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

Question

2705 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-25T09:33:23

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

#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)#

Answer 2 · 2016-07-25T09:46:41

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

#ln y = xe^x#

#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)#