Topic 3.3 Notes – Differentiating Inverse Functions

What the Derivative of an Inverse Function Is

Suppose $f$ is differentiable and one-to-one (so it has an inverse). Let $g = f^{-1}$. Then

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

or equivalently

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

Where this comes from

If two functions are inverses, then
$$f(f^{-1}(x)) = x.$$

Differentiate both sides using the chain rule:

$$f'(f^{-1}(x)) \cdot (f^{-1})'(x) = 1.$$

Solve for $(f^{-1})'(x)$ and you get the formula above. The chain rule is doing all the work.

Geometric Meaning

If $f(a) = b$, then:

• $(a,b)$ is on $f$
• $(b,a)$ is on $f^{-1}$

The graphs reflect across the line $y=x$, and their slopes at corresponding points are reciprocals.

Function and inverse reflected across $y = x$

In the diagram, notice how the tangent line to $f$ at $(f^{-1}(a), a)$ has slope $q/p$, while the tangent line to $f^{-1}$ at $(a, f^{-1}(a))$ has slope $p/q$. They are reciprocals, exactly as the formula predicts.

Important conditions:

• $f$ must be one-to-one (so the inverse exists).
• $f'(a) \neq 0$. If the slope of $f$ is 0, the inverse has a vertical tangent there and its derivative does not exist.

How to Find $(f^{-1})'(a)$ in Practice

Most problems ask for the derivative at a specific number, not the full formula.

If you’re asked for $(f^{-1})'(a)$, use:

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

Here’s the process you’ll use over and over:

1. Find $f^{-1}(a)$
   Solve $f(x) = a$.
   This gives the input where the original function outputs $a$.
2. Evaluate $f'$ at that input.
3. Take the reciprocal.

That’s it.

Example Idea (Formula Given)

Suppose $f(x) = x^3 + x$. Find $(f^{-1})'(2)$.

• First solve $f(x) = 2$:
  $x^3 + x = 2$.
  $x=1$ works. So $f^{-1}(2)=1$.

• Compute derivative:
  $f'(x)=3x^2+1$

• Evaluate at 1:
  $f'(1)=4$

• Take reciprocal:
  $(f^{-1})'(2)=\frac{1}{4}$

Notice we never found the actual inverse formula.

Table Setup (Very Common on FRQs)

You might get a table of $f(x)$ and $f'(x)$.

To find $(f^{-1})'(a)$:

• Look for where $f(x)=a$.
• Use that row’s $f'(x)$.
• Take the reciprocal.

Students often plug $a$ directly into $f'(x)$. That’s wrong. You must match the input that produces $a$.

Tangent Line to an Inverse Function

If you need the tangent line to $y=f^{-1}(x)$ at $x=a$:

1. The point is $(a, f^{-1}(a))$.
2. The slope is $\frac{1}{f'(f^{-1}(a))}$.
3. Use point-slope form:
   $$y - y_1 = m(x - x_1)$$

Careful with coordinates. If $f(4)=7$, then:

• On $f$: $(4,7)$
• On $f^{-1}$: $(7,4)$

AP graders look closely at this swap.

Inverse Trigonometric Derivatives

These come directly from the inverse rule. You are expected to know them:

$$\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}$$

$$\frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1-x^2}}$$

$$\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$$

And if there’s an inner function, use the chain rule:

$$\frac{d}{dx}[\arctan(3x)] = \frac{3}{1+9x^2}$$

The negative on $\arccos x$ is a common mistake.