If $x\sqrt {1 + y} + y\sqrt {1 + x} = 0$, then $\frac{{dy}}{{dx}} = $

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

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Application +4 | -1

x\sqrt {1 + y} + y\sqrt {1 + x} = 0

, then

\frac{{dy}}{{dx}} =

View answer and explanation

Correct option: B

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

Option B (

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

is correct because after simplifying the given equation and differentiating, the result is

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

Practice All Chapter MCQs →

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f^{\prime}\left(\frac{-\pi}{6}\right)

x

y

x>0

x y=1

x+y

f(x)=[x] \sin (\pi x)

[x]

x

x=k, k

f(x)

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

\frac{{dy}}{{dx}}

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

z = y + f\left( v \right),

v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}