How do you find (dy)/(dx) given lny=xe^x?

2 Answers
Jul 25, 2016

y'=e^(x(e^x+1) ) (x+1)

Explanation:

lny=xe^x

using the product rule on the RHS

1/y \ y'=e^x + x e^x

the rest is algebra

y'=ye^x( 1 + x )

y'=e^(x e^x)e^x( 1 + x )

y'=e^(x(e^x+1) ) (x+1)

Jul 25, 2016

dy/dx=(x+1) e^(xe^x +x)

Explanation:

ln y = xe^x

By implicit differentiation:

1/y dy/dx = xe^x + e^x (Standard differential and Product rule)

dy/dx = (x+1) e^x * y

But since ln y = xe^x -> y = e^(xe^x)

Therefore dy/dx = (x+1) e^x * e^(xe^x)

dy/dx = (x+1) e^(xe^x+x)