Q.If $x^2 + xy + y^2 = 7$, find $\dfrac{dy}{dx}$, and its value at the point $(1, 2)$.

The equation is not solved for $y$ and has a mixed $xy$-term, so differentiate implicitly (product rule on $xy$).

The relation mixes $x$ and $y$ and includes the product term $xy$, so use implicit differentiation (§6), treating $y$ as a function of $x$.

Differentiate term by term.

• $\dfrac{d}{dx}(x^2) = 2x$.
• $\dfrac{d}{dx}(xy) = x\dfrac{dy}{dx} + y$ (product rule — $x$ times $\tfrac{dy}{dx}$ plus $y$ times $1$).
• $\dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}$ (chain rule).
• $\dfrac{d}{dx}(7) = 0$.

So the differentiated equation is

Collect $\dfrac{dy}{dx}$ terms. Group them on one side:

Solve. $\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$.

Evaluate at $(1, 2)$. First confirm the point lies on the curve: $1 + 2 + 4 = 7$ ✓. Then $\dfrac{dy}{dx} = -\dfrac{2(1)+2}{1+2(2)} = -\dfrac{4}{5}$. …

The two typical errors are: differentiating $xy$ as just $y'$ or just $y$ instead of the full product-rule result $x\,y' + y$; and differentiating $y^2$ as $2y$ (forgetting the $\tfrac{dy}{dx}$). Also verify the give …