What is the implicit derivative of #1= x/y^3-e^(x-y)#?

1 Answer
Mar 7, 2018

#dy/dx = (y^-3 - e^(x-y))/(3xy^-4 - e^(x-y))#

Explanation:

We can use product rule and chain rule to fully expand this out:
#1 = x/y^3 - e^(x-y) #
Differentiating both sides,
#0 = d(x/y^3) - d(e^(x-y)) #
#0 = 1/y^3\ dx + x\ d(1/y^3) - e^xde^(-y) - e^(-y)de^x #
#0 = y^(-3) dx -3 xy^(-4)dy + e^x e^(-y)dy - e^(-y)e^x dx #
#0 = (y^-3 - e^(x-y))dx - (3xy^-4 - e^(x-y))dy #
#(3xy^-4 - e^(x-y))dy = (y^-3 - e^(x-y))dx #
#dy = ((y^-3 - e^(x-y)))/((3xy^-4 - e^(x-y))) dx #

#dy/dx = (y^-3 - e^(x-y))/(3xy^-4 - e^(x-y))#

If we want to make this a little prettier, we can multiply both sides by #y^4e^y#,

#dy/dx = (ye^y - y^4e^(x))/(3xe^y - y^4e^(x))#

but that's just my preference.