Teddy Dillard

2021-12-17

Find the derivatives of the function y defined implicity by each of the following equation

${x}^{4}+{y}^{4}-{a}^{2}xy=0$

Thomas Lynn

Beginner2021-12-18Added 28 answers

Step 1

Given

${x}^{4}+{y}^{4}-{a}^{2}xy=0$

Step 2

differentiating wrt "x"

$\frac{d}{dx}\left({x}^{4}\right)+\frac{d}{dx}\left({y}^{4}\right)-{a}^{2}\frac{d}{dx}\left(xy\right)=0$

$4{x}^{3}+4{y}^{3}\frac{dy}{dx}-{a}^{2}(x\frac{dy}{dx}+y\cdot 1)=0$

$4{x}^{3}+4{y}^{3}\frac{dy}{dx}-{a}^{2}x\frac{dy}{dx}-{a}^{2}y=0$

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

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

Given

Step 2

differentiating wrt "x"

John Koga

Beginner2021-12-19Added 33 answers

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:

$\frac{d}{dx}({x}^{4}+{y}^{4}-{a}^{2}xy)=\frac{d}{dx}\left(0\right)$

$\frac{d}{dx}\left({x}^{4}\right)+\frac{d}{dx}\left({y}^{4}\right)-{a}^{2}\frac{d}{dx}\left(xy\right)=0$

apply chain rule and product rule of differentiation:

$4{x}^{3}+4{y}^{3}\frac{dy}{dx}-{a}^{2}(x\frac{dy}{dx}+y\frac{dx}{dx})=0$

$4{x}^{3}+4{y}^{3}{y}^{\prime}-{a}^{2}(x{y}^{\prime}+y)=0$

$4{x}^{3}+4{y}^{3}{y}^{\prime}-{a}^{2}x{y}^{\prime}-{a}^{2}y=0$

$y}^{\prime}(4{y}^{3}-{a}^{2}x)={a}^{2}y-4{x}^{3$

$y}^{\prime}=\frac{{a}^{2}y-4{x}^{3}}{4{y}^{3}-{a}^{2}x$

Final answer:

$y}^{\prime}=\frac{{a}^{2}y-4{x}^{3}}{4{y}^{3}-{a}^{2}x$

In implicit differentiation all the differentiation rules are applicable.

Step 2

differentiate both sides wrt x:

apply chain rule and product rule of differentiation:

Final answer:

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