Question:

<b>Implicit Differentiation</b><br>$x = y\ln(xy)$. Differentiate both sides w.r.t. $y$:<br>$\frac{dx}{dy} = \ln(xy) + y \cdot \frac{1}{xy}\left(x + y\frac{dx}{dy}\right)$<br>$= \ln(xy) + \frac{1}{x}\left(x + y\frac{dx}{dy}\right)$<br>$= \ln(xy) + 1 + \frac{y}{x}\frac{dx}{dy}$<br>Since $x = y\ln(xy)$, we have $\ln(xy) = x/y$. Substituting:<br>$\frac{dx}{dy} = \frac{x}{y} + 1 + \frac{y}{x}\frac{dx}{dy}$<br>$\frac{dx}{dy}\left(1 - \frac{y}{x}\right) = \frac{x+y}{y}$<br>$\frac{dx}{dy} = \frac{x+y}{y} \cdot \frac{x}{x-y} = \frac{x(x+y)}{y(x-y)}$<br><b>Key concept:</b> Implicit differentiation + substituting the original equation.<br><b>Trap:</b> Forgetting that $\ln(xy) = x/y$ from the original equation.