WAEC 2013 · Paper 2 · Q2

(a)
Calculate the gradient of the curve $x^3 + y^3 - 2xy = 11$ at the point $(2, -1)$ .

Worked solution (try it first)

Differentiate each term with respect to $x$ : $3x^2 + 3y^2\dfrac{dy}{dx} - 2\left(y + x\dfrac{dy}{dx}\right) = 0$ .
Collect the $\dfrac{dy}{dx}$ terms: $(3y^2 - 2x)\dfrac{dy}{dx} = 2y - 3x^2$ .
So $\dfrac{dy}{dx} = \dfrac{2y - 3x^2}{3y^2 - 2x}$ .
At $(2, -1)$ : $\dfrac{dy}{dx} = \dfrac{-2 - 12}{3 - 4}$
$= \dfrac{-14}{-1}$
$= 14$ .