Derivative of Inverse Functions- Rules and Examples

What You're Actually Learning Here

Inverse functions sound complicated until you realize they're just "undoing" operations. If f(x) takes you forward, the inverse f⁻¹(x) takes you back. The derivative of an inverse function tells you how fast that return trip changes.

This isn't theoretical garbage. Engineers use this to model feedback systems. Economists use it to find marginal rates. You need it for calculus exams. Let's get to it.

The Core Formula

Here's what you need to know:

If y = f⁻¹(x), then dy/dx = 1 / (f'(y))

That's it. The derivative of the inverse at x equals 1 divided by the derivative of the original function evaluated at y, where y = f⁻¹(x).

The condition is simple: f'(y) ≠ 0. If the original derivative is zero, the inverse derivative either doesn't exist or goes to infinity. That's not a loophole—it's just math telling you the function isn't invertible there.

Why This Formula Works

You don't need to prove it to use it, but understanding helps when you get stuck.

If y = f⁻¹(x), then f(y) = x by definition. Take derivatives of both sides using the chain rule:

f'(y) · dy/dx = 1

Solve for dy/dx:

dy/dx = 1 / f'(y)

The proof is three lines. The confusion usually comes from tracking which variable goes where.

Worked Examples

Example 1: Square Root Function

Find the derivative of f⁻¹(x) = √x.

Your original function is f(x) = x². Its inverse is f⁻¹(x) = √x (restricted to x ≥ 0).

Apply the formula: f'(x) = 2x

So d/dx [√x] = 1 / (2√x)

Check it. The derivative of √x is indeed 1/(2√x). It works.

Example 2: Natural Logarithm

Find the derivative of f⁻¹(x) = ln(x).

Your original function is f(x) = eˣ. Its inverse is f⁻¹(x) = ln(x).

Apply the formula: f'(x) = eˣ

So d/dx [ln(x)] = 1 / e^(ln(x)) = 1 / x

You already know the derivative of ln(x) is 1/x. This confirms it.

Example 3: Inverse Trigonometric Function

Find the derivative of f⁻¹(x) = arctan(x).

Your original function is f(x) = tan(x), restricted to (-π/2, π/2) where it's one-to-one.

Apply the formula: f'(x) = sec²(x)

So d/dx [arctan(x)] = 1 / sec²(arctan(x))

You need to simplify. If y = arctan(x), then x = tan(y). Using the identity 1 + tan²(y) = sec²(y):

d/dx [arctan(x)] = 1 / (1 + tan²(y)) = 1 / (1 + x²)

This is the standard result.

Common Inverse Functions and Their Derivatives

Commit the general pattern to memory: inverse derivatives involve a reciprocal and evaluation at the inverse point. The table above is your reference sheet.

Getting Started: Step-by-Step Process

When you need to find the derivative of an inverse function, follow this:

• Identify the original function f(x). What did this inverse come from? If you're given arctan(x), the original is tan(x).
• Find f'(x). Take the derivative of the original function normally.
• Express y = f⁻¹(x). This is just renaming the inverse function you're working with.
• Substitute into the formula. You need f'(y), which means evaluate f' at the point y instead of x.
• Rewrite in terms of x. Replace y with f⁻¹(x) to get your final answer in terms of the original variable.
• Simplify. Use algebraic tricks or trig identities to clean up the expression.

Practice this sequence until it becomes automatic. Most mistakes happen at step 4 or 5—people forget to evaluate at the right point.

Where Students Go Wrong

Forgetting the domain restriction. The formula only works where f'(y) ≠ 0. Inverse trig functions have specific domains for this reason. You can't just apply it everywhere.

Swapping x and y incorrectly. When you set up f(y) = x, stick with it. Some students flip back to x = f(y) and then get confused about which derivative is which.

Not simplifying the final expression. You're often left with something like 1/(2√y) when you need 1/(2√x). The answer must be expressed in terms of the original variable.

Confusing notation. f⁻¹(x) is the inverse function, not 1/f(x). This trips people up constantly. The exponent -1 means inverse, not reciprocal.

When You'll Actually Use This

Beyond exams, this shows up in:

• Implicit differentiation problems where you need dy/dx but y is defined implicitly as a function of x
• Related rates where one quantity is the inverse of another
• Differential equations where you're solving for inverse relationships
• Statistics when working with inverse cumulative distribution functions

The formula is a tool. You learn it once, you use it whenever the situation calls for it.