Derivatives of Inverse Trig Functions for MATH 139
Exam Relevance for MATH 139
Know derivatives of arcsin, arctan, and arcsec at minimum. Often appears in implicit differentiation.
Derivatives of Inverse Trigonometric Functions
The inverse trig functions "undo" the regular trig functions. Their derivatives are algebraic expressions—no trig functions appear in the answers!
The Three Most Important
These appear constantly on exams:
$$\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$$
$$\frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}}$$
$$\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}$$
🔑 Key observations:
- Arcsine and arccosine have the same denominator $\sqrt{1-x^2}$, but arccosine is negative
- Arctangent has the simplest formula: $\frac{1}{1+x^2}$
- These derivatives are defined where the original functions are defined
The Other Three (Less Common)
$$\frac{d}{dx}[\text{arccot } x] = -\frac{1}{1+x^2}$$
$$\frac{d}{dx}[\text{arcsec } x] = \frac{1}{|x|\sqrt{x^2-1}}$$
$$\frac{d}{dx}[\text{arccsc } x] = -\frac{1}{|x|\sqrt{x^2-1}}$$
Pattern: Just like regular trig, the "co-" functions have negative derivatives.
Complete Reference Table
| Function | Derivative | Domain of derivative |
|---|---|---|
| $\arcsin x$ | $\dfrac{1}{\sqrt{1-x^2}}$ | $-1 < x < 1$ |
| $\arccos x$ | $-\dfrac{1}{\sqrt{1-x^2}}$ | $-1 < x < 1$ |
| $\arctan x$ | $\dfrac{1}{1+x^2}$ | all real $x$ |
| $\text{arccot } x$ | $-\dfrac{1}{1+x^2}$ | all real $x$ |
| $\text{arcsec } x$ | $\dfrac{1}{\|x\|\sqrt{x^2-1}}$ | $\|x\| > 1$ |
| $\text{arccsc } x$ | $-\dfrac{1}{\|x\|\sqrt{x^2-1}}$ | $\|x\| > 1$ |
Alternative Notation
You may see inverse trig functions written as:
- $\arcsin x = \sin^{-1} x$
- $\arccos x = \cos^{-1} x$
- $\arctan x = \tan^{-1} x$
These mean the same thing! (Note: $\sin^{-1} x \neq \frac{1}{\sin x}$)
Problem: Find $\frac{d}{dx}[\arctan x]$
This is a direct formula: $$\frac{d}{dx}[\arctan x] = \boxed{\frac{1}{1+x^2}}$$
Problem: Find $\frac{d}{dx}[\arcsin(2x)]$
Step 1: Use the chain rule. If $u = 2x$: $$\frac{d}{dx}[\arcsin(2x)] = \frac{1}{\sqrt{1-(2x)^2}} \cdot \frac{d}{dx}[2x]$$
Step 2: Simplify: $$= \frac{1}{\sqrt{1-4x^2}} \cdot 2 = \boxed{\frac{2}{\sqrt{1-4x^2}}}$$
Problem: Find $\frac{d}{dx}[\arctan(3x)]$
Step 1: Apply the chain rule: $$\frac{d}{dx}[\arctan(3x)] = \frac{1}{1+(3x)^2} \cdot \frac{d}{dx}[3x]$$
Step 2: Simplify: $$= \frac{1}{1+9x^2} \cdot 3 = \boxed{\frac{3}{1+9x^2}}$$
Problem: Find $\frac{d}{dx}[x \cdot \arcsin x]$
Step 1: Apply the product rule: $(fg)' = f'g + fg'$
Let $f(x) = x$ and $g(x) = \arcsin x$.
Step 2: Calculate: $$\frac{d}{dx}[x \cdot \arcsin x] = 1 \cdot \arcsin x + x \cdot \frac{1}{\sqrt{1-x^2}}$$
$$= \boxed{\arcsin x + \frac{x}{\sqrt{1-x^2}}}$$
Problem: Find $f'(0)$ if $f(x) = \arctan(x^2 + 1)$
Step 1: Find the general derivative using chain rule: $$f'(x) = \frac{1}{1+(x^2+1)^2} \cdot 2x = \frac{2x}{1+(x^2+1)^2}$$
Step 2: Evaluate at $x = 0$: $$f'(0) = \frac{2(0)}{1+(0+1)^2} = \frac{0}{1+1} = \boxed{0}$$
Problem: Prove that $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$
Step 1: Let $y = \arcsin x$, which means $\sin y = x$.
Step 2: Differentiate implicitly: $$\cos y \cdot \frac{dy}{dx} = 1$$ $$\frac{dy}{dx} = \frac{1}{\cos y}$$
Step 3: Express $\cos y$ in terms of $x$.
Since $\sin y = x$ and $\cos^2 y + \sin^2 y = 1$: $$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$$
(We take the positive root because $y = \arcsin x$ is in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, where cosine is positive.)
Step 4: Substitute: $$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$$
$$\boxed{\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}}$$
Common Mistakes and Misunderstandings
❌ Mistake: Confusing $\sin^{-1} x$ with $\frac{1}{\sin x}$
Wrong: $\frac{d}{dx}[\sin^{-1} x] = \frac{d}{dx}\left[\frac{1}{\sin x}\right] = -\frac{\cos x}{\sin^2 x}$
Why it's wrong: $\sin^{-1} x$ means the inverse sine function (arcsin), NOT one over sine.
Correct: $\frac{d}{dx}[\sin^{-1} x] = \frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$
❌ Mistake: Forgetting the chain rule
Wrong: $\frac{d}{dx}[\arctan(5x)] = \frac{1}{1+(5x)^2}$
Why it's wrong: You must multiply by the derivative of the inside function.
Correct: $\frac{d}{dx}[\arctan(5x)] = \frac{1}{1+25x^2} \cdot 5 = \frac{5}{1+25x^2}$
❌ Mistake: Forgetting that arccosine's derivative is negative
Wrong: $\frac{d}{dx}[\arccos x] = \frac{1}{\sqrt{1-x^2}}$
Why it's wrong: Arccosine has a negative derivative (like all "co-" inverse trig functions).
Correct: $\frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}}$
Memory Tips
- Arcsin and arctan are the "friendly" ones — their derivatives are positive
- All "co-" functions are negative — arccos, arccot, arccsc
- Arctan is the simplest — no square root, just $\frac{1}{1+x^2}$
- Arcsin/arccos share a denominator — $\sqrt{1-x^2}$, just differ by sign
Derivative of Arcsine
The derivative of arcsine (inverse sine). Domain: -1 < x < 1. One of the three most common inverse trig derivatives.
Variables:
- $x$:
- value between -1 and 1
Derivative of Arccosine
The derivative of arccosine (inverse cosine). Same as arcsin but NEGATIVE. Domain: -1 < x < 1.
Variables:
- $x$:
- value between -1 and 1
Derivative of Arctangent
The derivative of arctangent (inverse tangent). The simplest inverse trig derivative—no square root, defined for all real x.
Variables:
- $x$:
- any real number
Derivative of Arccotangent
The derivative of arccotangent. Same as arctan but NEGATIVE (all 'co-' functions have negative derivatives).
Variables:
- $x$:
- any real number
Derivative of Arcsecant
The derivative of arcsecant. Note the absolute value on x. Domain: |x| > 1.
Variables:
- $x$:
- value where |x| > 1
Derivative of Arccosecant
The derivative of arccosecant. Same as arcsec but NEGATIVE. Domain: |x| > 1.
Variables:
- $x$:
- value where |x| > 1
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