Question

If it is given that

${e}^{x} + {e}^{y} = {e}^{x + y} \:$

$prove \: that \: \frac{dy}{dx} = - {e}^{y - x}$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: {e}^{x} + {e}^{y} = {e}^{x + y}$

can be rewritten as

$\rm \: {e}^{x} + {e}^{y} = {e}^{x} \: {e}^{y}$

can be further rewritten as

$\rm \: \dfrac{{e}^{x} + {e}^{y}}{{e}^{x} \: {e}^{y}} = 1$

$\rm \: \dfrac{{e}^{x}}{{e}^{x} \: {e}^{y}} + \dfrac{{e}^{y}}{{e}^{x} \: {e}^{y}} = 1$

$\rm \: \dfrac{1}{{e}^{y}} + \dfrac{1}{{e}^{x}} = 1$

$\rm \: {e}^{ - y} + {e}^{ - x} = 1$

On differentiating both sides w. r. t. x, we get

$\rm \:\dfrac{d}{dx}( {e}^{ - y} + {e}^{ - x}) = \dfrac{d}{dx}1$

We know,

$\boxed{\sf{ \:\dfrac{d}{dx}{e}^{x} = {e}^{x} \: }}$

and

$\boxed{\sf{ \:\dfrac{d}{dx}k \: = \: 0 \: }}$

So, using these results, we get

$\rm \: {e}^{ - y}\dfrac{d}{dx}( - y) + {e}^{ - x}\dfrac{d}{dx}( - x) = 0$

$\rm \: - {e}^{ - y}\dfrac{dy}{dx} - {e}^{ - x} = 0$

$\rm \: {e}^{ - y}\dfrac{dy}{dx} = - {e}^{ - x}$

$\rm \: \dfrac{dy}{dx} \: = \: - \: \dfrac{{e}^{ - x}}{{e}^{ - y}}$

$\rm\implies \:\dfrac{dy}{dx} \: = \: - \: {e}^{y - x}$

Hence, Proved

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Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Refer the given attachment