Home/Class 12/Maths/

## QuestionMathsClass 12

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

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

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