---
title: "Derivatives"
description: "A derivative is a local rate of change. Watch the ratios for x² approach 6 at x = 3, derive 2x from the limit, and use it to estimate small changes."
canonical_url: "https://fanout.sh/ai/lessons/math-fundamentals-derivatives"
md_url: "https://fanout.sh/ai/lessons/math-fundamentals-derivatives.md"
access: "public"
---

# Derivatives

A derivative is a local rate of change. Watch the ratios for x² approach 6 at x = 3, derive 2x from the limit, and use it to estimate small changes.

## Public overview

Take the rule f(x) = 2x + 1: double the input, then add 1. Increasing x from 3 to 4 changes the output from 7 to 9. Wherever we start, adding 1 to the input adds 2 to the output.

For g(x) = x², the rate depends on where you start. Around x = 3, a small change in x produces roughly six times that change in the output. Let's inspect the numbers.

Input change Output change Output change ÷ input change --- --- --- 3 → 3.1 9 → 9.61, a change of 0.61 6.1 3 → 3.01 9 → 9.0601, a change of 0.0601 6.01 3 → 3.001 9 → 9.006001, a change of 0.006001 6.001

These ratios approach 6. That limiting local rate is the derivative at x = 3. We write it as g′(3) = 6. The prime mark means “derivative.”

Use the rate before memorizing rules

If x moves from 3 to 3.02, the derivative predicts an output change of approximately 6 × 0.02 = 0.12. The exact change is 3.02² − 3² = 0.1204. Close, but not identical: the curve's slope changes as we move.

Here h is a small input change. The derivative is the rate; multiplying by h estimates the output change. This approximation applies where the function is differentiable, and becomes useful over a sufficiently small step.

Try predicting: At x = −2, the derivative of x² is −4. If x increases to −1.99, should the output increase or decrease? Estimate the change.

answer It should decrease. The input change is +0.01, so the estimated output change is −4 × 0.01 = −0.04. The exact change is (−1.99)² − (−2)² = −0.0399.

A negative derivative means increasing the input locally decreases the output. It does not mean the output itself is negative.

Where the 6 came from

For x², compare the outputs at x and x + h, then divide by the nonzero input change:

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