Question

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

• A. $\frac{(2^x+2^y)}{(2^x–2^y)}$
• B. $\frac{(2^x+2^y)}{(1+2^{x+y})}$
• C. $\frac{2^{x-y}(2^y-1)}{(1–2^x)}$
• D. None of these

Answer 1

The correct option is C

$\frac{2^{x-y}(2^y-1)}{(1–2^x)}$

Explanation for the correct option:

Step 1. Find the value of $\frac{dy}{dx}$:

Given, $2^x+2^y=2^{x+y}$

$\Rightarrow$ $2^x+2^y=2^x{2}^y$

Step 2. Differentiate it with respect to $x$

$2^x\log 2+2^y\log 2\frac{dy}{dx}=2^x{2}^y\log 2\frac{dy}{dx}+2^y{2}^x\log 2$

$\Rightarrow$$\left(\frac{dy}{dx}\right)(2^y\log 2–2^x{2}^y\log 2)=2^y{2}^x\log 2–2^x\log 2$

$\Rightarrow$ $\left(\frac{dy}{dx}\right)2^y\log 2(1–2^x)=2^x\log 2(2^y-1)$

$\Rightarrow$ $\begin{array}{rcl}\frac{dy}{dx}& =& \frac{2^x\log 2(2^y-1)}{2^y\log 2(1–2^x)}\\ & =& \frac{2^x(2^y-1)}{2^y(1–2^x)}\\ & =& \frac{{\mathbf{2}}^{\mathbf{x}\mathbf{-}\mathbf{y}}\mathbf{(}{\mathbf{2}}^{\mathbf{y}}\mathbf{}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{(}\mathbf{}\mathbf{1}\mathbf{}\mathbf{–}\mathbf{}{\mathbf{2}}^{\mathbf{x}}\mathbf{}\mathbf{)}}\end{array}$

Hence, Option ‘C’ is Correct.