f(x) = √3 + (4x-2)⁵ where x > 1. Find an expression for f'(x), giving your answer as a simplified algebraic fraction.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [B2] 10(4x-2)⁴ / √(3+(4x-2)⁵)
2. [B2] for correct unsimplified form e.g. (1/2)(3+(4x-2)⁵)^(-1/2) × 5(4x-2)⁴ × 4
3. [B1] for 5(4x-2)⁴ × 4 soi
4. [B1] or (1/2)(3+(4x-2)⁵)^(-1/2) × g(x)
5. [B1] or 10(3+(4x-2)⁵)^(-1/2)(4x-2)⁴
6. [B1] dy/dx = (5(3x+2) - 3(5x)) / (3x+2)² oe isw
7. [B1] or dy/dx = 5x(-(3x+2)⁻² * 3) + 5(3x+2)⁻¹ oe isw
8. [B1] [y = 10] x = -0.8
9. [M1] 0.01 / (their dy/dx at x=-0.8) oe FT their x providing their x ≠ 10 or 0.01and their genuine attempt at a derivative using product or quotient rule
10. [A1] 1/6250 or 0.00016 or 1.6 × 10⁻⁴ isw dep on all previous marks awarded
11. [B1] Alternative method x = -2y / (3y-5) oe
12. [B1] dx/dy = (-2(3y-5) - 3(-2y)) / (3y-5)² oe isw
13. [M1] δx = 0.01 * (their dx/dy at y=10) oe FT their genuine attempt at a derivative using product or quotient rule
14. [A1] 1/6250 or 0.00016 or 1.6 × 10⁻⁴ isw dep on all previous marks awarded
15. [B2] 3x²lnx + x³ × (1/x) or better, isw
16. [B1] for (their 3x²)lnx + x³ × (their 1/x)
17. [B3] x³lnx/6 - x³/18 + c oe, isw must have arbitrary constant
18. [B2] for x³lnx/6 - x³/18 or x³lnx/6 + kx³/18 + c, k > 0 nfww
19. [B1] for (1/6)∫(3x²lnx + x²)dx + (1/6)∫x² dx soi
20. [B1] or ∫(3x²lnx)dx = x³lnx - ∫x² dx soi
21. [B1] or ∫(3x²lnx)dx = x³lnx - x³/3 soi