Question:

<b>Implicit Differentiation with Exponentials</b><br> Differentiate $2^x+2^y = 2^{x+y}$ w.r.t. $x$:<br> $2^x\ln 2 + 2^y\ln 2\cdot\dfrac{dy}{dx} = 2^{x+y}\ln 2\left(1+\dfrac{dy}{dx}\right)$<br> $2^x + 2^y\dfrac{dy}{dx} = 2^{x+y} + 2^{x+y}\dfrac{dy}{dx}$<br> $\dfrac{dy}{dx}(2^y-2^{x+y}) = 2^{x+y}-2^x = 2^x(2^y-1)$<br> $\dfrac{dy}{dx} = \dfrac{2^x(2^y-1)}{2^y(1-2^x)} = \dfrac{2^x-2^{x+y}}{2^{x+y}-2^y}$<br> All four options simplify to the same expression through algebraic manipulation — all are equivalent forms.<br> <b>Key concept:</b> One implicit differentiation result has many equivalent algebraic forms; all of ABCD are the same answer written differently.<br> <b>Trap:</b> Selecting only one form; check that all four are algebraically equivalent.