Question

If\[{2^{\text{x}}} + {2^{\text{y}}} = {2^{{\text{x + y}}}}\], then $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$is equal to
$
  {\text{A}}{\text{. }}\dfrac{{{{\text{2}}^{\text{x}}} + {2^{\text{y}}}}}{{{2^{\text{x}}} - {2^{\text{y}}}}} \\
  {\text{B}}{\text{. }}\dfrac{{{{\text{2}}^{\text{x}}} + {2^{\text{y}}}}}{{1 + {2^{{\text{x + y}}}}}} \\
  {\text{C}}{\text{. }}{{\text{2}}^{{\text{x - y}}}}\left( {\dfrac{{{{\text{2}}^{\text{y}}} - 1}}{{1 - {2^{\text{x}}}}}} \right) \\
  {\text{D}}{\text{. }}\dfrac{{{{\text{2}}^{{\text{x + y}}}} - {2^{\text{x}}}}}{{{2^{\text{y}}}}} \\
$

Answer 1

Hint: In order to find the value of$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$, we differentiate the given equation with respect to x and bring all $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$terms to the left hand side of the equation. $\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{a}}^{\text{x}}} = {{\text{a}}^{\text{x}}}{\text{loga}}} \right)$.

Complete step-by-step answer:

Given Data, \[{2^{\text{x}}} + {2^{\text{y}}} = {2^{{\text{(x + y)}}}}\]
Differentiating the given equation w.r.t x, we get
\[{2^{\text{x}}}{\text{log2}} + {2^{\text{y}}}{\text{log2}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = {2^{{\text{(x + y)}}}}{\text{log2}}\dfrac{{{\text{d}}}}{{{\text{dx}}}}\left( {{\text{x + y}}} \right)\] ___________________ $\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{a}}^{\text{y}}} = {{\text{a}}^{\text{y}}}{\text{loga}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(y)}}$
$ \Rightarrow {2^{\text{x}}}{\text{log2}} + {2^{\text{y}}}{\text{log2}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = {2^{{\text{(x + y)}}}}{\text{log2}}\left( {{\text{1 + }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)$
Bringing all $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$terms to the left hand side, we get

$
   \Rightarrow \left( {{{\text{2}}^{\text{y}}} - {{\text{2}}^{{\text{(x+y)}}}}} \right)\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = {{\text{2}}^{{\text{x + y}}}} - {{\text{2}}^{\text{x}}} \\
   \Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{{{\text{2}}^{{\text{(x + y)}}}} - {{\text{2}}^{\text{x}}}}}{{{{\text{2}}^{\text{y}}} - {{\text{2}}^{{\text{x + y}}}}}} = \dfrac{{{{\text{2}}^{\text{x}}}\left( {{{\text{2}}^{\text{y}}} - 1} \right)}}{{{{\text{2}}^{\text{y}}}\left( {{\text{1 - }}{{\text{2}}^{\text{x}}}} \right)}} = \dfrac{{{{\text{2}}^{{\text{(x - y)}}}}\left( {{{\text{2}}^{\text{y}}} - 1} \right)}}{{{\text{1 - }}{{\text{2}}^{\text{x}}}}} \\
$
Hence Option C is the correct answer.

Note: In order to solve questions of this type the key is to differentiate the given equation and bring all the required terms to one side of the equation to easily find its answer. Basic knowledge of differentiations of common terms is necessary in solving equations like these.