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)

$2^{x}+2^{y}=2^{(x+y)} \ldots . .(1)$
Differentiating both sides w.r.t. $x$, we get
$2^{x} \ln 2+2^{y} y' \ln 2=2^{(x+y)} \ln 2\left(1+y'\right)$
$\Rightarrow 2^{x}+2^{y} y'=2^{(x+y)}\left(1+y'\right)$
$\Rightarrow 2^{x}+2^{y} \cdot y'=2^{(x+y)}+2^{(x+y)} \cdot y'$
$\Rightarrow 2^{x}-2^{(x+y)}=y'\left(2^{(x+y)}-2^{y}\right)$
$\Rightarrow 2^{x}-2^{x}-2^{y}=y'\left(2^{x}+2^{y}-2^{y}\right)[$ From eqn. $(1)]$
$\Rightarrow -2^{y}=y'2^{x}$
$\Rightarrow y'=d y / d x=-2^{y-x}$