Derivatives of Inverse Functions

You’ll be able to

Use the inverse-function relationship to compute a derivative at a point
State and apply the derivatives of the inverse trigonometric functions
Relate the slope of a function and its inverse as reciprocals at matching points

Slopes of inverses are reciprocals

A function and its inverse are reflections across the line y = x, and that reflection swaps rise and run — so their slopes at corresponding points are reciprocals. If g is the inverse of f, then g′(x) = 1/f′(g(x)). To find the inverse’s slope at x = a, locate the matching input b = g(a) on f, compute f′(b), and take its reciprocal. You never need a formula for the inverse itself.

The inverse trig derivatives

The inverse trigonometric functions have clean algebraic derivatives worth memorizing: d/dx[arcsin x] = 1/√(1 − x²), d/dx[arctan x] = 1/(1 + x²), and d/dx[arcsec x] = 1/(|x|√(x² − 1)). Their "co-" partners are just the negatives: d/dx[arccos x] = −1/√(1 − x²), and d/dx[arccot x] = −1/(1 + x²). Combine any of these with the chain rule when the input is not a bare x.

Derivative of an inverse function

If g = f⁻¹, then g′(a) = 1 / f′(g(a))

The inverse’s slope at a is the reciprocal of f’s slope at the matching point b = g(a) where f(b) = a.

Key inverse trig derivatives

d/dx[arcsin x] = 1/√(1 − x²) · d/dx[arctan x] = 1/(1 + x²)

And d/dx[arccos x] = −1/√(1 − x²). Chain rule applies for arcsin(u): u′/√(1 − u²).

Worked example

Let f(x) = x³ + 2x + 1, and let g = f⁻¹. Given that f(1) = 4, find g′(4).

1.Since f(1) = 4, the matching point is g(4) = 1.
2.Compute f′(x) = 3x² + 2.
3.Evaluate at the matching input: f′(1) = 3(1)² + 2 = 5.
4.The inverse derivative is the reciprocal: g′(4) = 1/f′(1) = 1/5.

Answer: g′(4) = 1/5. Notice you never had to find a formula for g — only the reciprocal of f′ at the matching point.

Tip

For g′(a), always find the matching point first: the b with f(b) = a. Then g′(a) = 1/f′(b). Plugging a straight into f′ instead of b is the usual slip — remember to cross over to the other graph.

Checkpoint

What is d/dx[arctan(x)]?

Checkpoint

f is invertible with f(2) = 5 and f′(2) = 3. If g = f⁻¹, what is g′(5)?

On the exam

The formula g′(a) = 1/f′(g(a)) shows up on the AP exam most often in table form: you are handed values of f and f′ and asked for the inverse’s derivative. Identify the matching point from the table, then take the reciprocal.

Answer the 2 checkpoints as you read.