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