Differentiate x^4+y^4=8xy?

2 Answers
Sep 15, 2017

dy/dx = (2y-x^3)/(y^3-2x)

Explanation:

x^4+y^4=8xy

Apply implicit differentiation

4x^3+4y^3dy/dx = 8(xdy/dx+y*1) [Power rule and Product rule]

4(x^3+y^3dy/dx) = 8(xdy/dx+y)

x^3+y^3dy/dx = 2(xdy/dx+y)

(y^3-2x)dy/dx = (2y-x^3)

dy/dx = (2y-x^3)/(y^3-2x)

Sep 15, 2017

(x^3-2y)/(2x-y^3)

Explanation:

Using the chain rule on the left side and product rule on the right... we get...
4x^3dx + 4y^3dy=8(dx*y+dy*x)
Divide both sides by 4
x^3dx+y^3dy=2ydx+2xdy
Divide both sides by dx
x^3+y^3dy/dx=2y+2xdy/dx
Isolate dy/dxs and non-dy/dxs
dy/dx(y^3-2x)=2y-x^3
Divide both sides by y^3-2x to isolate just the dy/dx.
dy/dx=(2y-x^3)/(y^3-2x)
Multiply numerator and denominator by -1.
dy/dx = (x^3-2y)/(2x-y^3)
And you're done!