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Find dy/dx of ln(xy) = e^(x+y)
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Take derivative both sides
solve for dy/dx,
y'/(xy) = e^(x+y)*(1-y') ? Is this correct
(1+y')*****
yes
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Apparently I'm making a dumb error solving for y' then :(
\(\dfrac{y'}{xy}=e^{x+y}(1+y')=e^{x+y}+e^{x+y}y'\)
\(y'=xy(e^{x+y}+e^{x+y}y')= xye^{x+y}+xye^{x+y}y'\)
Move y' to the same side \(y'-xye^{x+y}y'=xye^{x+y}\) factor y'
@tjb69812
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\(y' (1-xye^{x+y})=xye^{x+y}\)
isolate y'
\(\dfrac{dy}{dx}= \dfrac{xye^{x+y}}{1-xye^{x+y}}\)
ok?
Yes! Thanks a ton
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yw
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