Question

How do you differentiate `12=ye^(x-y^3)+e^x-y`?

Answer 1

`(-e^x-ye^(x-y^3))/(e^(x-y^3)-3y^3e^(x-y^3)-1) = dy/dx`

Explanation:

Use implicit differentiation:

`d/dx12 = d/dx(ye^(x-y^3)+e^x-y)`

`0 = dy/dxe^(x-y^3) + ye^(x-y^3)(1-3y^2*dy/dx) + e^x-dy/dx`

`0 = dy/dxe^(x-y^3) + ye^(x-y^3) - 3y^3e^(x-y^3)dy/dx + e^x - dy/dx`

`-e^x-ye^(x-y^3) = dy/dx (e^(x-y^3)-3y^3e^(x-y^3)-1)`

`(-e^x-ye^(x-y^3))/(e^(x-y^3)-3y^3e^(x-y^3)-1) = dy/dx`