How do you implicitly differentiate #6=ylny-e^x-e^y#?

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Answer 1 · 2016-01-08T20:17:19

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

Find the derivative of each part.

#d/dx(6)=0#

#d/dx(ylny)=y'lny+y(1/y)y'=y'lny+y'#

#d/dx(e^x)=e^x#

Chain rule:

#d/dx(e^y)=y'e^y#

Thus, the derivative of the entire implicit equation is

#0=y'lny+y'-e^x-y'e^y#

Solve for #y'#.

#e^x=y'(lny+1-e^y)#

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