WAEC 2019 · Paper 2 · Q4

Differentiate, with respect to $x$ , $y = 3x^2 + 4x - 1$ , from first principles.

The curve and its gradient function. Where does the gradient function cross zero, and what is the curve doing there?

Worked solution (try it first)

$y + \delta y = 3(x + \delta x)^2 + 4(x + \delta x) - 1$
$= 3x^2 + 6x\,\delta x + 3(\delta x)^2 + 4x + 4\,\delta x - 1$ .
Take away $y = 3x^2 + 4x - 1$ : $\delta y = 6x\,\delta x + 3(\delta x)^2 + 4\,\delta x$ .
Divide by $\delta x$ : $\dfrac{\delta y}{\delta x} = 6x + 3\,\delta x + 4$ .
Let $\delta x \to 0$ : $\dfrac{dy}{dx} = 6x + 4$ .