Question

How do you differentiate `2=ye^(x-y^3)+xe^(x-y)`?

Answer 1

`y'=(y*e^(x-y^3)+x*e^(x-y)+e^(x-y))/(3y^3*e^(x-y^3)-e^(x-y^3)+x*e^(x-y))`

Explanation:

Use Implicit differentiation

from the given `2=y*e^(x-y^3)+x*e^(x-y)`

`d/dx(2)=d/dx(y*e^(x-y^3)+x*e^(x-y))`

`0=y*e^(x-y^3)*(1-3y^2*y')+e^(x-y^3)*y'+x*e^(x-y)*(1-y')+e^(x-y)*1`

Solving for `y'` in terms of x and y then simplification

`y'=(y*e^(x-y^3)+x*e^(x-y)+e^(x-y))/(3y^3*e^(x-y^3)-e^(x-y^3)+x*e^(x-y))`

I hope the explanation is useful....God bless...