Partial Derivatives

Main idea

A partial derivative asks how $f$ changes when you nudge just one input and hold the rest still. That makes it an ordinary derivative in disguise: treat the other variables as constants and differentiate the way you always have.

f_x(x, y) = {\htmlData{sym=0}{\frac{\partial f}{\partial x}}} = \lim_{{\htmlData{sym=1}{h}}\to0}\frac{f(x + {\htmlData{sym=1}{h}}, y) - f(x, y)}{\htmlData{sym=1}{{h}}}

What the symbols mean

$\frac{\partial f}{\partial x}$
Partial with respect to x
How fast f changes as x moves, with y held fixed. The curly ∂ signals that the other variables are held constant.
$h$
Step in x
Only x moves; y stays put.

Explanation

A partial derivative is simply the rate of change of a function when you change just one of its inputs and freeze all the others. It's called "partial" because it only tells part of the story, the change in one direction. It's how you'd ask "how much warmer does it get if I step east, without stepping north at all?" The formula looks like the usual derivative definition, and it is: $f(x + h, y) - f(x, y)$ (the change in height when only $x$ moves by $h$ ) divided by $h$ (the size of the step). The curly $\partial$ is just a reminder that other variables are being held still.

So why does this work so smoothly? Freeze $y$ at some value, say $y = 2$ , and $f(x, 2)$ is a function of $x$ alone. Its derivative is plain single-variable calculus. That means every rule you know, power, product, quotient and chain, still works. The only new skill is keeping track of what's constant:

In $\frac{\partial}{\partial x}\left(x^2y^3\right)$ , $y^3$ is just a constant factor, like the $5$ in $5x^2$ : $2xy^3$ .
In $\frac{\partial}{\partial y}\left(x^2y^3\right)$ , now $x^2$ is the constant: $3x^2y^2$ .

Notation: $f_x$ , $\frac{\partial f}{\partial x}$ , $\partial_x f$ and $D_x f$ all mean the same thing.

With three or more variables it's the same idea: hold all the others fixed. There's an animation further down showing what this looks like on a surface.

See it

The hill z = 3 − 0.5x² − 0.25y², sliced along y = 0.8. Freezing y leaves an ordinary curve, and the arrow is its slope, ∂f/∂x = −x, as the point slides across. Notice the slope is +1.8 on the left, 0 at the top of the slice, and −1.8 on the right: it's just one-variable calculus on that curve.

Worked examples

Example 1

Find $f_x$ and $f_y$ for $f(x, y) = x^2y^3 + \sin(xy)$ .

1
For $f_x$ , treat $y$ as a constant. The first term gives $2xy^3$ . For $\sin(xy)$ use the chain rule: the inside, $xy$ , has $x$ -derivative $y$ . So $f_x = 2xy^3 + y\cos(xy)$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

For $f(x, y) = x^2y^3 + 4x$ , find $f_x(1, 2)$ .

Problem 2+5 XP

For the same $f$ , find $f_y(1, 2)$ .

Problem 3+5 XP

Find $\frac{\partial}{\partial x}e^{xy}$ .

Problem 4+5 XP

For $f(x, y) = \ln(x^2 + y^2)$ , find $f_x(1, 1)$ .

Problem 5+5 XP

Find $\frac{\partial}{\partial y}\left(x\sin y\right)$ .

Problem 6+5 XP

A circuit board's temperature is $T(x, y) = 60 + 20e^{-(x^2 + 2y^2)}$ °C, with a hot chip at the origin ( $x$ and $y$ in cm). At $(1, 0.5)$ , how fast does the temperature change as you move in the $+x$ direction?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

For $f(x, y) = x^y$ , find $f_x(2, 3)$ and $f_y(2, 3)$ .

Where it's used

Every machine-learning model is trained with partial derivatives: the loss depends on millions of weights, and each update needs $\frac{\partial L}{\partial w}$ for every one of them.

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