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

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Answer 1 · 2016-09-01T18:27:19

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

#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!