Derivatives of Inverse Trigonometric Functions
Derive d/dx[sin⁻¹x] = 1/√(1−x²), d/dx[cos⁻¹x] = −1/√(1−x²), d/dx[tan⁻¹x] = 1/(1+x²) by implicit differentiation. Apply the chain rule extension and identify these as integral formulas in reverse.
A lesson of AP Calculus AB, in Differentiation: Composite, Implicit, and Inverse, about 10 minutes.
By the end of this lesson, you will be able to find derivatives of inverse trigonometric functions, apply them in combination with other differentiation rules, and recognize when inverse trig derivatives appear in integration problems.
The Inverse Trig Functions
The inverse trig functions arise when we "undo" trigonometry — given a ratio, find the angle.
- arcsin(x): the angle whose sine is x; domain [-1,1]; range [-π/2, π/2]
- arccos(x): the angle whose cosine is x; domain [-1,1]; range [0, π]
- arctan(x): the angle whose tangent is x; domain all reals; range (-π/2, π/2)
Notation: arcsin(x) = sin⁻¹(x); arctan(x) = tan⁻¹(x); etc.
Derivatives of Inverse Trig Functions
These are derived using implicit differentiation. You should memorize the results:
| Function | Derivative |
|---|---|
| d/dx[arcsin(x)] | 1/√(1-x²) |
| d/dx[arccos(x)] | -1/√(1-x²) |
| d/dx[arctan(x)] | 1/(1+x²) |
| d/dx[arccsc(x)] | -1/(x√(x²-1)) |
| d/dx[arcsec(x)] | 1/(x√(x²-1)) |
| d/dx[arccot(x)] | -1/(1+x²) |
For AP Calculus AB, focus on arcsin, arccos, and arctan. The others appear less frequently.
Pairs: Notice that arcsin and arccos derivatives are negatives of each other; arctan and arccot are negatives. This reflects the fact that arcsin(x) + arccos(x) = π/2 (constant), so their derivatives must sum to 0.
Deriving d/dx[arctan(x)]
Method: Let y = arctan(x). Then tan(y) = x.
Differentiate both sides with respect to x (implicit differentiation):
sec²(y) · dy/dx = 1
dy/dx = 1/sec²(y) = cos²(y)
Converting back to x: Since tan(y) = x, draw a right triangle: opposite = x, adjacent = 1, hypotenuse = √(1+x²).
cos(y) = 1/√(1+x²), so cos²(y) = 1/(1+x²).
Therefore: d/dx[arctan(x)] = 1/(1+x²) ✓