Unit 3.4 – Differentiating Inverse Trigonometric Functions
AP® Calculus AB & BC | Essential Derivative Formulas for Arcsin, Arccos, Arctan, etc.
Inverse trig derivatives are heavily tested on AP® Calculus. These rules let you differentiate arcsin, arccos, arctan, arcsec, arccsc, arccot and their compositions. Memorize the patterns: they're fast, predictable, and all built from the inverse function rule or implicit differentiation.
🔑 Master Formulas
All Six Classic Inverse Trig Derivatives:
- \( \frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}} \)
- \( \frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}} \)
- \( \frac{d}{dx}[\arctan x] = \frac{1}{1+x^2} \)
- \( \frac{d}{dx}[\arccot x] = -\frac{1}{1+x^2} \)
- \( \frac{d}{dx}[\arcsec x] = \frac{1}{|x|\sqrt{x^2-1}} \)
- \( \frac{d}{dx}[\arccsc x] = -\frac{1}{|x|\sqrt{x^2-1}} \)
Composite Version:
For \( y = \arcsin(u) \), \( \frac{d}{dx}[y] = \frac{1}{\sqrt{1-u^2}} \cdot u' \)
Apply Chain Rule for all forms with \(u(x)\).
For \( y = \arcsin(u) \), \( \frac{d}{dx}[y] = \frac{1}{\sqrt{1-u^2}} \cdot u' \)
Apply Chain Rule for all forms with \(u(x)\).
📐 How to Derive (Proof Sketch)
- Example: If \( y = \arcsin x \), then \( x = \sin y \), so \( 1 = \cos y \frac{dy}{dx} \) → \( \frac{dy}{dx} = \frac{1}{\cos y} \).
- Convert using \( \cos y = \sqrt{1- \sin^2 y} = \sqrt{1-x^2} \), so final answer: \( \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \)
- All results for other inverse trigs work similarly—swap, differentiate, solve!
💡 Memory Tricks, Shortcuts, and Patterns
- Squared under the root for arcsin/arccos: \( \sqrt{1-x^2} \)
- Plus \(x^2\) for arctan/arccot: \( 1 + x^2 \)
- Absolute value and root for arcsec/arccsc: \( |x|\sqrt{x^2-1} \)
- Minus sign: All "co"-functions' derivatives are negative: arccos, arccot, arccsc
- For \(u(x)\): Chain rule: Always multiply by derivative of inner \(u'\)
- Saying: “Arcsin root 1 minus, Arctan one plus, Arcsec absolute root, Chain rule at last!”
📖 Worked Examples
Example 1: \( \frac{d}{dx}[\arcsin(3x)] \)
\( u = 3x \Rightarrow \frac{d}{dx}[\arcsin(3x)] = \frac{1}{\sqrt{1-(3x)^2}} \cdot 3 = \frac{3}{\sqrt{1-9x^2}} \)
Example 2: \( \frac{d}{dx}[\arctan(x^2)] \)
\( u = x^2 \Rightarrow \frac{1}{1+(x^2)^2} \cdot 2x = \frac{2x}{1+x^4} \)
Example 3: \( \frac{d}{dx}[\arcsec(2x)] \)
\( u = 2x \Rightarrow \frac{1}{|2x|\sqrt{(2x)^2-1}} \cdot 2 = \frac{2}{2|x|\sqrt{4x^2-1}} = \frac{1}{|x|\sqrt{4x^2-1}} \)
❌ Common Mistakes
- Forgetting to multiply by inner derivative (\(u'\)) for composites!
- Mixing up which formulas use plus and which use minus
- Leaving out the absolute value and root for arcsec/arccsc
- Sign errors: “co”-functions are negative
- Plug in values: Watch out for domains! Not all \(x\) are allowed for arcsec/arccsc/arcsin/arccos
📝 Practice Problems
Try These Yourself:
- \( \frac{d}{dx}[ \arccos(5x) ] \)
- \( \frac{d}{dx}[ \arctan(x^4) ] \)
- \( \frac{d}{dx}[ \arccsc(2x) ] \)
- \( \frac{d}{dx}[ \arcsin(x^2) ] \)
Answers:
- \( -\frac{5}{\sqrt{1-(5x)^2}} \)
- \( \frac{4x^3}{1+x^8} \)
- \( -\frac{2}{|2x|\sqrt{(2x)^2-1}} = -\frac{1}{|x|\sqrt{4x^2-1}} \)
- \( \frac{2x}{\sqrt{1-x^4}} \)
✏️ AP® Exam Success – Inverse Trig Derivatives
- Always state the chain rule used for composites: show \(u(x)\) and \(u'\)
- Box the final answer, especially in FRQs
- Check domain! No real answer for arcsin or arccos outside \([-1, 1]\)
- Watch which sign is positive or negative—memorize the pattern
- For log/arcsec/arccsc, always include absolute value when required
- Show all work for full credit; AP® graders want clear step-by-step logic!