If y= ^{-1}≤ft({1}{{x+1}}) then d y/d x=

Correct Answer is : −12x(1+x)-{1}{2 {x}(1+x)}−2x(1+x)1

Correct Answer is : −12x(1+x)-{1}{2 {x}(1+x)}−2x(1+x)1

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COMEDK 2025 Morning Solved Paper

Question : 158 of 180

Marks: +1, -0

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

\frac{d y}{d x}=

Solution:

Concept:

Differentiate an inverse sine function using the chain rule and simplify the result.

Explanation:

Let

y = \sin^{-1} u

where

u = \frac{1}{\sqrt{x+1}}

First, find

\frac{dy}{du} = \frac{1}{\sqrt{1-u^2}}

Next, rewrite

u = (x+1)^{-1/2}

and differentiate:

\frac{du}{dx} = -\frac12 (x+1)^{-3/2}

Apply the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \left(-\frac12 (x+1)^{-3/2}\right)

Simplify

1-u^2

1 - \frac{1}{x+1} = \frac{x}{x+1}

, so

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

Substitute:

\frac{dy}{dx} = -\frac12 \cdot \frac{\sqrt{x+1}}{\sqrt{x}} \cdot \frac{1}{(x+1)^{3/2}} = -\frac{1}{2\sqrt{x}\,(x+1)}

This matches the form

-\frac{1}{2\sqrt{x}(1+x)}

Answer:

-\frac{1}{2\sqrt{x}(1+x)}

(Option D).

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