AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/3/3-4)
Unit 3 · Topic 3.4
3.4 Differentiating Inverse Trigonometric Functions
The inverse trig functions, like arcsin x and arctan x, have derivatives that don't contain any trig at all, such as 1/√(1 − x²) and 1/(1 + x²). They come from the inverse function idea in 3.3 and usually appear combined with the chain rule.
Key terms
- inverse trigonometric function
- arcsin x
- arccos x
- arctan x
- derivative formulas
The formulas
arcsin x is also written sin⁻¹x, and the −1 means inverse, not reciprocal. These four are the ones to know well. The “co” versions are the negatives of their partners:
| Function | Derivative | Domain of the derivative |
|---|---|---|
| arcsin x | 1/√(1 − x²) | −1 < x < 1 |
| arccos x | −1/√(1 − x²) | −1 < x < 1 |
| arctan x | 1/(1 + x²) | All x |
| arccot x | −1/(1 + x²) | All x |
Where 1/√(1 − x²) comes from
Let y = arcsin x. Then sin y = x, with y between −π/2 and π/2. Differentiate implicitly: cos y·dy/dx = 1, so dy/dx = 1/cos y.
Now write cos y in terms of x. From sin²y + cos²y = 1, cos y = ±√(1 − sin²y) = ±√(1 − x²). Because y is between −π/2 and π/2, cos y ≥ 0, so take the positive root. That gives dy/dx = 1/√(1 − x²).
The same steps with tan y = x give sec²y·dy/dx = 1, and since sec²y = 1 + tan²y = 1 + x², dy/dx = 1/(1 + x²).
The less common two
arcsec x and arccsc x show up rarely. Their derivatives are d/dx arcsec x = 1/(|x|√(x² − 1)) and d/dx arccsc x = −1/(|x|√(x² − 1)), for |x| > 1. (Some books define arcsec with a different range and drop the absolute value, so focus on arcsin, arccos and arctan.)
If you forget a formula, rebuild it with the method above: write the inverse as a regular trig equation, differentiate implicitly and use a Pythagorean identity to rewrite the result in terms of x.
With the chain rule
When the input is something other than x, multiply by the derivative of the inside:
- d/dx arcsin(u) = u′/√(1 − u²)
- d/dx arctan(u) = u′/(1 + u²)
- Example: d/dx arctan(3x) = 3/(1 + 9x²). Note that (3x)² = 9x², not 3x².
Values to know
You'll need exact values of the inverse trig functions themselves, which are angles: arcsin(1/2) = π/6, arccos(1/2) = π/3, arctan 1 = π/4, arctan 0 = 0, arcsin 1 = π/2. Remember the output ranges: arcsin gives angles in [−π/2, π/2], arccos in [0, π], and arctan in (−π/2, π/2).
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Chain rule with arcsin
Differentiate y = arcsin(x²).
Show the solutionHide the solution
- Step 1: Inside: u = x², with u′ = 2x.
- Step 2: d/dx arcsin(u) = u′/√(1 − u²) = 2x/√(1 − (x²)²).
- Step 3: Simplify: (x²)² = x⁴.
Answer: dy/dx = 2x/√(1 − x⁴)
- Example 2
Product rule with arctan
Find the derivative of f(x) = x arctan x, and evaluate f′(1).
Show the solutionHide the solution
- Step 1: Product rule: f′(x) = 1·arctan x + x·1/(1 + x²).
- Step 2: At x = 1: arctan 1 = π/4, and 1/(1 + 1) = 1/2.
- Step 3: f′(1) = π/4 + 1/2.
Answer: f′(x) = arctan x + x/(1 + x²); f′(1) = π/4 + 1/2
- Example 3
Trap: forgetting the inner derivative
A student writes d/dx arctan(3x) = 1/(1 + 9x²). Find the correct derivative.
Show the solutionHide the solution
- Step 1: The student used the arctan formula with u = 3x but didn't multiply by u′ = 3.
- Step 2: Correct: 3/(1 + (3x)²) = 3/(1 + 9x²).
- Step 3: Check at x = 0: the slope of arctan(3x) at 0 should be 3 times the slope of arctan x at 0 (which is 1), so 3. The student's answer gives 1.
Answer: d/dx arctan(3x) = 3/(1 + 9x²)
Common mistakes
- Writing the derivative of arcsin x as 1/√(1 + x²) or 1/(1 − x²). Learn which formula has the root and which has the plus sign.
- Treating sin⁻¹x as 1/sin x. That would be csc x.
- Forgetting the minus sign on arccos x.
On the exam
- Inverse trig derivatives appear mostly in multiple choice. In Unit 6, these same formulas run in reverse to find antiderivatives like ∫ 1/(1 + x²) dx = arctan x + C.
- Be ready to evaluate them at simple points such as x = 0, 1/2 or 1.
Connected topics
Videos
Check yourself
4 questions on 3.4 Differentiating Inverse Trigonometric Functions. Pick an answer to see if you got it, and why.
What is the derivative of y = arctan(3x)?
If f(x) = arcsin(x²), then f′(x) =
If f(x) = arccos x, what is the value of f′(1/2)?
To find the derivative of y = arcsin x for −1 < x < 1, a student writes sin y = x and differentiates both sides to get cos y · (dy/dx) = 1. The student then replaces cos y with √(1 − x²), the positive square root. Which of the following justifies using the positive square root?
0 of 4 answered