WAEC 2008 · Paper 2 · Q3

Worked solution (try it first)

Let $y = x^3 + 2x$ .
Increase $x$ by a small amount $\Delta x$ : $y + \Delta y = (x + \Delta x)^3 + 2(x + \Delta x)$ .
Expand: $y + \Delta y = x^3 + 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3 + 2x + 2\Delta x$ .
Subtract $y = x^3 + 2x$ : $\Delta y = 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3 + 2\Delta x$ .
Divide by $\Delta x$ : $\dfrac{\Delta y}{\Delta x} = 3x^2 + 3x\Delta x + (\Delta x)^2 + 2$ .
Let $\Delta x \to 0$ : the terms with $\Delta x$ vanish, so $\dfrac{dy}{dx} = 3x^2 + 2$ .