Topic 3.3 Notes – Differentiating Inverse Functions

1. What the Derivative of an Inverse Function Is

If $f$ is differentiable and has an inverse $f^{-1}$, then

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

Read that slowly. The derivative of the inverse at $x$ equals 1 divided by the derivative of the original function, evaluated at the corresponding input.

Where this comes from

You already know that
$$f(f^{-1}(x)) = x.$$

Differentiate both sides:

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

Solve for $(f^{-1})'(x)$ and you get the formula above. It’s just the chain rule in action.

Geometric meaning

A function and its inverse reflect across the line $y = x$. In the graph below, the dark curve is $y=f(x)$ and the lighter curve is $y=f^{-1}(x)$, mirrored across that line.

Notice the labeled points $(-1, \tfrac{1}{2})$ and $(\tfrac{1}{2}, -1)$. The coordinates swap, which is exactly what an inverse does.

At corresponding points:

• If $f'(a) = m$,
• Then $(f^{-1})'(f(a)) = \frac{1}{m}$.

So their slopes are reciprocals.

Important conditions:

• The function must be one-to-one (so the inverse exists).
• It must be differentiable.
• And $f'(a) \neq 0$. If $f'(a) = 0$, the inverse has a vertical tangent there.

That reciprocal relationship is the whole story.

2. How to Find the Derivative of an Inverse at a Point

This is the most common quiz and FRQ setup.

You’re asked for something like
$$(f^{-1})'(4).$$

You cannot just plug 4 into $f'$. You first have to find the matching input.

The process

1. Find the corresponding value.
   Solve
   $$f(a) = 4.$$
   That means $f^{-1}(4) = a$.

2. Use the formula.
   $$(f^{-1})'(4) = \frac{1}{f'(a)}.$$

That’s it.

Example idea (table setup)

Suppose a table tells you:

• $f(2) = 4$
• $f'(2) = 6$

Then:

• $f^{-1}(4) = 2$
• $(f^{-1})'(4) = \frac{1}{6}$

On FRQs, this often becomes a tangent line question. If the problem asks for the tangent line to $y = f^{-1}(x)$ at $x = 4$:

• Point: $(4, 2)$
• Slope: $\frac{1}{6}$

Use point-slope form:

$$y - 2 = \frac{1}{6}(x - 4)$$

The most common mistake is plugging 4 into $f'$ instead of finding the matching input first.

3. Derivatives of Inverse Trigonometric Functions

You are expected to know these.

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

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

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

These come from the same inverse-function idea, but you don’t have to re-derive them on the exam.

When there’s an inside function

If you see something like:

$$\arctan(5x),$$

you must use the chain rule:

$$\frac{1}{1 + (5x)^2} \cdot 5.$$

Same pattern for $\arcsin(g(x))$ and $\arccos(g(x))$. Always multiply by $g'(x)$.

A common multiple-choice trap is forgetting that extra factor.

4. Common Mistakes and Exam Traps

• Mixing up inverse and reciprocal.
  $\sin^{-1}x$ means arcsin, not $1/\sin x$.

• Using
  $$\frac{1}{f'(b)}$$
  instead of
  $$\frac{1}{f'(a)}$$
  where $f(a)=b$.

• Ignoring when $f'(a)=0$. That means the inverse derivative does not exist there.

• Dropping the negative sign for $\arccos x$.

On no-calculator sections, inverse trig derivatives show up inside larger chain rule problems. On FRQs, inverse derivatives often appear in table-based reasoning where you must clearly show the reciprocal relationship.