How do you implicitly differentiate #-1=y^2+(xy-e^y)/(x)#?

1 Answer
Dec 24, 2015

The key to answering implicit differentiation problems is to derive portions of the equation involving a #y# the same as you would the portions with #x#, remembering to multiply by a #dy/dx#, which you will later isolate to solve.

Explanation:

Rewriting the problem,

#-1 = y^2 + (xy)/x - (e^y)/x#
#-1 = y^2 + y - (e^y)/x#

We now derive both sides of the equation, remembering to include a #dy/dx# when deriving any functions of #y#:

#d/dx( -1 ) = d/dx( y^2 + y - e^y x^-1 )#
#0 = 2y dy/dx + dy/dx - (e^y dy/dx x^-1 + (-1)e^y x^-2)#
#0 = 2y dy/dx + dy/dx - e^y x^-1 dy/dx + e^y x^-2#

We now move all terms with #dy/dx# to one side of the equations in order to isolate the term:

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

We can now isolate #dy/dx#:

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

The #x# terms with negative powers can now be rewritten to simplify:

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

Giving us the simplified answer:

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