JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

For

x \sqrt{1+y}+y \sqrt{1+x}=0, \frac{d y}{d x}=\ldots \ldots . \quad(x \neq 0,-1)

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because simplifying the given equation and then differentiating yields dy/dx =

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

More Questions from DIFFERENTIATION

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