Ask Me Help Desk - Implicit differentation

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implicit differentation

How do I differentate ln(x + y) = y?

I start with
[ln(x + y)]' = [y]'
[ln(x)]' + [ln(y)]' = y'

I know [ln(x)]' = 1/x
But I'm not sure what [ln(y)]' should be.
I thought it was (1/y)y' (or y'/y), but I can't
get to the answer, shown as:
1
y' = -------
x+y-1

We have to use a little chain rule action.

$ln(x+y)=y$

$\underbrace{\frac{1}{x+y}\cdot\left(1+\frac{dy}{dx }\right)}_{\text{chain rule}} \;\ = \;\ \frac{dy}{dx}$

$\frac{1}{x+y}=\frac{dy}{dx}-\frac{1}{x+y}\cdot\frac{dy}{dx}$

$\frac{1}{x+y}=\frac{dy}{dx}\left(1-\frac{1}{x+y}\right)$

$\frac{dy}{dx}=\frac{1}{x+y-1}$

Thank you vary much! I rated your answer "very clear and very quick response - outstanding!"

Haven't messed with math for a while, Does X = 0? 0+Y=Y?

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