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.

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- Dec 8th 2012, 06:50 PMBucklrPartial 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

e^{x}=x+y

y=e^{x}-x

(dy/dx)=e^{x}-1

Many thanks. - Dec 8th 2012, 07:11 PMMarkFLRe: 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 $\displaystyle \ln(x+y)=x$ 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:

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