June 2022 Paper 1 Q6
6 \(\quad y = 2x^4\left(x^3 + 2 - \dfrac{3}{x}\right)\)
Work out \(\quad \dfrac{\mathrm{d}y}{\mathrm{d}x}\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| (\(y =\)) \(2x^7 + 4x^4 - 6x^3\) | M1 | for any 2 terms correct |
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 14x^6 + 16x^3 - 18x^2\) | A2 | oe eg \(2x^2(7x^4 + 8x - 9)\) A1 for any correct term correctly differentiated |
| Alternative method 2 (product rule) | ||
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right)\) \(8x^3\left(x^3 + 2 - \dfrac{3}{x}\right) + 2x^4(3x^2 + 3x^{-2})\) | M1 | for either \(2x^4\) differentiated correctly multiplied by the bracket or the bracket differentiated correctly multiplied by \(2x^4\) eg \(8x^3\left(x^3 + 2 - \dfrac{3}{x}\right)\) |
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 14x^6 + 16x^3 - 18x^2\) | A2 | oe eg \(2x^2(7x^4 + 8x - 9)\) A1 for any term correct |
Additional guidance
| Ignore subsequent incorrect factorisation Condone incorrect use of \(y =\) on the answer line |