If $2^x+2^y=2^{x+y}$, then $\displaystyle \frac {dy}{dx}$ has the value equal to

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

Differentiating both the sides

$\Rightarrow \displaystyle \dfrac { d }{ dx } \left( 2^{ x }+2^{ y } \right) =\displaystyle \dfrac { d }{ dx } \left(2^{ x+y }\right)$

$\Rightarrow { 2 }^{ x }\log { 2+{ 2 }^{ y }\log { 2\displaystyle \dfrac { dy }{ dx } ={ 2 }^{ x+y }\log { 2 } \left( 1+\displaystyle \dfrac { dy }{ dx } \right) } }$

$\Rightarrow \log { 2\left( { 2 }^{ x }+{ 2 }^{ y }\displaystyle \dfrac { dy }{ dx } \right) =\log { 2 } \left( { 2 }^{ x+y }+{ 2 }^{ x+y }\displaystyle \dfrac { dy }{ dx } \right) }$

$\Rightarrow \left( { 2 }^{ y }-{ 2 }^{ x+y } \right)\displaystyle \dfrac { dy }{ dx } ={ 2 }^{ x+y }-{ 2 }^{ x }$

$\Rightarrow \displaystyle \dfrac { dy }{ dx } =\displaystyle \dfrac { { 2 }^{ x }{ 2 }^{ y }-{ 2 }^{ x } }{ { 2 }^{ y }-{ 2 }^{ x }{ 2 }^{ y } } =\displaystyle \dfrac { { 2 }^{ x }\left( { 2 }^{ y }-1 \right) }{ { 2 }^{ y }\left( 1-{ 2 }^{ x } \right) }$

$=\displaystyle \dfrac { { 2 }^{ x }\left( 1-{ 2 }^{ y } \right) }{ { 2 }^{ y }\left( { 2 }^{ x }-1 \right) }$