If $x\sqrt{1+y}+y\sqrt{1+x}=0$ and $x\neq y$, prove that $\dfrac{dy}{dx}=-\dfrac{1}{(x+1)^{2}}$.

CBSE Class 12 · 2019 · DifferentiationLong Answer

x 1 + y + y 1 + x = 0

and

x \neq = y

, prove that

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

Answer: Proved

From

x 1 + y = - y 1 + x

, square both sides:

x^{2} (1 + y) = y^{2} (1 + x)

Rearrange and factor:

(x - y) (x + y) + x y (x - y) = 0 \Rightarrow (x - y) (x + y + x y) = 0

Since

x \neq = y

x + y + x y = 0 \Rightarrow y = \frac{- x}{1 + x}

.
Differentiate by the quotient rule:

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

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

CBSE Class 12 Board · 2019 · Subjective