How do you find the derivatives of $y=ln(x+y)$ ?

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Answer 1 · 2017-05-29T03:09:22

$dy/dx = 1/(x + y - 1)$

Here's another method.

$y = ln(x+ y)$

$e^y = e^ln(x + y)$

$e^y = x + y$

The derivative of $e^a$ is $e^a$ . Therefore:

$e^y(dy/dx) = 1 + dy/dx$

$e^y(dy/dx) - dy/dx = 1$

$dy/dx(e^y - 1) = 1$

$dy/dx= 1/(e^y - 1)$

$dy/dx= 1/(e^ln(x+ y) - 1)$

$dy/dx = 1/(x + y - 1)$

Hopefully this helps!

How do you find the derivatives of y=ln(x+y)y=ln(x+y)?