## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x^2\tan 2x$ | M1 | Product rule: $u \times$ their $v' + v \times$ their $u'$ attempted |
| | M1 | $\frac{d}{du}(\tan u) = \sec^2 u$ soi; M0 if $\frac{d}{dx}(\tan 2x) = (2)\sec^2 x$ |
| $\Rightarrow \frac{dy}{dx} = 2x^2\sec^2 2x + 2x\tan 2x$ | A1cao | or $2x^2/\cos^2 2x + 2x\tan 2x$ isw |
| **OR** $y = x^2\dfrac{\sin 2x}{\cos 2x}$ | | see additional notes for complete solution |
| $\dfrac{dy}{dx} = x^2\dfrac{\cos 2x \cdot 2\cos 2x - \sin 2x(-2\sin 2x)}{\cos^2 2x} + 2x\dfrac{\sin 2x}{\cos 2x}$ | M1 | Product rule: $u \times$ their $v' + v \times$ their $u'$ attempted |
| | A1 | Correct expression |
| $= \ldots = 2x^2\sec^2 2x + 2x\tan 2x$ | A1cao | or $2x^2/\cos^2 2x + 2x\tan 2x$ (isw); or $(2x^2 + 2x\sin 2x\cos 2x)/\cos^2 2x$; or $2x^2/\cos^2 2x + 2x\sin 2x/\cos 2x$ |
| **OR** $y = \dfrac{x^2\sin 2x}{\cos 2x}$ | | see additional notes for complete solution |
| $\dfrac{dy}{dx} = \dfrac{\cos 2x(2x\sin 2x + x^2 2\cos 2x) - x^2\sin 2x(-2\sin 2x)}{\cos^2 2x}$ | M1 | Quotient rule: $(v \times$ their $u' - u \times$ their $v')/v^2$ attempted |
| | A1 | Correct expression |
| $= \ldots = 2x^2\sec^2 2x + 2x\tan 2x$ | A1cao | or $2x^2/\cos^2 2x + 2x\tan 2x$ (isw) |
| **[3]** | | |

---