Find $\dfrac{dy}{dx}$ if $xy-4=0$.

Solution (a) To differentiate implicitly, we assume $y$ is a differentiable function of $x$ and differentiate both sides with respect to $x$. $$\begin{eqnarray*} \begin{array}{rl@{\qquad}l} \dfrac{d}{dx}(xy-4) &=\dfrac{d}{dx}0 & {\color{#0066A7}{\hbox{Differentiate both sides with respect to}\ {x}.}}\\ \dfrac{d}{dx}(xy) -\dfrac{d}{dx}4 &= 0 & {\color{#0066A7}{\hbox{Use the Difference Rule.}}}\\ x\cdot \dfrac{d}{dx}y+\left(\dfrac{d}{dx}x\right) y-0 &= 0 & {\color{#0066A7}{\hbox{Use the Product Rule.}}}\\ x\dfrac{dy}{dx}+y & =0 & {\color{#0066A7}{\hbox{Simplify}.}} \\ \dfrac{dy}{dx} & = -\dfrac{y}{x} & {\color{#0066A7}{\hbox{Solve for}\, \dfrac{dy}{dx}.}} & (1) \end{array} \end{eqnarray*}$$

(b) Solve $xy-4=0$ for $y$, obtaining $y=\dfrac{4}{x}=4x^{-1}$. Then $$\begin{equation*} \dfrac{dy}{dx}=\dfrac{d}{dx}(4x^{-1}) =-4x^{-2}=-\dfrac{4}{x^{2}}\tag{2} \end{equation*}$$

(c) At first glance, the results in (1) and (2) appear to be different. However, since $xy-4=0,$ or equivalently, $y=\dfrac{4}{x},$ the result from (1) becomes $$\begin{eqnarray*} && \dfrac{dy}{dx}\underset{\underset{\color{#0066A7}{(1)}}{\color{#0066A7}{\uparrow}}}{=}-\dfrac{y}{x}\underset{\underset{\color{#0066A7}{y=\tfrac{4}{x}}}{\color{#0066A7}{\uparrow}}}{=} -{\dfrac{\dfrac{4}{x}}{x}}=-\dfrac{4}{x^{2}} \end{eqnarray*}$$ which is the same as (2).