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MHTCET 11 Apr 2026 Shift 2 Paper

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Question : 120 of 150

Marks: +1, -0

The derivative of

\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)

with respect to

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

is...

$\frac{1}{2}$
$\frac{1}{4}$
1
2

Solution:

Concept:

We use

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

and simplify inverse trigonometric functions.

Explanation:

Let

y=\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)

and

z=\sin^{-1}\left(\frac{2x}{1+x^2}\right)

.

Rationalising the expression in

y

, we get:

\frac{\sqrt{1+x^2}-1}{x} = \frac{x}{\sqrt{1+x^2}+1}

Using the standard identity:

\tan^{-1}\left(\frac{x}{\sqrt{1+x^2}+1}\right) = \frac{1}{2}\tan^{-1}x

Therefore:

y = \frac{1}{2}\tan^{-1}x

Differentiating with respect to

x

:

\frac{dy}{dx} = \frac{1}{2(1+x^2)}

Now simplify

z

using the identity:

\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x

So:

z = 2\tan^{-1}x

Differentiating with respect to

x

:

\frac{dz}{dx} = \frac{2}{1+x^2}

Hence:

\frac{dy}{dz} = \frac{\frac{1}{2(1+x^2)}}{\frac{2}{1+x^2}} = \frac{1}{4}

Thus, the required derivative is

\frac{1}{4}

.

Answer:

\frac{1}{4}

(Option B).

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