Q.2 (A) (ii) If e^x+e^y=e^((x+y)) then show that dy/dx=-e^(y-x) .

Answer: Shown below; dy/dx=-e^(y-x) . e^x+e^y=e^(x+y) ...(1) Differentiating both sides w.r.t. x: e^x+e^ydy/dx=e^(x+y)(1+dy/dx) (e^y-e^(x+y))dy/dx=e^(x+y)-e^x Using (1), e^(x+y)-e^x=e^y and e^y-e^(x+y)=-e^x . Therefore -e^xdy/dx=e^y⇒ dy/dx=-e^y/e^x=-e^(y-x) .

If \(e^x+e^y=e^{(x+y)}\) then show that \(\frac{dy}{dx}=-e^{y-x}\).

Q.2 (A) (ii) If \(e^x+e^y=e^{(x+y)}\) then show that \(\frac{dy}{dx}=-e^{y-x}\).

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Answer: Shown below; \(\frac{dy}{dx}=-e^{y-x}\).

\(e^x+e^y=e^{x+y}\) ...(1)

Differentiating both sides w.r.t. x:

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

\(\left(e^y-e^{x+y}\right)\frac{dy}{dx}=e^{x+y}-e^x\)

Using (1), \(e^{x+y}-e^x=e^y\) and \(e^y-e^{x+y}=-e^x\). Therefore

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

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If \(e^x+e^y=e^{(x+y)}\) then show that \(\frac{dy}{dx}=-e^{y-x}\).

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