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Question

# If ${2}^{x}+{2}^{y}={2}^{x+y},$ then $\frac{dy}{dx}$ is equal to

A

$\frac{\left({2}^{x}+{2}^{y}\right)}{\left({2}^{x}–{2}^{y}\right)}$

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B

$\frac{\left({2}^{x}+{2}^{y}\right)}{\left(1+{2}^{x+y}\right)}$

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C

$\frac{{2}^{x-y}\left({2}^{y}-1\right)}{\left(1–{2}^{x}\right)}$

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D

None of these

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Solution

## The correct option is C $\frac{{2}^{x-y}\left({2}^{y}-1\right)}{\left(1–{2}^{x}\right)}$Explanation for the correct option:Step 1. Find the value of $\frac{dy}{dx}$:Given, ${2}^{x}+{2}^{y}={2}^{x+y}$$⇒$ ${2}^{x}+{2}^{y}={2}^{x}{2}^{y}$Step 2. Differentiate it with respect to $x$ ${2}^{x}\mathrm{log}2+{2}^{y}\mathrm{log}2\frac{dy}{dx}={2}^{x}{2}^{y}\mathrm{log}2\frac{dy}{dx}+{2}^{y}{2}^{x}\mathrm{log}2$$⇒$$\left(\frac{dy}{dx}\right)\left({2}^{y}\mathrm{log}2–{2}^{x}{2}^{y}\mathrm{log}2\right)={2}^{y}{2}^{x}\mathrm{log}2–{2}^{x}\mathrm{log}2$$⇒$ $\left(\frac{dy}{dx}\right){2}^{y}\mathrm{log}2\left(1–{2}^{x}\right)={2}^{x}\mathrm{log}2\left({2}^{y}-1\right)$$⇒$ $\begin{array}{rcl}\frac{dy}{dx}& =& \frac{{2}^{x}\mathrm{log}2\left({2}^{y}-1\right)}{{2}^{y}\mathrm{log}2\left(1–{2}^{x}\right)}\\ & =& \frac{{2}^{x}\left({2}^{y}-1\right)}{{2}^{y}\left(1–{2}^{x}\right)}\\ & =& \frac{{\mathbf{2}}^{\mathbf{x}\mathbf{-}\mathbf{y}}\mathbf{\left(}{\mathbf{2}}^{\mathbf{y}}\mathbf{}\mathbf{-}\mathbf{1}\mathbf{\right)}}{\mathbf{\left(}\mathbf{}\mathbf{1}\mathbf{}\mathbf{–}\mathbf{}{\mathbf{2}}^{\mathbf{x}}\mathbf{}\mathbf{\right)}}\end{array}$Hence, Option ‘C’ is Correct.

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