Further derivatives and antiderivatives (arcsin etc.) practice questions.

Determine d/dx (arcsin 2x).

• 1/√(1 - 4x²)
• 2/√(1 - 4x²)
• 2/√(1 - 2x²)
• -2/√(1 - 4x²)

Answer: 2/√(1 - 4x²)

By the chain rule, d/dx arcsin u = u′/√(1 - u²). With u = 2x this gives 2/√(1 - (2x)²) = 2/√(1 - 4x²).

Determine d/dx (arctan(x/3)).

• 1/(9 + x²)
• 1/(3(1 + x²))
• 3/(9 + x²)
• 9/(9 + x²)

Answer: 3/(9 + x²)

d/dx arctan u = u′/(1 + u²), so with u = x/3 we get (1/3)/(1 + x²/9). Multiplying top and bottom by 9 gives 3/(9 + x²).

Determine ∫ 1/(4 + x²) dx.

• (1/2) arctan(x/2) + C
• arctan(x/2) + C
• (1/4) arctan(x/4) + C
• (1/2) arcsin(x/2) + C

Answer: (1/2) arctan(x/2) + C

∫ 1/(a² + x²) dx = (1/a) arctan(x/a) + C. Here a = 2, so the factor 1/2 in front is essential.

For a > 0, a ≠ 1, determine d/dx (logₐ x).

• 1/x
• ln a / x
• 1/(x ln a)
• x ln a

Answer: 1/(x ln a)

Since logₐ x = ln x / ln a, differentiating gives (1/x)·(1/ln a) = 1/(x ln a).

For a > 0, determine d/dx (aˣ).

• x·aˣ⁻¹
• aˣ/ln a
• aˣ
• aˣ ln a

Answer: aˣ ln a

Write aˣ = e^(x ln a); the chain rule gives e^(x ln a)·ln a = aˣ ln a. Applying the power rule x·aˣ⁻¹ is the classic error.

Determine ∫ 1/√(9 - x²) dx.

• (1/3) arcsin(x/3) + C
• arcsin(3x) + C
• (1/3) arctan(x/3) + C
• arcsin(x/3) + C

Answer: arcsin(x/3) + C

∫ 1/√(a² - x²) dx = arcsin(x/a) + C with no 1/a factor, unlike the arctan result. Here a = 3.