---
title: "Partial derivatives"
description: "Compute first partial derivatives by holding every other variable fixed, use the limit definition at a piecewise point, and work a two-variable example."
canonical_url: "https://fanout.sh/ml-math/lessons/calculus-introduction-to-partial-derivatives-t-1txalv9"
md_url: "https://fanout.sh/ml-math/lessons/calculus-introduction-to-partial-derivatives-t-1txalv9.md"
access: "public"
---

# Partial derivatives

Compute first partial derivatives by holding every other variable fixed, use the limit definition at a piecewise point, and work a two-variable example.

A function can change when either of its inputs changes. Start by finding valid inputs and equal outputs, then use coordinate slopes to build a local approximation and follow changes through a composition.

## Change one input while holding the other fixed

Choose the real function $F(x,y)=x^2-xy+2y^2$. At (2,1) its output is 4. We want to know how that output changes when x moves a little while y stays at 1. A derivative measures the limiting output change per unit input change.

### Start with one input

1. For q(x)=x², change x to x+h, with h≠0. The output change is (x+h)²−x²=2xh+h². Divide by the input change to obtain $\frac{q(x+h)-q(x)}h=2x+h$.
2. As h approaches zero from either side, 2x+h approaches 2x. We write q′(x)=2x. At x=2 the derivative is 4. This does not say a finite change is exactly 4h: the exact output change there is 4h+h².

For two inputs, hold one fixed and apply the same definition to the other. The partial derivative with respect to x is written Fₓ or ∂F/∂x. The subscript names the input that changes; y has one fixed value throughout this limit.

$$\begin{aligned}F_x(x,y)&=\lim_{h\to0}\frac{F(x+h,y)-F(x,y)}h,\\F_y(x,y)&=\lim_{h\to0}\frac{F(x,y+h)-F(x,y)}h.\end{aligned}$$

Only the selected input receives the increment h. A finite two-sided limit is required.

### Find both slopes from the actual increments

1. Changing x gives $F(x+h,y)-F(x,y)=2xh+h^2-hy$. Dividing by h gives 2x+h−y, so Fₓ=2x−y. The term 2y² cancels because y did not change.
2. Changing y gives $F(x,y+h)-F(x,y)=-xh+4yh+2h^2$. Dividing by h gives −x+4y+2h, so Fᵧ=−x+4y. Now x² cancels because x did not change.
3. These are derivative functions. Evaluate them at (2,1) only afterward: Fₓ(2,1)=3 and Fᵧ(2,1)=2. Small x-only changes produce about three times that change in the output; small y-only changes produce about twice the change.

A constant term has derivative zero because it has no output change. Derivatives of sums add, and a fixed multiplier stays outside the derivative: both facts follow by splitting the difference quotient before taking its limit. Together with the square calculation, these rules reproduce both derivatives above. Holding y fixed makes it a constant for this calculation, even though F also accepts other y-values.

One partial answers one question

Fₓ describes changes with y held fixed. If y also changes as x changes, Fₓ alone does not give the whole output change. We will account for both paths when we reach the chain rule.

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