**(a)**

| Applies $\frac{vu'-uv'}{v^2}$ to $y = \frac{\ln(x^2+1)}{x^2+1}$ with $u = \ln(x^2+1)$ and $v = x^2+1$ | M1 A1 |
| $\frac{dy}{dx} = \frac{(x^2+1) \times \frac{2x}{x^2+1} - 2x\ln(x^2+1)}{(x^2+1)^2}$ | M1 A1 |
| $\frac{dy}{dx} = \frac{2x - 2x\ln(x^2+1)}{(x^2+1)^2}$ | A1 |

(3 marks)

**(b)**

| Sets $2x - 2x\ln(x^2+1) = 0$ | M1 |
| $2x(1-\ln(x^2+1)) = 0 \Rightarrow x = \pm\sqrt{e-1}$ | M1,A1 |
| Sub $x = \pm\sqrt{e-1}, 0$ into $f(x) = \frac{\ln(x^2+1)}{x^2+1}$ | dM1 |
| Stationary points $\left(\sqrt{e-1}, \frac{1}{e}\right), \left(-\sqrt{e-1}, \frac{1}{e}\right), (0,0)$ | A1 B1 |

(6 marks)
(9 marks total)

**Guidance notes:**

**(a)**

- M1: Attempts the quotient or product rule to achieve an expression in the correct form
  
Using the quotient rule achieves an expression of the form: $\frac{dy}{dx} = \frac{(x^2+1) \times ... - 2x\ln(x^2+1)}{(x^2+1)^2}$

or the form: $\frac{dy}{dx} = ... - 2x\ln(x^2+1) \times ...$ where $... = A$ or $Ax$

or using the product rule achieves and an expression: $\frac{dy}{dx} = (x^2+1)^{-1} \times ... - 2x(x^2+1)^{-2}\ln(x^2+1)$

You may condone the omission of brackets ........especially on the denominator

- A1: A correct un-simplified expression for $\frac{dy}{dx}$
  
$\frac{dy}{dx} = \frac{(x^2+1) \times \frac{2x}{x^2+1} - 2x\ln(x^2+1)}{(x^2+1)^2}$ or $\frac{dy}{dx} = (x^2+1)^{-1} \times \frac{2x}{x^2+1} - 2x(x^2+1)^{-2}\ln(x^2+1)$

- A1: $\frac{dy}{dx} = \frac{2x - 2x\ln(x^2+1)}{(x^2+1)^2}$ or exact simplified equivalent such as $\frac{dy}{dx} = \frac{2x(1-\ln(x^2+1))}{(x^2+1)^2}$

**Condone:** $\frac{dy}{dx} = \frac{2x - \ln(x^2+1)2x}{(x^2+1)^2}$ which may be a little ambiguous. The lhs $\frac{dy}{dx} =$ does not need to be seen. You may assume from the demand in the question that is what they are finding.

**ISW** can be applied here.

**(b)**

- M1: Sets the numerator of their $\frac{dy}{dx}$, which must contain at least two terms, equal to 0
- M1: For solving an equation of the form $\ln(x^2+1)=k, k > 0$ to get at least one non-zero value of $x$. Accept decimal answers. $x = awrt \pm 1.31$. The equation must be legitimately obtained from a numerator = 0
- A1: Both $x = \pm\sqrt{e-1}$ scored from $\pm$ a correct numerator. Condone $x = \pm e^1 - 1$

**dM1:** Substitutes any of their non-zero solutions to $\frac{dy}{dx} = 0$ into $f(x) = \frac{\ln(x^2+1)}{x^2+1}$ to find at least one 'y' value. It is dependent upon both previous M's

- A1: Both $\left(\sqrt{e-1}, \frac{1}{e}\right), \left(-\sqrt{e-1}, \frac{1}{e}\right)$ oe or the equivalent with $x = ..., y = ...$. $\ln e$ must be simplified. Condone $\sqrt{e^1-1}, \frac{1}{e^1}$ but the $y$ coordinates must be simplified as shown.

**Condone:** $\left(\pm\sqrt{e-1}, \frac{1}{e}\right)$ but the $y$ coordinates must be simplified as shown. Condone extra solutions to these apart from $(0,0)$

- B1: $(0,0)$ or the equivalent $x = 0, y = 0$

**Notes:**

(1) A candidate can "recover" and score all marks in (b) when they have an incorrect denominator in part (a) or a numerator the wrong way around in (a)

(2) A candidate who differentiates $\ln(x^2+1) \rightarrow \frac{1}{x^2+1}$ will probably only score (a) 100 (b) 100000 (though they would have $k < 0$)

(3) A candidate who has $\frac{vu'+uv'}{v^2}$ cannot score anything more than (a) 000 (b) 100001 as they would have $k < 0$

(4) A candidate who attempts the product rule to get $\frac{dy}{dx} = (x^2+1)^{-1} \times \frac{1}{x^2+1}(x^2+1)^{-2}\ln(x^2+1) = \frac{1-\ln(x^2+1)}{(x^2+1)^2}$ can score (a) 000 (b) 110100 even though they may obtain the correct non-zero coordinates.

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