The curve C has equation $y = \frac{\ln(x^2 + 1)}{x^2 + 1}, \quad x \in \mathbb{R}$ (a) Find $\frac{dy}{dx}$ as a single fraction, simplifying your answer - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 5

To find the stationary points, we set $\frac{dy}{dx} = 0$:

$2x (1 - \ln(x^2 + 1)) = 0$

This gives two possible cases:

1. $2x = 0 \Rightarrow x = 0$
2. $1 - \ln(x^2 + 1) = 0 \Rightarrow \ln(x^2 + 1) = 1 \Rightarrow x^2 + 1 = e \Rightarrow x^2 = e - 1 \Rightarrow x = \pm \sqrt{e - 1}$

Now we find the corresponding y-values:

• For $x = 0$:
  $y(0) = \frac{\ln(0^2 + 1)}{0^2 + 1} = \frac{\ln(1)}{1} = 0$
• For $x = \sqrt{e - 1}$:
  $y = \frac{\ln(e)}{e} = \frac{1}{e}$
• For $x = -\sqrt{e - 1}$:
  $y = \frac{\ln(e)}{e} = \frac{1}{e}$

Thus, the exact coordinates of the stationary points are:

• $(0, 0)$
• $\left( \sqrt{e - 1}, \frac{1}{e} \right)$
• $\left( -\sqrt{e - 1}, \frac{1}{e} \right)$.