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in Derivatives by (15 points)

ex+ey=ex+y

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1 Answer

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Given,

ex + e= ex+y = exey

⇒ e−y + e−x = 1

Differentiating both sides with respect to x, we get

\(-e^{-y} \frac{dy}{dx} - e^{-x} = 0\)

⇒ \(\frac{dy}{dx} = \frac{e^{-x}}{-e^{-y}}\)

⇒ \(\frac{dy}{dx} = -e^{y-x}\)

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