Question 765977
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Just another application of the sum rule and the chain rule.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\left(e^x\ +\ e^y\ =\ e^{x\,+\,y}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^x\ +\ e^y\,\cdot\,\frac{dy}{dx}\ =\ e^{x\,+\,y}\,\cdot\,\frac{d\left(x\,+\,y\right)}{dx}]


Then second application of the sum and chain rules:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^x\ +\ e^y\,\cdot\,\frac{dy}{dx}\ =\ e^{x\,+\,y}\,\cdot\,\left(1\ +\ \frac{dy}{dx}\right)]


And the rest is just a little algebra:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^y\,\cdot\,\frac{dy}{dx}\ -\ e^{x\,+\,y}\,\cdot\,\frac{dy}{dx}\ =\ e^{x\,+\,y}\ -\ e^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\left(e^y\ -\ e^{x\,+\,y}\right)\ =\ e^{x\,+\,y}\ -\ e^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \frac{e^{x\,+\,y}\ -\ e^x}{e^y\ -\ e^{x\,+\,y}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \frac{e^x\left(e^y\ -\ 1\right)}{e^y\left(1\ -\ e^x\right)}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \frac{e^x\left(1\ -\ e^{-y}\right)}{1\ -\ e^x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \frac{e^x\left(e^{-y}\ -\ 1\right)}{e^x\ -\ 1}]


Either of the last two forms are acceptable for an answer.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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