Calculus practice questions.

Using the product rule, what is the derivative of y = x²eˣ?

• 2x eˣ
• (x² + 2x)eˣ
• x² eˣ
• (x² − 2x)eˣ

Answer: (x² + 2x)eˣ

Product rule: (uv)′ = u′v + uv′ = 2x·eˣ + x²·eˣ = (x² + 2x)eˣ. Factor out the common eˣ to tidy the answer.

Using the quotient rule, what is the derivative of y = x/(x + 1)?

• 1/(x + 1)²
• −1/(x + 1)²
• x/(x + 1)²
• 1

Answer: 1/(x + 1)²

Quotient rule: (u′v − uv′)/v² = ((x + 1) − x)/(x + 1)² = 1/(x + 1)². The numerator order matters — 'low d-high minus high d-low'.

Using the substitution u = x², which is ∫ 2x cos(x²) dx?

• sin(x²) + C
• −sin(x²) + C
• cos(x²) + C
• x² sin(x²) + C

Answer: sin(x²) + C

With u = x², du = 2x dx, the integral becomes ∫ cos u du = sin u + C = sin(x²) + C. Spot the derivative of the inner function sitting outside.

Using integration by parts, which is ∫ x eˣ dx?

• (x − 1)eˣ + C
• (x + 1)eˣ + C
• x eˣ + C
• ½x²eˣ + C

Answer: (x − 1)eˣ + C

Take u = x, dv = eˣ dx: ∫ x eˣ dx = x eˣ − ∫ eˣ dx = (x − 1)eˣ + C. Differentiate the answer to check: (x − 1)eˣ + eˣ = x eˣ.

Which are the first three terms of the Maclaurin series for eˣ?

• 1 + x + x²
• 1 + x + x²/2
• x + x²/2 + x³/6
• 1 − x + x²/2

Answer: 1 + x + x²/2

Maclaurin: f(0) + f′(0)x + f″(0)x²/2!. Every derivative of eˣ is eˣ, equal to 1 at x = 0, so the series is 1 + x + x²/2! + x³/3! + …

The derivative of y = 5x⁴ is

• 5x³
• 20x³
• 20x⁴
• 4x³

Answer: 20x³

Power rule: d/dx(axⁿ) = anxⁿ⁻¹, so d/dx(5x⁴) = 5 × 4 × x³ = 20x³. Multiply by the old power, then reduce the power by 1.