WAEC 2014 · Paper 2 · Q4

(a)

1. Differentiate each term with the product rule: $\frac{d}{dx}(xy^2) = y^2 + 2xy\frac{dy}{dx}$.
2. $\frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx}$ and $\frac{d}{dx}(4xy) = 4y + 4x\frac{dy}{dx}$.
3. So $y^2 + 2xy\frac{dy}{dx} + 2xy + x^2\frac{dy}{dx} = 4y + 4x\frac{dy}{dx}$.
4. Collect the $\frac{dy}{dx}$ terms: $(2xy + x^2 - 4x)\frac{dy}{dx} = 4y - y^2 - 2xy$.
5. Put $x = 1$, $y = 3$: $(6 + 1 - 4)\frac{dy}{dx} = 12 - 9 - 6$, so $3\frac{dy}{dx} = -3$.
6. The gradient at $(1, 3)$ is $-1$.