then differentiate using the chain rule.

Explanation:

Let

y = \sin^{-1}\left(2x\sqrt{1-x^2}\right)

and

z = \sin^{-1}\left(3x - 4x^3\right)

.

Put

x = \sin \theta

, so

\theta = \sin^{-1}x

.

Then

y = \sin^{-1}\left(2\sin\theta \sqrt{1-\sin^2\theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x

.

Similarly,

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

.

Differentiate both with respect to

x

:

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

and

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

.

Therefore,

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

.

Answer:

\frac{2}{3}

(Option A)

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