Question

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

Answer 1

`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)`

Answer 2

`(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/dx`s and non-`dy/dx`s
`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!