# Find the derivatives of the function y defined implicity by

Find the derivatives of the function y defined implicity by each of the following equation
${x}^{4}+{y}^{4}-{a}^{2}xy=0$
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Thomas Lynn
Step 1
Given
${x}^{4}+{y}^{4}-{a}^{2}xy=0$
Step 2
differentiating wrt "x"
$\frac{d}{dx}\left({x}^{4}\right)+\frac{d}{dx}\left({y}^{4}\right)-{a}^{2}\frac{d}{dx}\left(xy\right)=0$
$4{x}^{3}+4{y}^{3}\frac{dy}{dx}-{a}^{2}\left(x\frac{dy}{dx}+y\cdot 1\right)=0$
$4{x}^{3}+4{y}^{3}\frac{dy}{dx}-{a}^{2}x\frac{dy}{dx}-{a}^{2}y=0$
$\left(4{y}^{3}-{a}^{2}x\right)\frac{dy}{dx}={a}^{2}y-4{x}^{3}$
$\therefore \frac{dy}{dx}=\frac{{a}^{2}y-4{x}^{2}}{4{y}^{3}-{a}^{2}x}$

John Koga
Step 1
In implicit differentiation all the differentiation rules are applicable.
Step 2
${x}^{4}+{y}^{4}-{a}^{2}xy=0$
differentiate both sides wrt x:
$\frac{d}{dx}\left({x}^{4}+{y}^{4}-{a}^{2}xy\right)=\frac{d}{dx}\left(0\right)$
$\frac{d}{dx}\left({x}^{4}\right)+\frac{d}{dx}\left({y}^{4}\right)-{a}^{2}\frac{d}{dx}\left(xy\right)=0$
apply chain rule and product rule of differentiation:
$4{x}^{3}+4{y}^{3}\frac{dy}{dx}-{a}^{2}\left(x\frac{dy}{dx}+y\frac{dx}{dx}\right)=0$
$4{x}^{3}+4{y}^{3}{y}^{\prime }-{a}^{2}\left(x{y}^{\prime }+y\right)=0$
$4{x}^{3}+4{y}^{3}{y}^{\prime }-{a}^{2}x{y}^{\prime }-{a}^{2}y=0$
${y}^{\prime }\left(4{y}^{3}-{a}^{2}x\right)={a}^{2}y-4{x}^{3}$
${y}^{\prime }=\frac{{a}^{2}y-4{x}^{3}}{4{y}^{3}-{a}^{2}x}$
${y}^{\prime }=\frac{{a}^{2}y-4{x}^{3}}{4{y}^{3}-{a}^{2}x}$