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GUJCET 2026 Math Paper

Question : 2 of 40

Marks: +1, -0

e^y(x+1)=1

\frac{d^2y}{dx^2} - \left(\frac{dy}{dx}\right)^2 =

. . . . . . .

Solution:

Concept:

Differentiate the given relation twice and substitute into the required expression.

Explanation:

Given

e^y(x+1)=1

Take natural logarithm on both sides:

y+\ln(x+1)=0

So,

y=-\ln(x+1)

Differentiate with respect to

x

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

Differentiate again:

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

Now compute:

\frac{d^2y}{dx^2}-\left(\frac{dy}{dx}\right)^2

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

=\frac{1}{(x+1)^2}-\frac{1}{(x+1)^2}=0

Answer:

The value is

0

, so the correct option is D.

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