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If ex+ey=ex+ye^x + e^y = e^{x+y}, then dydx\frac{dy}{dx} equals

Solution

✅ Correct Option: 1

Given the implicit equation:

ex+ey=ex+ye^x + e^y = e^{x+y}

  1. Differentiate both sides with respect to xx:

ddx(ex)+ddx(ey)=ddx(ex+y)\frac{d}{dx}(e^x) + \frac{d}{dx}(e^y) = \frac{d}{dx}(e^{x+y})

ex+eydydx=ex+y(1+dydx)e^x + e^y\frac{dy}{dx} = e^{x+y}\left(1 + \frac{dy}{dx}\right)

  1. Expand the right-hand side:

ex+eydydx=ex+y+ex+ydydxe^x + e^y\frac{dy}{dx} = e^{x+y} + e^{x+y}\frac{dy}{dx}

  1. Substitute ex+y=ex+eye^{x+y} = e^x + e^y back into the right side to simplify:

ex+eydydx=(ex+ey)+(ex+ey)dydxe^x + e^y\frac{dy}{dx} = (e^x + e^y) + (e^x + e^y)\frac{dy}{dx}

ex+eydydx=ex+ey+exdydx+eydydxe^x + e^y\frac{dy}{dx} = e^x + e^y + e^x\frac{dy}{dx} + e^y\frac{dy}{dx}

  1. Cancel out identical terms on both sides:

Subtract exe^x and eydydxe^y\frac{dy}{dx} from both sides:

0=ey+exdydx0 = e^y + e^x\frac{dy}{dx}

  1. Isolate dydx\frac{dy}{dx}:

exdydx=−eye^x\frac{dy}{dx} = -e^y

dydx=−eyex=−ey−x\frac{dy}{dx} = -\frac{e^y}{e^x} = -e^{y-x}

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