WAEC 2014 · Paper 2 · Q4

(a)
Find the equation of the tangent to the curve $y = \dfrac{1}{x + 1}$ when $x = 1$ .
Show the answer
$x + 4y - 3 = 0$

Worked solution (try it first)

At $x = 1$ : $y = \dfrac12$ , so the point is $\left(1, \frac12\right)$ .
Write $y = (x + 1)^{-1}$ .
By the chain rule, $\dfrac{dy}{dx} = -(x + 1)^{-2}$
$= -\dfrac{1}{(x + 1)^2}$ .
At $x = 1$ the gradient is $-\dfrac14$ .
The tangent: $y - \frac12 = -\frac14(x - 1)$ .
Multiply by 4: $4y - 2 = -x + 1$ .
Rearrange: $x + 4y - 3 = 0$ .

Report a problem with this question