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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) \(\displaystyle \frac{-4x^{3}}{4y^{3}-a^{2}x}\)

Answer

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