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

1 Answer
May 29, 2017

#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!