Derivatives of Inverse Functions
- Apply the inverse-function derivative formula
- Differentiate inverse trigonometric functions
- Relate the slopes of a function and its inverse at corresponding points
Slopes of inverses are reciprocals
If g is the inverse of f, then their graphs are reflections across the line y = x, and at corresponding points their tangent slopes are reciprocals. Concretely, if f(a) = b, then g′(b) = 1 / f′(a). You never need a formula for g itself — you only need the point correspondence and the derivative of f. This lets you find the slope of an inverse even when the inverse cannot be written explicitly.
Inverse trig derivatives
Applying this idea to the inverse trig functions yields formulas worth memorizing. The three most tested are: d/dx[arcsin x] = 1/√(1 − x²), d/dx[arctan x] = 1/(1 + x²), and d/dx[arccos x] = −1/√(1 − x²) (the negative of arcsin’s). Combined with the chain rule, these appear throughout differentiation and — reversed — as key antiderivatives in integration. The arctan form in particular is the antiderivative you reach for whenever you see 1/(1 + x²).
Let f(x) = x³ + 2x + 1, so f(1) = 4. If g is the inverse of f, find g′(4).
- 1.The inverse-derivative rule gives g′(4) = 1 / f′(1), using the correspondence f(1) = 4.
- 2.Differentiate f: f′(x) = 3x² + 2.
- 3.Evaluate at the matching input x = 1: f′(1) = 3(1)² + 2 = 5.
- 4.Take the reciprocal: g′(4) = 1/5.
For g′(b) you must evaluate f′ at the input a where f(a) = b — not at b. Plugging b directly into f′ is the standard trap. Always find the matching input first: solve f(a) = b, then use that a.
f is invertible with f(2) = 7 and f′(2) = 4. If g = f⁻¹, what is g′(7)?
What is the derivative of y = arctan(x)?
Inverse-function problems on the AP exam almost always give you a table or a point pairing. The move is mechanical: find where f equals the target output, evaluate f′ there, take the reciprocal. Practice it until it is automatic.
Answer the 2 checkpoints as you read.
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