Inverse Function Derivatives (AP Calculus AB Unit 3): General Inverses + Inverse Trig

Differentiating General Inverse Functions

What an inverse function really is

An inverse function “undoes” what the original function does. If a function takes an input and produces an output, then its inverse takes that output and returns the original input.

Formally, if a function has an inverse, the undoing relationships are:

f(f^{-1}(x))=x

f^{-1}(f(x))=x

Not every function has an inverse that is itself a function. To have an inverse function, the original function must be one-to-one on the domain you are using (it must pass the horizontal line test). In AP Calculus, it’s common to restrict the domain of a function (like a quadratic) so that it becomes one-to-one and therefore invertible.

Why inverse derivatives matter

On many AP problems you’re given information about a function and its derivative, but you’re asked about the inverse function’s slope. This comes up because:

• You might be asked for the slope of the inverse at a particular point.
• Inverse trig derivatives (like the derivative of arcsine) are special cases of inverse-function differentiation.
• Many exam questions provide a table or graph of a function and ask for the value of the inverse derivative at a point, without ever giving (or requiring) a formula for the inverse.

The key conceptual idea is that inverse functions swap inputs and outputs, and their tangent line slopes become reciprocals (as long as the original slope is not zero).

The geometric picture: reflection across the line y=x

Geometrically, the graph of an inverse function is the reflection of the original function across the line:

y=x

If a point lies on the original function, then the swapped point lies on the inverse. That same “swap” happens to tangent lines: reflecting a line across y=x changes its slope to its reciprocal (when defined).

The core derivative relationship (Inverse Function Derivative Theorem)

Suppose f is differentiable and one-to-one, and let its inverse be:

g=f^{-1}

Because they undo each other:

f(g(x))=x

Differentiate both sides with respect to x. The left side uses the Chain Rule:

f'(g(x))\cdot g'(x)=1

Solve for the inverse derivative:

g'(x)=\frac{1}{f'(g(x))}

Written directly in inverse-function notation:

(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}

Key interpretation (what the formula is saying): to find the derivative of the inverse at an input value (which was an output of the original function), you first find the corresponding input on the original function, evaluate the original derivative there, then take the reciprocal.

The most practical point-swap version

A very AP-friendly version of the theorem is the “matching point” form. If:

f(a)=b

then the inverse satisfies:

f^{-1}(b)=a

and the slopes satisfy:

(f^{-1})'(b)=\frac{1}{f'(a)}

This is exactly why tables and graphs are so useful: you often don’t need an explicit formula for the inverse.

Conditions and “when things break”

This reciprocal slope idea works cleanly only when the original derivative is not zero at the corresponding point.

• If:

f'(a)\neq 0

then the inverse is differentiable at b=f(a) and the reciprocal formula applies.

• If:

f'(a)=0

then the original function has a horizontal tangent at %%LATEX18%%, and the inverse would have a vertical tangent at %%LATEX19%%. In that case, the inverse derivative is undefined there.

Also remember: if %%LATEX20%% is not one-to-one on the domain you’re using, then %%LATEX21%% is not a function unless you restrict the domain of f.

Two reliable methods on AP problems

There are two standard approaches you’ll use repeatedly.

Method A: Use the inverse derivative formula

Use either form:

(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}

g'(x)=\frac{1}{f'(g(x))}

This is best when you can identify %%LATEX25%%, either because you can compute the inverse explicitly or because you can solve %%LATEX26%%.

Method B: Use implicit differentiation after swapping %%LATEX27%% and %%LATEX28%%

If:

y=f^{-1}(x)

then by definition:

x=f(y)

Differentiate implicitly with respect to x and solve for:

\frac{dy}{dx}

This is especially useful when an explicit inverse is messy or impossible to find.

Step-by-step problem solving strategy (4-step process)

When asked to find:

(f^{-1})'(b)

use this reliable routine:

1. Identify the goal: you need the slope of the inverse at input x=b.
2. Find the corresponding coordinate: solve or look up the value %%LATEX35%% such that %%LATEX36%%.
3. Find the derivative of the original at that input: compute or look up f'(a).
4. Reciprocate:

(f^{-1})'(b)=\frac{1}{f'(a)}

Worked Example 1: Derivative of an inverse at a point (typical AP style)

Let:

f(x)=x^3+5

and let:

g=f^{-1}

Find:

g'(13)

First, find the matching input on f by solving:

f(a)=13

a^3+5=13

a^3=8

a=2

So:

f^{-1}(13)=2

Differentiate the original function:

f'(x)=3x^2

Evaluate at the matching input:

f'(2)=12

Take the reciprocal:

g'(13)=\frac{1}{12}

Interpretation: at the matching point on the original function the slope is large, so the slope on the inverse is small.

Worked Example 2: Using a table/graph idea (reciprocal slopes)

Suppose you are told:

f(4)=10

f'(4)=-2

Let:

h=f^{-1}

Find:

h'(10)

Because f(4)=10, the inverse satisfies:

h(10)=4

Then the reciprocal slope rule gives:

h'(10)=\frac{1}{f'(4)}=-\frac{1}{2}

Worked Example 3: Using a table (additional AP-style practice)

Let f be differentiable and suppose:

f(3)=7

f'(3)=-2

Find:

(f^{-1})'(7)

Since %%LATEX62%%, the corresponding inverse point uses input %%LATEX63%% and matches original input 3. Therefore:

(f^{-1})'(7)=\frac{1}{f'(3)}=-\frac{1}{2}

Worked Example 4: Implicit differentiation approach (no explicit inverse needed)

Find:

\frac{d}{dx}f^{-1}(x)

if:

f(x)=x+\cos x

Express your answer in terms of:

f^{-1}(x)

Let:

y=f^{-1}(x)

Then:

x=f(y)=y+\cos y

Differentiate with respect to x:

1=\frac{dy}{dx}-\sin y\cdot \frac{dy}{dx}

Factor out \frac{dy}{dx}:

1=\frac{dy}{dx}(1-\sin y)

Solve:

\frac{dy}{dx}=\frac{1}{1-\sin y}

Replace %%LATEX76%% with %%LATEX77%%:

(f^{-1})'(x)=\frac{1}{1-\sin(f^{-1}(x))}

Common misconceptions to watch for

A lot of errors come from confusing similar-looking notations or evaluating at the wrong point.

• Confusing the inverse function with the reciprocal. The notation can be misleading, but:

f^{-1}(x)\neq \frac{1}{f(x)}

• Forgetting to find the matching input. If you want:

(f^{-1})'(b)

you must find the value a such that:

f(a)=b

and then use f'(a).

• Trying to “invert the derivative” directly. In general:

(f^{-1})'(x)\neq \frac{1}{f'(x)}

because the derivative must be evaluated at the corresponding swapped point.

Exam Focus

Typical question patterns include (1) being given values like %%LATEX85%% and %%LATEX86%% from a table/graph and being asked for %%LATEX87%%, (2) being given a formula for %%LATEX88%% and a value %%LATEX89%% and being asked to compute %%LATEX90%% by first solving %%LATEX91%%, and (3) deriving a general expression for %%LATEX92%% in terms of f^{-1}(x) using implicit differentiation.

Common mistakes include using %%LATEX94%% instead of %%LATEX95%%, solving the matching equation %%LATEX96%% incorrectly and therefore using the wrong point, ignoring the requirement that %%LATEX97%% (so you report a number where the inverse actually has a vertical tangent), and evaluating the derivative at the wrong value (for example, computing %%LATEX98%% when asked for %%LATEX99%% instead of first finding the input that produces output 5).

Differentiating Inverse Trigonometric Functions

What inverse trig functions are (and why they need special care)

An inverse trigonometric function answers questions like “what angle has this trig value?” For example, arcsine returns the angle whose sine is a given number.

A subtle but crucial point is that trig functions like sine and cosine are not one-to-one over all real numbers, so they do not have inverses on their full domains. To make inverse trig functions into true functions, we use principal-value definitions by restricting the original trig functions to intervals where they are one-to-one.

The standard principal-value restrictions you’re expected to know are:

y=\arcsin(x)\text{ means }\sin(y)=x\text{ with }-\frac{\pi}{2}\le y\le \frac{\pi}{2}

y=\arccos(x)\text{ means }\cos(y)=x\text{ with }0\le y\le \pi

y=\arctan(x)\text{ means }\tan(y)=x\text{ with }-\frac{\pi}{2}