# Partial differentiation ln(x+y)=x

• Dec 8th 2012, 06:50 PM
Bucklr
Partial differentiation ln(x+y)=x
Hello guys,
wondering if anyone can help me out with this problem.
find dy/dx:
ln(x+y)=x

why is this solution invalid:
ln(x+y)=x
ex=x+y
y=ex-x
(dy/dx)=ex-1

Many thanks.
• Dec 8th 2012, 07:11 PM
MarkFL
Re: Partial differentiation ln(x+y)=x
I believe you meant to title your topic "Implicit differentiation..."

Your result is valid, it is just in a different form than that which is obtained by implicit differentiation.

Given $\ln(x+y)=x$ then implicit differentiation yields:

$\frac{1+y'}{x+y}=1$

$y'=x+y-1$

Now, if we substitute for $y$, we then find:

$y'=x+e^x-x-1=e^x-1$