Differentiating
Inverse Trig Functions
Master the derivatives of \(\arcsin\), \(\arccos\), \(\arctan\), and more: the essential formulas and chain rule techniques for AP Calculus.
AP Calculus BC
What Are Inverse Trig Functions?
An inverse trigonometric function reverses the role of input and output. If \(\sin\theta = x\), then \(\arcsin x = \theta\). These functions answer: "What angle produces this ratio?"
\(\arcsin x\)
Angle whose sine is \(x\).
Range: \(\bigl[-\tfrac{\pi}{2},\,\tfrac{\pi}{2}\bigr]\)
Range: \(\bigl[-\tfrac{\pi}{2},\,\tfrac{\pi}{2}\bigr]\)
\(\arccos x\)
Angle whose cosine is \(x\).
Range: \([0,\,\pi]\)
Range: \([0,\,\pi]\)
\(\arctan x\)
Angle whose tangent is \(x\).
Range: \(\bigl(-\tfrac{\pi}{2},\,\tfrac{\pi}{2}\bigr)\)
Range: \(\bigl(-\tfrac{\pi}{2},\,\tfrac{\pi}{2}\bigr)\)
Why Differentiate Inverse Trig?
Inverse trig derivatives appear throughout calculus: related rates, implicit differentiation, and especially integration (they are the antiderivatives of key rational and radical expressions).
On the AP exam, these formulas are not provided: you must memorize them.
Graph of \(y = \arcsin x\)
The Big Three Derivative Formulas
\(\arcsin x\)
\(\displaystyle\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}}\)
\(\arccos x\)
\(\displaystyle\frac{d}{dx}[\arccos x] = \frac{-1}{\sqrt{1 - x^2}}\)
\(\arctan x\)
\(\displaystyle\frac{d}{dx}[\arctan x] = \frac{1}{1 + x^2}\)
The Other Three Formulas
\(\text{arccot}\, x\)
\(\displaystyle\frac{d}{dx}[\text{arccot}\, x] = \frac{-1}{1 + x^2}\)
\(\text{arcsec}\, x\)
\(\displaystyle\frac{d}{dx}[\text{arcsec}\, x] = \frac{1}{|x|\sqrt{x^2 - 1}}\)
\(\text{arccsc}\, x\)
\(\displaystyle\frac{d}{dx}[\text{arccsc}\, x] = \frac{-1}{|x|\sqrt{x^2 - 1}}\)
\(\arcsin\) vs \(\arctan\): Key Distinction
| \(\arcsin x\) | \(\arctan x\) | |
|---|---|---|
| Derivative | \(\dfrac{1}{\sqrt{1-x^2}}\) | \(\dfrac{1}{1+x^2}\) |
| Denominator type | Square root | Polynomial |
| Sign in denominator | \(1 - x^2\) (minus) | \(1 + x^2\) (plus) |
| Domain of \(f'\) | \((-1, 1)\) | \((-\infty, \infty)\) |
| Mnemonic | "Sine Subtract" | "Tan aDd" |
Deriving \(\frac{d}{dx}[\arcsin x]\) by Implicit Differentiation
1
Let \(y = \arcsin x\). Then \(\sin y = x\).
2
Differentiate both sides: \(\cos y \cdot \frac{dy}{dx} = 1\).
3
Solve: \(\frac{dy}{dx} = \frac{1}{\cos y}\).
4
Rewrite \(\cos y\): Since \(\sin^2 y + \cos^2 y = 1\) and \(\sin y = x\), we get \(\cos y = \sqrt{1 - x^2}\).
5
Result: \(\displaystyle\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}}\). ✓
Chain Rule Extensions
When the argument is not simply \(x\), apply the chain rule. If \(u = g(x)\):
\(\arcsin(u)\)
\(\displaystyle\frac{d}{dx} = \frac{u'}{\sqrt{1 - u^2}}\)
\(\arctan(u)\)
\(\displaystyle\frac{d}{dx} = \frac{u'}{1 + u^2}\)
\(\text{arcsec}(u)\)
\(\displaystyle\frac{d}{dx} = \frac{u'}{|u|\sqrt{u^2 - 1}}\)
Step-by-Step Strategy
1
Identify the inverse trig function: Is it \(\arcsin\), \(\arccos\), \(\arctan\), or one of the others?
2
Identify the inner function \(u\): What is inside the inverse trig? (e.g., \(u = 3x^2\))
3
Write the base derivative formula for that inverse trig function.
4
Multiply by \(u'\) (chain rule). Replace \(x\) with \(u\) in the formula.
5
Simplify. Look for common factors and algebraic cleanup.
Example 1: \(\frac{d}{dx}[\arctan(3x)]\)
1
Inverse trig: \(\arctan\). Formula: \(\frac{1}{1+u^2}\).
2
Inner function: \(u = 3x\), so \(u' = 3\).
3
Apply chain rule: \(\displaystyle\frac{u'}{1 + u^2} = \frac{3}{1 + (3x)^2}\).
4
Simplify: \(\displaystyle\frac{d}{dx}[\arctan(3x)] = \frac{3}{1 + 9x^2}\). ✓
A ceiling camera follows a person beneath it
For a camera 4 m above the path, tan θ = 4/x. At x = 3 m and walking speed 2 m/s, the angular-speed magnitude is 8/25 rad/s; the sign depends on direction.Angle rate depends on the sign of the distance rate.
For a camera 4 m above the path, tan θ = 4/x. At x = 3 m and walking speed 2 m/s, the angular-speed magnitude is 8/25 rad/s; the sign depends on direction.Example 2: \(\frac{d}{dx}[\arcsin(x^2)]\)
1
Inverse trig: \(\arcsin\). Formula: \(\frac{1}{\sqrt{1-u^2}}\).
2
Inner function: \(u = x^2\), so \(u' = 2x\).
3
Apply chain rule: \(\displaystyle\frac{2x}{\sqrt{1 - (x^2)^2}}\).
4
Simplify: \(\displaystyle\frac{d}{dx}[\arcsin(x^2)] = \frac{2x}{\sqrt{1 - x^4}}\). ✓
Example 3: \(\frac{d}{dx}\bigl[x \cdot \arctan x\bigr]\)
This requires the product rule combined with the inverse trig derivative.
1
Product rule: \(\frac{d}{dx}[f \cdot g] = f' g + f g'\). Here \(f = x\) and \(g = \arctan x\).
2
Compute pieces: \(f' = 1\), \(g' = \frac{1}{1+x^2}\).
3
Combine: \(\displaystyle (1)(\arctan x) + (x)\!\left(\frac{1}{1+x^2}\right)\).
4
Result: \(\displaystyle\frac{d}{dx}[x\arctan x] = \arctan x + \frac{x}{1+x^2}\). ✓
Example 4: \(\frac{d}{dx}\!\left[\arcsin\!\left(\frac{x}{3}\right)\right]\)
1
Inner function: \(u = \frac{x}{3}\), so \(u' = \frac{1}{3}\).
2
Chain rule: \(\displaystyle\frac{1/3}{\sqrt{1 - (x/3)^2}} = \frac{1/3}{\sqrt{1 - x^2/9}}\).
3
Simplify: \(\displaystyle = \frac{1/3}{\sqrt{(9 - x^2)/9}} = \frac{1/3}{{\sqrt{9 - x^2}}/{3}} = \frac{1}{\sqrt{9 - x^2}}\). ✓
Common Mistakes to Avoid
✕ Forgetting the chain rule
Writing \(\frac{1}{1+(3x)^2}\) instead of \(\frac{3}{1+9x^2}\). Always multiply by \(u'\).
✕ Wrong sign for \(\arccos\)
The derivative of \(\arccos x\) has a negative sign: \(\frac{-1}{\sqrt{1-x^2}}\). Don't omit it.
✕ Confusing \(+\) and \(-\)
\(\arcsin\) uses \(1 - x^2\) under the radical. \(\arctan\) uses \(1 + x^2\). Mixing them up is the most common error.
Complete Reference Table
| Function | Derivative | Domain |
|---|---|---|
| \(\arcsin x\) | \(\frac{1}{\sqrt{1-x^2}}\) | \(|x| < 1\) |
| \(\arccos x\) | \(\frac{-1}{\sqrt{1-x^2}}\) | \(|x| < 1\) |
| \(\arctan x\) | \(\frac{1}{1+x^2}\) | All \(\mathbb{R}\) |
| \(\text{arccot}\, x\) | \(\frac{-1}{1+x^2}\) | All \(\mathbb{R}\) |
| \(\text{arcsec}\, x\) | \(\frac{1}{|x|\sqrt{x^2-1}}\) | \(|x| > 1\) |
| \(\text{arccsc}\, x\) | \(\frac{-1}{|x|\sqrt{x^2-1}}\) | \(|x| > 1\) |
Question 1 of 5
What is \(\dfrac{d}{dx}[\arctan(5x)]\)?
The chain rule gives \(\frac{u'}{1+u^2}\) where \(u=5x\) and \(u'=5\). So the answer is \(\frac{5}{1+(5x)^2} = \frac{5}{1+25x^2}\). Choice (A) forgets the chain rule factor of 5.
Question 2 of 5
What is \(\dfrac{d}{dx}[\arccos(2x)]\)?
For \(\arccos(u)\), the derivative is \(\frac{-u'}{\sqrt{1-u^2}}\). With \(u=2x\) and \(u'=2\): \(\frac{-2}{\sqrt{1-(2x)^2}} = \frac{-2}{\sqrt{1-4x^2}}\). Choice (A) forgets the negative. Choice (D) uses \(+4x^2\) instead of \(-4x^2\).
Question 3 of 5
If \(f(x) = \arcsin(x^3)\), what is \(f'(x)\)?
With \(u = x^3\) and \(u' = 3x^2\), the chain rule gives \(\frac{3x^2}{\sqrt{1-(x^3)^2}} = \frac{3x^2}{\sqrt{1-x^6}}\). Choice (B) forgets the chain rule. Choice (C) fails to square the inner function. Choice (D) uses the \(\arctan\) formula instead.
Question 4 of 5
What is \(\dfrac{d}{dx}\!\left[\arcsin\!\left(\dfrac{x}{4}\right)\right]\)?
Let \(u = x/4\), so \(u' = 1/4\). Then \(\frac{1/4}{\sqrt{1 - x^2/16}} = \frac{1/4}{\sqrt{(16-x^2)/16}} = \frac{1/4}{(\sqrt{16-x^2})/4} = \frac{1}{\sqrt{16-x^2}}\). This is the standard form that appears in integration problems.
Question 5 of 5
If \(g(x) = x^2 \arctan x\), what is \(g'(x)\)?
Use the product rule: \((x^2)'\arctan x + x^2(\arctan x)' = 2x\arctan x + \frac{x^2}{1+x^2}\). Choice (C) forgets the first term. Choice (D) uses the \(\arcsin\) derivative instead of \(\arctan\).