Question

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

• A. $2^{y-x}$
• B. $-2^{y-x}$
• C. $2^{x-y}$
• D. $\frac{2^y-1}{2^x-1}$

Answer 1

Answer: (B)

$2^x+2^y=2^{(x+y)}\dots ..(1)$
Differentiating both sides w.r.t. $x$, we get
$2^x\ln 2+2^y{y}^{\prime }\ln 2=2^{(x+y)}\ln 2\left(1+y^{\prime }\right)$
$\Rightarrow 2^x+2^y{y}^{\prime }=2^{(x+y)}\left(1+y^{\prime }\right)$
$\Rightarrow 2^x+2^y\cdot y^{\prime }=2^{(x+y)}+2^{(x+y)}\cdot y^{\prime }$
$\Rightarrow 2^x-2^{(x+y)}=y^{\prime }\left(2^{(x+y)}-2^y\right)$
$\Rightarrow 2^x-2^x-2^y=y^{\prime }\left(2^x+2^y-2^y\right)[$ From eqn. $(1)]$
$\Rightarrow -2^y=y^{\prime }{2}^x$
$\Rightarrow y^{\prime }=dy/dx=-2^{y-x}$