Partial Derivatives

Partial derivatives tell you how a function with more than one input changes when you change just one variable and hold the others constant. If you searched for how to find partial derivatives, that is the rule: differentiate with respect to one variable, and treat the rest like constants.

For a function $f(x,y)$, the two most common first partial derivatives are $f_x$ and $f_y$:

$$f_x = \frac{\partial f}{\partial x}, \qquad f_y = \frac{\partial f}{\partial y}.$$

The symbol $\frac{\partial f}{\partial x}$ means differentiate with respect to $x$ while treating $y$ as fixed. The symbol $\frac{\partial f}{\partial y}$ means do the same thing with respect to $y$ while treating $x$ as fixed.

What partial derivatives mean

A regular derivative measures change for a one-variable function. A partial derivative does the same job for a function of several variables, one direction at a time.

For example, if temperature is modeled by $T(x,y)$, then $\frac{\partial T}{\partial x}$ measures how temperature changes as you move in the $x$ direction while staying at the same $y$ value. That "same $y$ value" condition is the whole idea.

How to find a partial derivative

Use this checklist:

1. Pick the variable you want to differentiate with respect to.
2. Treat every other variable as a constant.
3. Apply the usual derivative rules.
4. Substitute a point only after you have found the derivative formula.

Worked example: find $f_x$ and $f_y$

Let

$$f(x,y) = x^2y + 3y^2.$$

Find the first partial derivatives with respect to $x$ and $y$.

Step 1: Find $f_x$

Hold $y$ constant. Then $x^2y$ acts like a constant multiple of $x^2$, and $3y^2$ is just a constant with respect to $x$:

$$\frac{\partial}{\partial x}(x^2y + 3y^2).$$ $$f_x(x,y) = 2xy.$$

Step 2: Find $f_y$

Now hold $x$ constant. The term $x^2y$ differentiates like $x^2 \cdot y$, where $x^2$ is a constant multiplier:

$$\frac{\partial}{\partial y}(x^2y + 3y^2).$$ $$f_y(x,y) = x^2 + 6y.$$

So the two first partial derivatives are

$$f_x(x,y) = 2xy, \qquad f_y(x,y) = x^2 + 6y.$$

If the problem asks for the values at $(1,2)$, substitute after differentiating:

$$f_x(1,2) = 2(1)(2) = 4, \qquad f_y(1,2) = 1^2 + 6(2) = 13.$$

This example shows the main pattern: the variable you are not using behaves like a number during that derivative.

Why "hold the other variable constant" matters

When you compute $\frac{\partial f}{\partial x}$, you are asking for change along the $x$ direction only. So every variable other than $x$ is fixed for that calculation.

That is why

$$\frac{\partial}{\partial x}(3y^2) = 0$$

in the example above. The expression $3y^2$ can depend on $y$, but it does not change as $x$ changes when $y$ is held fixed.

Common mistakes

1. Differentiating with respect to $x$ while still letting $y$ change.
2. Forgetting that a term with no chosen variable becomes a constant, so its derivative is $0$.
3. Mixing up $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. They answer different questions.
4. Plugging in a point before taking the derivative, which can hide the structure of the function.
5. Assuming partial derivatives automatically exist everywhere. They can fail to exist at points where the function is not well behaved.

When partial derivatives are used

Partial derivatives appear in multivariable calculus whenever an output depends on several inputs.

Common uses include gradients, tangent planes, optimization, differential equations, and models from physics, economics, and engineering. In each case, the practical question is similar: what happens if one input changes while the others stay fixed?

A mental picture that helps

Think of the graph of $z = f(x,y)$ as a surface. The partial derivative $\frac{\partial f}{\partial x}$ tells you the slope of that surface if you slice it in the direction where $y$ is fixed. The partial derivative $\frac{\partial f}{\partial y}$ does the same in the direction where $x$ is fixed.

That picture is often enough to make the idea click before you move on to gradients or tangent planes.

Try a similar problem

Try

$$g(x,y) = x^3 - 2xy + y^2.$$

Find $g_x$ and $g_y$, then evaluate both at $(2,1)$. If you want a next step, try your own version first and then compare it with a solver to check whether you really held the other variable constant each time.