WAEC 2020 · Paper 1 · Q9

Differentiate $\dfrac{x}{x + 1}$ with respect to $x$ .

Worked solution (try it first)

Use the quotient rule with $u = x$ and $v = x + 1$ : $\dfrac{dy}{dx} = \dfrac{v\,u' - u\,v'}{v^2}$ .
Here $u' = 1$ and $v' = 1$ , so the top is $(x + 1)(1) - x(1) = 1$ .
So $\dfrac{dy}{dx} = \dfrac{1}{(x + 1)^2}$ , option D.