How do you implicitly differentiate $6 = y ln y - e^{x} - e^{y}$ $6=ylny-e^x-e^y$ ?

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Answer 1 · 2016-01-08T20:17:19

$\frac{d y}{d x} = \frac{e^{x}}{ln y + 1 - e^{y}}$ $dy/dx=e^x/(lny+1-e^y)$

Find the derivative of each part.

$\frac{d}{d x} (6) = 0$ $d/dx(6)=0$

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

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

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 $y'$ .

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

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

How do you implicitly differentiate 6=ylny-e^x-e^y6=ylny−ex−ey6=ylny-e^x-e^y?