How do you find #(dy)/(dx)# given #e^y=e^(3x)+5#?

1 Answer
Nov 2, 2016

# dy/dx = (3e^(3x))/(e^(3x)+5) #

Explanation:

To differentiate implicitly we use an implicit application of the chain rule. We can't differentiate a function of #y# wrt #x# but we can differentiate it wrt #y#.

So, # e^y = e^(3x)+5 #

Differentiate wrt #x#:
So, # d/dxe^y = 3e^(3x) #

And, with practice you should be able to skip this line:
# :. dy/dxd/dye^y = 3e^(3x) # (by the chain rule)

# :. dy/dxe^y = 3e^(3x) #

# :. dy/dx = (3e^(3x))/e^y #

# :. dy/dx = (3e^(3x))/(e^(3x)+5) #