Question

If $x^4+y^4-a^{2}xy=0$ defines $y$ implicitly as function of $x$ , then $\frac{dy}{dx}=$

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

Answer 1

The correct option is B $-\left(\frac{4x^3-a^{2}y}{4y^3-a^{2}x}\right)$

Given, $x^4+y^4-a^{2}xy=0$
$x^4+y^4=a^{2}xy$
Differentiate both sides w.r.t $x$ we get,
$4x^3+4y^3\frac{dy}{dx}$

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

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