Question

If ${(x^2+y^2)}^2=xy,then\frac{dy}{dx}$=

• A. $\frac{4x^3+4x{y}^2-y}{x-4x^{2}y-4y^3}$
• B. $\frac{-4x^3+4x{y}^2-y}{4y^2+4x^{2}y-x}$
• C. $\frac{4x^3+4x{y}^2-y}{4y^2-4x^{2}y-x}$
• D. $\frac{(4x^3+4x{y}^2-y)}{(4y^3+4x^{2}y-x)}$

Answer 1

The correct option is A $\frac{4x^3+4x{y}^2-y}{x-4x^{2}y-4y^3}$

$(x^2+y^2{)}^2=xy,y^{\prime }=?$

on differentiating wrt $x$

$2(x^2+y^2)(2x+2y\frac{dy}{dx})=x\frac{dy}{dx}+y$

$(2x^2+2y^2)(2x+2y\frac{dy}{dx})=x\frac{dy}{dx}+y$

$4x^3+4x^{2}y\frac{dy}{dx}+4x{y}^2+4y^3\frac{dy}{dx}=x\frac{dy}{dx}+y$

$4x^{2}y\frac{dy}{dx}+4y^3\frac{dy}{dx}-x\frac{dy}{dx}=y-4x^3-4x{y}^2$

$(4x^{2}y+4y^3-x)\frac{dy}{dx}=y-4x^3-4x{y}^2$

$\Rightarrow \frac{dy}{dx}=\frac{y-4x^3-4x{y}^2}{4x^{2}y+4y^3-x}$

or

$\frac{dy}{dx}=\left(\frac{4x^3+4x{y}^2-y}{x-4x^{2}y-4y^3}\right)$