WAEC 2012 · Paper 2 · Q6

(a)
Find, from first principles, the derivative of $\left(3x - \dfrac{1}{2x}\right)$ with respect to $x$ .

Worked solution (try it first)

$f(x + h) = 3(x + h) - \dfrac{1}{2(x + h)}$ .
Take away $f(x)$ : $f(x + h) - f(x) = 3h - \left[\dfrac{1}{2(x + h)} - \dfrac{1}{2x}\right]$ .
Put the fractions over $2x(x + h)$ : $\dfrac{x - (x + h)}{2x(x + h)} = \dfrac{-h}{2x(x + h)}$ .
So $f(x + h) - f(x) = 3h + \dfrac{h}{2x(x + h)}$ .
Divide by $h$ : $3 + \dfrac{1}{2x(x + h)}$ .
Let $h \to 0$ : $f'(x) = 3 + \dfrac{1}{2x^2}$ .