These ten questions follow the order of the differentiation lessons. All numbers are original, and each answer shows the working and a check.
Work on paper first. Use the mistake log and paper-error review to record any step that went wrong.
Questions and answers
Question 1. Differentiate y = 5x⁴ − 3x² + 8.
Answer
dy/dx = 20x³ − 6x. The constant 8 disappears.
Question 2. Differentiate y = 2√x − 6 ÷ x².
Answer
Rewrite as y = 2x^(1/2) − 6x⁻². Then dy/dx = x^(−1/2) + 12x⁻³, which is 1 ÷ √x + 12 ÷ x³.
Question 3. Differentiate y = (2x − 3)⁵.
Answer
Outside: 5(2x − 3)⁴. Inside: 2. So dy/dx = 10(2x − 3)⁴.
Question 4. Differentiate y = (x² + 1)(2x − 5) using the product rule, and check by expanding.
Answer
u = x² + 1, v = 2x − 5, so dy/dx = 2x(2x − 5) + (x² + 1)(2) = 4x² − 10x + 2x² + 2 = 6x² − 10x + 2.
Check: expanding gives 2x³ − 5x² + 2x − 5, whose derivative is 6x² − 10x + 2.
Question 5. Differentiate y = x ÷ (x + 2).
Answer
u = x, v = x + 2. dy/dx = (1(x + 2) − x(1)) ÷ (x + 2)² = 2 ÷ (x + 2)².
Question 6. Find the equations of the tangent and normal to y = x² + 4 ÷ x at x = 2.
Answer
Point: y = 4 + 2 = 6, so (2, 6).
dy/dx = 2x − 4x⁻², which at x = 2 is 4 − 1 = 3.
Tangent: y − 6 = 3(x − 2), so y = 3x.
Normal: gradient −1/3, so y − 6 = −(1/3)(x − 2), which gives x + 3y = 20. Check: 2 + 18 = 20.
Question 7. Find the stationary points of y = x³ − 12x + 5 and determine their nature.
Answer
dy/dx = 3x² − 12 = 0, so x = 2 or x = −2.
y(2) = 8 − 24 + 5 = −11 and y(−2) = −8 + 24 + 5 = 21.
d²y/dx² = 6x. At x = 2 it is 12 > 0, so (2, −11) is a minimum. At x = −2 it is −12 < 0, so (−2, 21) is a maximum.
Question 8. A farmer has 40 m of fencing for three sides of a rectangular pen against a wall. Find the dimensions that give the greatest area.
Answer
Let each side perpendicular to the wall be x m. The side parallel to the wall is 40 − 2x, so A = x(40 − 2x) = 40x − 2x².
dA/dx = 40 − 4x = 0, so x = 10. The other side is 20 m.
d²A/dx² = −4 < 0, so the area is a maximum: 10 m by 20 m, area 200 m².
Question 9. Given y = x³ − x, use differentiation to estimate the change in y when x increases from 3 to 3.02.
Answer
dy/dx = 3x² − 1 = 26 at x = 3. The change δx = 0.02.
δy ≈ 26 × 0.02 = 0.52.
Question 10. The radius of a circular ripple increases at 0.5 cm per second. Find the rate at which the area is increasing when the radius is 6 cm.
Answer
A = πr², so dA/dr = 2πr. By the chain rule, dA/dt = 2πr × dr/dt.
At r = 6: dA/dt = 2π(6)(0.5) = 6π cm² per second, about 18.85 cm² per second.
If you got these wrong
- Questions 1 to 5 need differentiating powers, products and quotients and the chain rule.
- Question 6 needs tangents and normals.
- Questions 7 and 8 need stationary points and optimisation.
- Question 9 needs small changes and approximations.
For mixed problems, continue to recovering the method in a mixed calculus problem. To have a teacher watch your working, see online one-to-one Additional Mathematics tuition.