How do you implicitly differentiate #-1=y^2+(xy-e^y)/(x)#?
1 Answer
The key to answering implicit differentiation problems is to derive portions of the equation involving a
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
#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
#2y dy/dx + dy/dx - e^y x^-1 dy/dx = -e^y x^-2#
We can now isolate
#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
#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])#