How do you implicitly differentiate #6=lny/(e^x-e^y)-x#?

1 Answer
Sep 1, 2016

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

Explanation:

#d/dx(6) = d/dx((lny)/(e^x - e^y) - x)#

#d/dx(6) = d/dx(lny/(e^x - e^y)) + d/dx(-x)#

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

Isolate #dy/dx#:

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

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

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

Hopefully this helps!