Derivatives of inverse trigonometric functions
Learn the key results: d/dx sin⁻¹x = 1/√(1 − x²) and d/dx tan⁻¹x = 1/(1 + x²); cos⁻¹x and cot⁻¹x are the same with a minus sign.
Derivation: d/dx (sin⁻¹x) = 1/√(1 − x²)
- 1Let y = sin⁻¹x, where −1 < x < 1 and −π/2 < y < π/2.
- 2Then x = sin y.
- 3Differentiate with respect to x: 1 = cos y × dy/dx.
- 4So dy/dx = 1/cos y.
- 5cos y > 0 for −π/2 < y < π/2, so cos y = √(1 − sin²y) = √(1 − x²).
- 6★So, d/dx (sin⁻¹x) = 1/√(1 − x²).
★ key step
The results
| Function | Derivative | Valid for |
|---|---|---|
| sin⁻¹x | 1/√(1 − x²) | −1 < x < 1 |
| cos⁻¹x | −1/√(1 − x²) | −1 < x < 1 |
| tan⁻¹x | 1/(1 + x²) | all real x |
| cot⁻¹x | −1/(1 + x²) | all real x |
Example 1: direct use
Differentiate (i) y = tan⁻¹(x²), (ii) y = cos⁻¹(2x).
- (i) Use d/dx tan⁻¹(u) = u′/(1 + u²) with u = x², u′ = 2x.
- dy/dx = 2x/(1 + x⁴)
- (ii) Use d/dx cos⁻¹(u) = −u′/√(1 − u²) with u = 2x, u′ = 2.
- dy/dx = −2/√(1 − 4x²), for −1/2 < x < 1/2
(i) 2x/(1 + x⁴); (ii) −2/√(1 − 4x²)
Example 2: a substitution
Find dy/dx if y = tan⁻¹(2x/(1 − x²)), −1 < x < 1.
- Put x = tan θ. Since −1 < x < 1, −π/4 < θ < π/4.
- 2x/(1 − x²) = 2 tan θ/(1 − tan²θ) = tan 2θ
- So y = tan⁻¹(tan 2θ). Here −π/2 < 2θ < π/2, the principal value branch of tan⁻¹.
- So y = 2θ = 2 tan⁻¹x.
- dy/dx = 2/(1 + x²)
dy/dx = 2/(1 + x²)
Tips
- Learn the sin⁻¹ and tan⁻¹ results; cos⁻¹ and cot⁻¹ are the same with a minus sign.
- With a function inside, multiply by its derivative: d/dx sin⁻¹(u) = u′/√(1 − u²).
- For messy expressions, try x = tan θ when you see 1 + x², or 2x with 1 − x², and x = sin θ when you see √(1 − x²).
- After substituting, check that the angle lies in the principal value branch before writing tan⁻¹(tan A) = A.
A traffic camera on a pole of height h follows a car on the road below. The angle it turns through is θ = tan⁻¹(x/h), so the derivative of tan⁻¹ tells the motor how fast to turn as the car moves.
1 marks: “Find d/dx [sin⁻¹(x/2)], −2 < x < 2.” (1/2)/√(1 − x²/4) = 1/√(4 − x²).
2 marks: “Find dy/dx if y = tan⁻¹ √((1 − cos 2x)/(1 + cos 2x)), 0 < x < π/2.” (1 − cos 2x)/(1 + cos 2x) = (2 sin²x)/(2 cos²x) = tan²x, and tan x > 0. So y = tan⁻¹(tan x) = x, since 0 < x < π/2 lies in the principal value branch. dy/dx = 1.
2 marks: “Prove that d/dx (cos⁻¹x) = −1/√(1 − x²), −1 < x < 1.” Let y = cos⁻¹x, 0 < y < π, so x = cos y. Then 1 = −sin y × dy/dx. sin y > 0, so sin y = √(1 − x²). Hence dy/dx = −1/√(1 − x²).
sin⁻¹ has a root, tan⁻¹ has a plus; the 'co' ones carry a minus.