Questions: Derivatives of Inverse Trigonometric Functions

Question 1

A student needs to find ∫1/(1 + 9x²) dx. She rewrites the denominator as 1 + (3x)². What is the correct antiderivative?

• A. (1/3)arctan(3x) + C
• B. arctan(3x) + C
• C. (1/9)arctan(3x) + C
• D. 3·arctan(3x) + C

Answer: (A)

Use u = 3x, so du = 3dx, meaning dx = du/3. The integral becomes ∫1/(1 + u²) · (du/3) = (1/3)arctan(u) + C = (1/3)arctan(3x) + C. The factor of 1/3 comes from the substitution: when the argument of arctan is (3x) rather than x, the chain rule in reverse requires dividing by the inner derivative (3). Check by differentiating: d/dx[(1/3)arctan(3x)] = (1/3) · 1/(1+(3x)²) · 3 = 1/(1+9x²). ✓

Question 2

To derive d/dx[arcsin(x)], a student writes sin(y) = x and differentiates implicitly, obtaining cos(y)·(dy/dx) = 1. The correct next step is:

• A. Use the Pythagorean identity to write cos(y) = √(1 − x²), accounting for the restricted domain of arcsin, then solve for dy/dx
• B. Set cos(y) = x, since sin(y) = x implies cos(y) = x by symmetry
• C. Apply the quotient rule to 1/cos(y) to get the final derivative
• D. Substitute y = arcsin(x) back and leave the answer as 1/cos(arcsin(x))

Answer: (A)

After implicit differentiation, dy/dx = 1/cos(y), but the answer must be in terms of x. Since sin(y) = x, we have sin²(y) = x², and by the Pythagorean identity cos²(y) = 1 − x². Since the domain of arcsin is [−π/2, π/2] — where cosine is non-negative — cos(y) = +√(1 − x²). Therefore dy/dx = 1/√(1 − x²). The domain restriction is crucial: without it, we could not take the positive square root.

Question 3

d/dx[arccos(x)] = 1/√(1 − x²), the same formula as d/dx[arcsin(x)].

• T. True
• F. False

Answer: False

d/dx[arccos(x)] = −1/√(1 − x²) — the same magnitude but with a minus sign. This follows from the identity arcsin(x) + arccos(x) = π/2 (a constant). Differentiating both sides: d/dx[arcsin(x)] + d/dx[arccos(x)] = 0, so d/dx[arccos(x)] = −d/dx[arcsin(x)] = −1/√(1 − x²). The minus sign is the most common error when recalling this formula.

Question 4

d/dx[arctan(x²)] = 2x/(1 + x⁴), obtained by applying the arctan derivative formula to x² and multiplying by the derivative of x².

• T. True
• F. False

Answer: True

By the chain rule with u = x²: d/dx[arctan(x²)] = 1/(1 + (x²)²) · d/dx[x²] = 1/(1 + x⁴) · 2x = 2x/(1 + x⁴). The outer derivative is 1/(1 + u²) evaluated at u = x², giving 1/(1 + x⁴); the inner derivative is 2x. Their product is 2x/(1 + x⁴). The most common error is forgetting the inner derivative and writing 1/(1 + x⁴) alone.

Question 5

Explain how implicit differentiation is used to derive d/dx[arctan(x)] = 1/(1 + x²). What role does a Pythagorean identity play in the derivation?

Model answer: Let y = arctan(x), so tan(y) = x. Differentiate both sides with respect to x using the chain rule: sec²(y) · dy/dx = 1. Solve: dy/dx = 1/sec²(y). To express this in terms of x, use the identity sec²(y) = 1 + tan²(y) = 1 + x² (since tan(y) = x). Therefore dy/dx = 1/(1 + x²). The Pythagorean identity is the step that converts the expression from y-form to x-form, completing the derivation.

The same strategy applies to all inverse trig derivatives: (1) write the trig equation from the inverse function definition, (2) differentiate implicitly, (3) solve for dy/dx, (4) use a Pythagorean identity to re-express in terms of x. For arcsin: sin(y) = x → cos(y)·dy/dx = 1 → use cos²(y) = 1 − sin²(y) = 1 − x² → dy/dx = 1/√(1−x²). No new rules are needed — just implicit differentiation applied to a known inverse relationship, followed by an algebraic substitution using a trig identity.