Inverse trigonometric derivatives
\[(\sin^{-1}x)'=\frac{1}{\sqrt{1-x^2}},\ (\tan^{-1}x)'=\frac1{1+x^2},\ (\sec^{-1}x)'=\frac{1}{|x|\sqrt{x^2-1}}\]
Continuity and Differentiability
Derivatives of Inverse Trigonometric Functions
Understand
| \(f(x)\) | \(f\,'(x)\) | Valid for |
|---|---|---|
| \(\sin^{-1}x\) | \(\frac{1}{\sqrt{1-x^2}}\) | \(|x|<1\) |
| \(\cos^{-1}x\) | \(-\frac{1}{\sqrt{1-x^2}}\) | \(|x|<1\) |
| \(\tan^{-1}x\) | \(\frac{1}{1+x^2}\) | all \(x\) |
| \(\cot^{-1}x\) | \(-\frac{1}{1+x^2}\) | all \(x\) |
| \(\sec^{-1}x\) | \(\frac{1}{|x|\sqrt{x^2-1}}\) | \(|x|>1\) |
| \(\csc^{-1}x\) | \(-\frac{1}{|x|\sqrt{x^2-1}}\) | \(|x|>1\) |
Combine these with the chain rule: \(\dfrac{d}{dx}\tan^{-1}(x^2)=\dfrac{2x}{1+x^4}\).
Substitution trick. Expressions like \(\tan^{-1}\frac{6x}{1-9x^2}\) simplify with \(3x=\tan\theta\): it equals \(2\tan^{-1}3x\) for \(|x|<\frac13\), so the derivative is \(\frac{6}{1+9x^2}\).
Key Concepts
- Learn the six derivatives; the "co-" functions have a minus sign.
- Use the chain rule for inner functions.
- Try a trigonometric substitution to simplify first.
Formula Bank
Key Points
Substitution shortcuts
Recognise multiple-angle formulas: with \(x=\tan\theta\), \(\frac{2x}{1-x^2}=\tan2\theta\) and \(\frac{1-x^2}{1+x^2}=\cos2\theta\); with \(x=\sin\theta\), \(3x-4x^3=\sin3\theta\). Always check that the resulting angle lies in the principal branch.
Practice & Topic Test
Topic Practice
A fresh set of questions from this topic. Change answers freely – solutions appear after you submit.
Start PracticeTopic Summary
Six standard derivatives plus the chain rule; simplify with x = tan θ, sin θ or cos θ when the expression suggests a double-angle formula.
Ready to practise this topic?
Create Student Account
Create a free account to take practice sessions and tests, see detailed explanations and track your progress.