# How do you find #dy/dx# by implicit differentiation given #e^y=x^2+y#?

##### 1 Answer

Dec 16, 2016

Note that since

#(df)/(dx) = (df)/(dy)(dy)/(dx)# where

#(df)/(dy)# is the derivative of, say,#e^y# or#y# in your given function.

Thus:

#e^y (dy)/(dx) = 2x + 1(dy)/(dx)#

#=> (e^y - 1)(dy)/(dx) = 2x#

#=> color(blue)((dy)/(dx) = (2x)/(e^y - 1))#

If you know what