C3 January 2006 Q4
4.
(a) Differentiate with respect to \(x\)
(i) \(x^2\mathrm{e}^{3x + 2}\), (4)
(ii) \(\dfrac{\cos(2x^3)}{3x}\). (4)
(b) Given that \(x = 4\sin(2y + 6)\), find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\). (5)
| Scheme | Marks |
|---|---|
| (i) \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{3x + 2}\right) = 3\mathrm{e}^{3x + 2}\) \(\left(\text{or } 3\mathrm{e}^2\mathrm{e}^{3x}\right)\) At any stage | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2\mathrm{e}^{3x + 2} + 2x\mathrm{e}^{3x + 2}\) Or equivalent | M1 A1+A1 |
| (4) | |
| (ii) \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\cos(2x^3)\right) = -6x^2\sin(2x^3)\) At any stage | M1 A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-18x^3\sin(2x^3) - 3\cos(2x^3)}{9x^2}\) | M1 A1 |
| (4) |
Notes
(ii) Alternatively using the product rule for second M1 A1
\(y = (3x)^{-1}\cos(2x^3)\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3(3x)^{-2}\cos(2x^3) - 6x^2(3x)^{-1}\sin(2x^3)\)
Accept equivalent unsimplified forms
| Scheme | Marks |
|---|---|
| \(1 = 8\cos(2y + 6)\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 8\cos(2y + 6)\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{8\cos(2y + 6)}\) | M1 A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{8\cos\left(\arcsin\left(\dfrac{x}{4}\right)\right)}\) \(\left(= (\pm)\dfrac{1}{2\sqrt{(16 - x^2)}}\right)\) | M1 A1 |
| (5) | |
| (13 marks) |