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
e^{x}=x+y
y=e^{x}-x
(dy/dx)=e^{x}-1
Many thanks.
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 $\displaystyle \ln(x+y)=x$ then implicit differentiation yields:
$\displaystyle \frac{1+y'}{x+y}=1$
$\displaystyle y'=x+y-1$
Now, if we substitute for $\displaystyle y$, we then find:
$\displaystyle y'=x+e^x-x-1=e^x-1$