Differentiating
Inverse Trig Functions
Master the derivatives of \(\arcsin\), \(\arccos\), \(\arctan\), and their companions: the essential formulas and chain rule applications for AP Calculus.
AP Calculus AB
What Are Inverse Trig Functions?
A regular trig function takes an angle and returns a ratio. An inverse trig function does the opposite: it takes a ratio and returns an angle.
Forward
\(\sin(\theta) = x\)
Angle in, ratio out
Angle in, ratio out
Inverse
\(\arcsin(x) = \theta\)
Ratio in, angle out
Ratio in, angle out
Why Restricted Domains Matter
Trig functions are not one-to-one on their full domain. To define an inverse, we restrict each function to an interval where it passes the horizontal line test.
Why These Derivatives Matter
Inverse trig derivatives appear throughout calculus, especially in integration. Recognizing them lets you work both directions:
Differentiation
Find the rate of change of expressions involving \(\arcsin\), \(\arctan\), etc.
Integration
Recognizing \(\frac{1}{\sqrt{1-x^2}}\) as \(\frac{d}{dx}[\arcsin x]\) lets you antidifferentiate.
Modeling
Angles in physics, engineering, and geometry are often expressed as inverse trig functions.
The Three Core Formulas
\(\arcsin(x)\)
$$\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1 - x^2}}$$
\(\arccos(x)\)
$$\frac{d}{dx}[\arccos(x)] = \frac{-1}{\sqrt{1 - x^2}}$$
\(\arctan(x)\)
$$\frac{d}{dx}[\arctan(x)] = \frac{1}{1 + x^2}$$
The Reciprocal Trio
\(\text{arccsc}(x)\)
$$\frac{d}{dx}[\text{arccsc}(x)] = \frac{-1}{|x|\sqrt{x^2 - 1}}$$
\(\text{arcsec}(x)\)
$$\frac{d}{dx}[\text{arcsec}(x)] = \frac{1}{|x|\sqrt{x^2 - 1}}$$
\(\text{arccot}(x)\)
$$\frac{d}{dx}[\text{arccot}(x)] = \frac{-1}{1 + x^2}$$
\(\arcsin\) vs \(\arccos\): Key Distinction
| \(\arcsin(x)\) | \(\arccos(x)\) | |
|---|---|---|
| Domain | \([-1, 1]\) | \([-1, 1]\) |
| Range | \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) | \([0, \pi]\) |
| Derivative | \(\frac{1}{\sqrt{1-x^2}}\) | \(\frac{-1}{\sqrt{1-x^2}}\) |
| Sign of derivative | Always positive | Always negative |
| Sum identity | \(\arcsin(x) + \arccos(x) = \dfrac{\pi}{2}\) | |
Deriving \(\frac{d}{dx}[\arcsin x]\) with a Triangle
Let \(y = \arcsin(x)\), so \(\sin(y) = x\). Differentiate both sides implicitly:
Adding the Chain Rule
When the argument is not just \(x\) but some function \(u(x)\), apply the chain rule:
\(\arcsin(u)\)
$$\frac{d}{dx}[\arcsin(u)] = \frac{u'}{\sqrt{1 - u^2}}$$
\(\arctan(u)\)
$$\frac{d}{dx}[\arctan(u)] = \frac{u'}{1 + u^2}$$
Step-by-Step Strategy
1
Identify the inverse trig function: \(\arcsin\), \(\arccos\), \(\arctan\), etc.
2
Identify the inner function \(u\): What is the argument? Is it just \(x\), or something like \(3x^2\)?
3
Write the base derivative formula: e.g., \(\frac{1}{\sqrt{1-u^2}}\) for \(\arcsin(u)\).
4
Multiply by \(u'\): Apply the chain rule. Compute \(\frac{du}{dx}\) and multiply.
5
Simplify: Combine terms, simplify the expression under the radical if possible.
Example 1: Basic Derivative
Find \(\frac{d}{dx}[\arctan(x)]\).
1
Inverse trig function: \(\arctan\)
2
Inner function: \(u = x\), so \(u' = 1\)
3
Base formula: \(\frac{1}{1 + u^2}\)
4
Apply chain rule: \(\frac{1}{1 + x^2} \cdot 1\)
A light-and-camera studio with exact sightlines
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.Camera: angle above horizontal
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: Chain Rule Application
Find \(\frac{d}{dx}\!\left[\arcsin(3x)\right]\).
1
Inverse trig function: \(\arcsin\)
2
Inner function: \(u = 3x\), so \(u' = 3\)
3
Base formula: \(\frac{1}{\sqrt{1 - u^2}} = \frac{1}{\sqrt{1 - (3x)^2}} = \frac{1}{\sqrt{1 - 9x^2}}\)
4
Multiply by \(u'\): \(\frac{3}{\sqrt{1 - 9x^2}}\)
Example 3: More Complex Argument
Find \(\frac{d}{dx}\!\left[\arctan(x^2)\right]\).
1
Inverse trig: \(\arctan\), Inner function: \(u = x^2\), \(u' = 2x\)
2
Base formula: \(\frac{1}{1 + u^2} = \frac{1}{1 + (x^2)^2} = \frac{1}{1 + x^4}\)
3
Chain rule: \(\frac{1}{1 + x^4} \cdot 2x = \frac{2x}{1 + x^4}\)
Example 4: Product Rule Combination
Find \(\frac{d}{dx}\!\left[x \cdot \arcsin(x)\right]\).
1
Product rule: \(\frac{d}{dx}[f \cdot g] = f' g + f g'\)
2
Let \(f = x\), \(g = \arcsin(x)\). Then \(f' = 1\), \(g' = \frac{1}{\sqrt{1-x^2}}\).
3
Combine: \(1 \cdot \arcsin(x) + x \cdot \frac{1}{\sqrt{1-x^2}}\)
Common Mistakes to Avoid
✕ Forgetting the chain rule
Writing \(\frac{1}{\sqrt{1-9x^2}}\) instead of \(\frac{3}{\sqrt{1-9x^2}}\) for \(\frac{d}{dx}[\arcsin(3x)]\). Always multiply by \(u'\).
✕ Wrong sign for arccos
Using \(\frac{+1}{\sqrt{1-x^2}}\) for \(\arccos\). It must be \(\frac{-1}{\sqrt{1-x^2}}\): the negative sign is essential.
✕ Mixing up denominators
Using \(\sqrt{1-x^2}\) for \(\arctan\) (should be \(1+x^2\)) or \(1+x^2\) for \(\arcsin\) (should be \(\sqrt{1-x^2}\)).
Complete Reference Table
| Function | Derivative | Domain of Derivative |
|---|---|---|
| \(\arcsin(x)\) | \(\frac{1}{\sqrt{1-x^2}}\) | \(|x| < 1\) |
| \(\arccos(x)\) | \(\frac{-1}{\sqrt{1-x^2}}\) | \(|x| < 1\) |
| \(\arctan(x)\) | \(\frac{1}{1+x^2}\) | 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 \(\dfrac{d}{dx}[\arcsin(x)]\)?
The derivative of \(\arcsin(x)\) is \(\frac{1}{\sqrt{1-x^2}}\). Choice (a) is the derivative of \(\arccos(x)\), and choice (c) is the derivative of \(\arctan(x)\).
Question 2 of 5
Find \(\dfrac{d}{dx}[\arctan(5x)]\).
With \(u = 5x\) and \(u' = 5\): \(\frac{d}{dx}[\arctan(5x)] = \frac{5}{1+(5x)^2} = \frac{5}{1+25x^2}\). Choice (a) forgets the chain rule factor of 5.
Question 3 of 5
What is \(\dfrac{d}{dx}[\arccos(2x)]\)?
The derivative of \(\arccos(u)\) is \(\frac{-1}{\sqrt{1-u^2}} \cdot u'\). With \(u=2x\): \(\frac{-1}{\sqrt{1-4x^2}} \cdot 2 = \frac{-2}{\sqrt{1-4x^2}}\). Choice (b) uses the wrong sign, and choice (c) forgets the chain rule.
Question 4 of 5
Find \(\dfrac{d}{dx}\!\left[\arcsin(x^3)\right]\).
With \(u = x^3\), \(u' = 3x^2\), and \(u^2 = x^6\): the answer is \(\frac{3x^2}{\sqrt{1 - x^6}}\). Choice (a) forgets the chain rule. Choice (c) incorrectly squares as \(x^3\) instead of \(x^6\).
Question 5 of 5
If \(f(x) = x^2 \cdot \arctan(x)\), what is \(f'(x)\)?
Use the product rule: \(f'(x) = 2x \cdot \arctan(x) + x^2 \cdot \frac{1}{1+x^2}\). Choice (d) forgets the first term of the product rule. Choice (c) incorrectly differentiates \(x^2\) in the second term.