AP Calculus BC Unit 3 Notes: Derivatives Involving Inverses

Inverse function

A function that reverses the action of another function: if $f(x)=y$, then $f^{-1}(y)=x$ (when $f$ is one-to-one on its domain).

One-to-one function (injective)

A function where each output corresponds to exactly one input; required so the inverse is a function.

Inverse composition identities

The “undoing” equations for inverses: $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$ (on the appropriate domains).

Inverse notation $f^{-1}(x)$

Means “the inverse function of f evaluated at x,” not a reciprocal and not a power.

Reciprocal of a function value

1/$(f(x))$ (or [$f(x)$]^{-1} as a number); this is NOT the same as the inverse function $f^{-1}(x)$.

Input-output swap property

If f(a)=b, then f^{-1}(b)=a (inverse functions swap inputs and outputs).

Graphical relationship of inverses

Graphs of $f$ and $f^{-1}$ are reflections across the line $y=x$.

General inverse derivative formula

($f^{-1}$)'(x)=1 / $f'$($f^{-1}$($x$)) (obtained by differentiating $f(f^{-1}(x))=x$ using the chain rule).

Point-specific inverse derivative rule

If $f(a)=b$ and $f'(a)≠0$, then ($f^{-1}$)'($b$)=1/$f'(a)$.

Condition: invertibility for inverse derivatives

To use inverse-derivative ideas, $f$ must be one-to-one on an interval so $f^{-1}$ exists as a function.

Condition: nonzero derivative for inverse slope

If $f'(a)=0$, then the inverse has an undefined/infinite slope at $b=f(a)$, so a finite ($f^{-1}$)'($b$) cannot be found via 1/$f'(a)$.

Cusp/corner issue for inverse differentiability

If $f$ is not differentiable at a (cusp/corner), then $f^{-1}$ may fail to be differentiable at $b=f(a)$ as well.

Common mistake: wrong input for inverse derivative

Using ($f^{-1}$)'(a) instead of ($f^{-1}$)'(b) where b=$f(a)$; the inverse derivative is evaluated at the original output.

Why inverse derivatives are useful (AP context)

They let you find slopes of $f^{-1}$ using values of $f$ and $f'$ from a table/graph without explicitly solving for $f^{-1}$.

Inverse trigonometric function meaning

An inverse trig function returns an angle whose trig value equals the input (e.g., $\text{arcsin}(x)$ is the angle $y$ such that $\text{sin}(y)=x$).

Arcsine (arcsin) principal range

For $y=\text{arcsin}(x)$, y is restricted to $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$ so the inverse is a function.

Arccosine (arccos) principal range

For $y = \text{arccos}(x)$, $y$ is restricted to $0 \, \text{and} \, \frac{\text{}\text{1}}{\text{}} \, \text{and} \, y \\leq \, \text{π}$ so the inverse is a function.

Arctangent (arctan) principal range

For $y = \text{arctan}(x)$, $y$ is restricted to $-\frac{\text{π}}{2} < y < \frac{\text{π}}{2}$ so the inverse is a function.

Derivative of arcsin(x)

$\frac{d}{dx}[\text{arcsin}(x)] = \frac{1}{\sqrt{1-x^2}}$ (for $|x| < 1$; and defined appropriately at endpoints).

Derivative of arccos(x)

d/dx[arccos(x)] = −1/$\sqrt{1−x^2}$.

Derivative of arctan(x)

d/dx[arctan(x)] = 1/(1+x^2).

Derivative of arccot(x)

d/dx[arccot(x)] = −1/(1+x^2) (commonly used in AP Calculus BC).

Derivative of arcsec(x)

d/dx[arcsec(x)] = 1/(|x|$\sqrt{x^2−1}$); the absolute value is essential.

Derivative of arccsc(x)

d/dx[arccsc(x)] = −1/(|x|√(x^2−1)); the absolute value is essential.

Procedure selection principle (derivatives)

Choose the differentiation method based on structure: use inverse-derivative rules for $f^{-1}$, implicit differentiation when $y$ isn’t isolated (or when you have $f(y)=x$), and inverse-trig formulas + chain rule when the outer function is $\text{arcsin}$/$\text{arccos}$/$\text{arctan}$/etc.