If $(x^{2}+y^{2})^{2}=xy$, find $\dfrac{dy}{dx}$.

$\dfrac{y-4x(x^{2}+y^{2})}{4y(x^{2}+y^{2})-x}$

CBSE Class 12 · 2018 · DifferentiationLong Answer

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

, find

\frac{d y}{d x}

Answer: $\frac{y - 4 x ( x ^{2} + y ^{2} )}{4 y ( x ^{2} + y ^{2} ) - x}$

Differentiate both sides w.r.t.

x

(chain rule left, product rule right):

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

Expand:

4 x (x^{2} + y^{2}) + 4 y (x^{2} + y^{2}) \frac{d y}{d x} = x \frac{d y}{d x} + y

. Collect

\frac{d y}{d x}

[4 y (x^{2} + y^{2}) - x] \frac{d y}{d x} = y - 4 x (x^{2} + y^{2})

\frac{d y}{d x} = \frac{y - 4 x ( x ^{2} + y ^{2} )}{4 y ( x ^{2} + y ^{2} ) - x} = \frac{y - 4 x ^{3} - 4 x y ^{2}}{4 x ^{2} y + 4 y ^{3} - x}

CBSE Class 12 Board · 2018 · Subjective