Question

What is the implicit derivative of `1= x/y^3-e^(x-y)`?

Answer 1

`dy/dx = (y^-3 - e^(x-y))/(3xy^-4 - e^(x-y))`

Explanation:

We can use product rule and chain rule to fully expand this out:
`1 = x/y^3 - e^(x-y)`
Differentiating both sides,
`0 = d(x/y^3) - d(e^(x-y))`
`0 = 1/y^3\ dx + x\ d(1/y^3) - e^xde^(-y) - e^(-y)de^x`
`0 = y^(-3) dx -3 xy^(-4)dy + e^x e^(-y)dy - e^(-y)e^x dx`
`0 = (y^-3 - e^(x-y))dx - (3xy^-4 - e^(x-y))dy`
`(3xy^-4 - e^(x-y))dy = (y^-3 - e^(x-y))dx`
`dy = ((y^-3 - e^(x-y)))/((3xy^-4 - e^(x-y))) dx`

`dy/dx = (y^-3 - e^(x-y))/(3xy^-4 - e^(x-y))`

If we want to make this a little prettier, we can multiply both sides by `y^4e^y`,

`dy/dx = (ye^y - y^4e^(x))/(3xe^y - y^4e^(x))`

but that's just my preference.