Nếu 2^x + 2^y = 2^x + y. Tính \[\frac{{dy}}{{dx}}\].

\(\ln {2.2^x} + \ln {2.2^y}\frac{{dy}}{{dx}} = \ln {2.2^{x + y}}\left( {1 + \frac{{dy}}{{dx}}} \right)\)

\( \Rightarrow {2^x} + {2^y}\frac{{dy}}{{dx}} = {2^{x + y}}\left( {1 + \frac{{dy}}{{dx}}} \right)\)

\( \Rightarrow {2^x} + {2^y}\frac{{dy}}{{dx}} = {2^{x + y}} + {2^{x + y}}\frac{{dy}}{{dx}}\)

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

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{{2^{x + y}} - {2^x}}}{{{2^y} - {2^{x + y}}}}\)

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

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