Q88 / 170 · Manipal (MET) 2015 · Mathematics
If $2^x+2^y=2^{x+y}$, then $\frac{d y}{d x}$ is equal to
- A$\frac{2^x+2^y}{2^x-2^y}$
- B$\frac{2^x+2^y}{1+2^{x+y}}$
- Correct answerC$2^{x-y}\left(\frac{2^y-1}{1-2^x}\right)$
- D$\frac{2^{x+y}-2^x}{2^y}$
Solution
We have,
$2^x+2^y=2^{x+y}$
On differentiating both sides w.r.t. $x$, we get
$2^x \log 2+2^y \log 2 \frac{d y}{d x}$
$=2^{x+y} \log 2\left(1+\frac{d y}{d x}\right)$
$\begin{aligned} & \Rightarrow \quad\left(2^y-2^{x+y}\right) \frac{d y}{d x}=2^{x+y}-2^x \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{2^x\left(2^y-1\right)}{2^y\left(1-2^x\right)}=2^{x-y}\left(\frac{2^y-1}{1-2^x}\right)\end{aligned}$
Hence, the correct answer is option: 'C'.