If $x\sqrt{1+y}+y\sqrt{1+x}=0$ (with $x\neq y$), find $\dfrac{dy}{dx}$.

CBSE Class 12 · DifferentiationLong Answer

x 1 + y + y 1 + x = 0

(with

x \neq = y

), find

\frac{d y}{d x}

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

From

x 1 + y = - y 1 + x

, square both sides:

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

Rearrange:

x^{2} - y^{2} + x^{2} y - x y^{2} = 0 \Rightarrow (x - y) (x + y) + x y (x - y) = 0

Since

x \neq = y

, divide by

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

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

CBSE Class 12 Board · Subjective