Rates for inverse functions and inverse trigonometry

Try

A calibration function converts an input into an output. Near input 2, an increase of 0.01 produces approximately 0.04 extra output. If you reverse the calibration, how much input change corresponds to an output increase of 0.01? The reciprocal rate suggests 0.0025, provided the calibration can be inverted locally.

Predict

Reflect a graph across the line y=x. Predict how the slope changes, including the case of a negative slope. Then consider a horizontal tangent: does its reflected tangent have a finite slope? Invertibility alone does not guarantee a finite derivative for the inverse.

Experiment

Read each row as a correspondence, keeping the locations of the original and inverse derivatives distinct.

Original information	Inverse information	Rate
f(2)=5, f′(2)=4	f inverse at 5 gives 2	1/4
f(3)=7, f′(3)=−2	f inverse at 7 gives 3	−1/2
f(x)=x³ at 0	inverse is cube root	no finite slope at 0

The reciprocal is evaluated at the corresponding original input, not at the inverse’s input without conversion. Sketch a steep positive tangent and its reflected shallow tangent. Draw a second negative-slope pair to see that reflection does not change the sign.

Understand

Let y=f inverse of x, so f(y)=x. Differentiating gives f′(y)y′=1. When the local inverse exists with the required differentiability and f′(y) is nonzero, its derivative is 1/f′(y). A standard sufficient setting is a continuously differentiable function with nonzero derivative near the point. The nonzero condition cannot be discarded: x³ is one-to-one, but its inverse has a vertical tangent at zero.

Trigonometric functions require restricted branches before inversion. Arcsine returns an angle y in [−π/2,π/2] with sin y=x, for real x in [−1,1]. On the interior, cos y is positive, so differentiating gives y′=1/cos y=1/√(1−x²). The positive square root is justified by the chosen branch. It is not an arbitrary simplification of cos²y=1−sin²y.

Arccosine returns angles in [0,π]. Since differentiating cos y=x gives −sin y·y′=1 and sine is positive in the branch interior, its derivative is −1/√(1−x²). Both inverse functions exist at their domain endpoints, but neither has a finite derivative there.

Arctangent returns angles in (−π/2,π/2) and has every real input. Differentiating tan y=x gives sec²y·y′=1. Using sec²y=1+tan²y=1+x² yields (arctan x)′=1/(1+x²). Unlike the arcsine formula, this denominator never vanishes for real x.

The notation sin inverse is sometimes ambiguous on paper. Here arcsin means the inverse branch, while csc means the reciprocal 1/sin. They have different domains and different derivatives. Compositions still need chain factors: arccos(2x) has derivative −2/√(1−4x²), finite only for −1/2<x<1/2. Checking the inner argument’s range is part of the calculation.

Apply

Use the practice bank to compute local inverse rates, derive the arcsine formula and diagnose the cube-root exception. In a written derivation, state the selected branch before choosing a square-root sign. Separate the domain where the inverse exists from the smaller domain where its derivative is finite.

Continue

Next use repeated derivatives to analyze motion. The same care with variables and units will distinguish position, velocity, acceleration and speed, which are related but not interchangeable quantities.