Question

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

Answer 1

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

Explanation:

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!