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

1 Answer

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...