Quick Recall — Inverse Trig
An inverse trig function reverses the trig function. For example, sin⁻¹(x) gives the angle whose sine is x.
Inverse trig functions are only defined on restricted domains to ensure they are one-one and onto. The principal value branches are used for differentiation.
The Six Standard Derivatives
💡 Pattern to notice: sin⁻¹ & cos⁻¹ are negatives of each other. tan⁻¹ & cot⁻¹ are negatives. sec⁻¹ & cosec⁻¹ are negatives. So you really only need to memorise 3 formulas!
Chain Rule with Inverse Trig
When the argument is f(x) instead of just x, apply the chain rule:
d/dx [tan⁻¹(f(x))] = f'(x) / (1 + [f(x)]²)
d/dx[sin⁻¹(2x)] = 2 / √(1−4x²) = 2/√(1−4x²)
d/dx[tan⁻¹(x²)] = 2x/(1+x⁴) = 2x/(1+x⁴)
d/dx[cos⁻¹(√x)] = (1/2√x)·(−1/√(1−x)) = −1 / (2√x·√(1−x))
d/dx[tan⁻¹(sin x)] = cos x/(1+sin²x) = cos x / (1 + sin²x)
Substitution Technique
Many inverse trig expressions simplify greatly with a trig substitution before differentiating.
| Expression in y = | Use Substitution | Simplifies to |
|---|---|---|
| sin⁻¹(2x√(1−x²)) | x = sin θ | 2 sin⁻¹(x) |
| tan⁻¹(2x/(1−x²)) | x = tan θ | 2 tan⁻¹(x) |
| cos⁻¹(2x²−1) | x = cos θ | 2 cos⁻¹(x) |
| tan⁻¹((1−x)/(1+x)) | x = tan θ | π/4 − tan⁻¹(x) |
| sin⁻¹(1/√(1+x²)) | x = cot θ | cot⁻¹(x) |
Let x = sin θ, so 2x√(1−x²) = 2 sin θ · cos θ = sin 2θ
y = sin⁻¹(sin 2θ) = 2θ = 2 sin⁻¹(x)
dy/dx = d/dx[2 sin⁻¹(x)] = 2/√(1−x²)
💡 Exam Tip: Always check if the given expression matches a known double-angle or compound-angle formula. Substitution can convert a messy expression into something like 2sin⁻¹(x), making the derivative trivial.