WAEC 2018 · Paper 2 · Q11

(a)
Find, from first principles, the derivative of $5x - \dfrac{6}{x^2}$ with respect to $x$ .
(b)
Find $\displaystyle\int x\sqrt{1 - x}\,dx$ .

Worked solution (try it first)

(a)

$f(x) = 5x - 6x^{-2}$ , so $f(x + h) - f(x) = 5h - 6\left[\dfrac{1}{(x + h)^2} - \dfrac{1}{x^2}\right]$ .
Over one denominator: $\dfrac{1}{(x + h)^2} - \dfrac{1}{x^2} = \dfrac{x^2 - (x + h)^2}{x^2(x + h)^2}$
$= \dfrac{-h(2x + h)}{x^2(x + h)^2}$ .
So $f(x + h) - f(x) = 5h + \dfrac{6h(2x + h)}{x^2(x + h)^2}$ .
Divide by $h$ : $5 + \dfrac{6(2x + h)}{x^2(x + h)^2}$ .
Let $h \to 0$ : $f'(x) = 5 + \dfrac{12x}{x^4}$
$= 5 + \dfrac{12}{x^3}$ .

(b)

Let $u = 1 - x$ .
Then $x = 1 - u$ and $dx = -du$ .
$\displaystyle\int x\sqrt{1 - x}\,dx = -\int (1 - u)u^{\frac12}\,du$
$= -\int \left(u^{\frac12} - u^{\frac32}\right)du$ .
Integrate: $-\left(\frac23u^{\frac32} - \frac25u^{\frac52}\right) + c = \frac25u^{\frac52} - \frac23u^{\frac32} + c$ .
Put $u$ back: $\frac25(1 - x)^{\frac52} - \frac23(1 - x)^{\frac32} + c$ .

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