Calculus

# What is Partial Derivative? Definition, Rules and Examples

## Rachel McLean

Subject Matter Expert

## What is Partial Derivative?

We use partial differentiation to differentiate a function of two or more variables. For example,

$f(x, y) = xy + x^2y$

is a function of two variables.

If we want to find the partial derivative of a two-variable function with respect to $x$, we treat $y$ as a constant and use the notation $\frac{\partial{f}}{\partial{x}}$. If we want to find the partial derivative of a two-variable function with respect to $y$, we treat $x$ as a constant and use the notation $\frac{\partial{f}}{\partial{y}}$.

You can think of $\partial$ as the partial derivative symbol, sometimes called “del.” When you see this symbol, it shows that we’re taking a partial derivative.

This notation should look familiar — it’s just like the derivative of a function in Leibniz’s notation, expressed $\frac{dy}{dx}$, except that the letter “d” is replaced by $\partial$, a stylized curly $d$.

In calculus, derivatives measure the rate of change of a function with respect to a change in its input variable $x$. Since the input of a multivariable function is more than one variable, we call $\frac{\partial{f}}{\partial{x}}$ and $\frac{\partial{f}}{\partial{y}}$ partial derivatives because they only reveal the rate of change of $f$ when one variable changes, instead of both.

The partial derivative allows us to understand the behavior of a multivariable function when ​​we let just one of its variables change, while the rest stay constant.

## How to Do Partial Derivatives

How do partial derivatives work? To find a partial derivative, we find the derivative of a function of two or more variables by treating one of the variables as a constant.

To find $\frac{\partial{f}}{\partial{x}}$:

1. Treat $y$ as a constant.

2. Differentiate the function normally.

To find $\frac{\partial{f}}{\partial{y}}$:

1. Treat $x$ as a constant.

2. Differentiate the function normally.

After we designate one variable as a constant, we can use the derivative rules that are already familiar to us to differentiate the function.

### Partial Derivative Rules

Derivative rules help us differentiate more complicated functions by breaking them into pieces. Here are some of the most common derivative rules to know:

#### Constant Rule

$\frac{d}{dx}c = 0$

#### Power Rule

$\frac{d}{dx}x^n = nx^{n-1}$

#### Chain Rule

$\frac{d}{dx}f(g(x)) = f’(g(x))g’(x)$

#### Product Rule

$\frac{d}{dx}f(x) \cdot g(x) = f’(x) \cdot g(x) + f(x)\cdot g’(x)$

#### Quotient Rule

$\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{g(x)f’(x)-f(x)g’(x)}{(g(x))^2}$

#### Sum/Difference Rule

$\frac{d}{dx}(f(x) \pm g(x)) = f’(x) \pm g’(x)$

#### Trigonometry Rules

$\frac{d}{dx}(\sin{(x)}) = \cos{(x)}$

$\frac{d}{dx}(\cos{(x)}) = -\sin{(x)}$

$\frac{d}{dx}(\tan{(x)}) = \sec ^2 (x)$

#### Logarithmic and Exponential Rules

$\frac{d}{dx} (\ln{x}) = \frac{1}{x}$

$\frac{d}{dx}(e^x) = e^x$

You can view more about these rules in an explanation by one of our instructors Dr. Tim Chartier.

For example, let’s take another look at the function $f(x, y) = xy + x^2y$. Suppose we want to find $\frac{\partial{f}}{\partial{x}}$, the partial derivative with respect to $x$. The first thing to do is treat $y$ as a constant.

What does it mean to treat $y$ as a constant? A constant is a fixed, unchanging value. Examples of constants are 1, 3.5, 17, and 100,000. To treat $y$ as a constant, we imagine that $y$ is any non-zero constant value you choose.

We can do this because of the constant rule, which states that the derivative of any constant is 0.

To make it easier to imagine $y$ as a constant, we can replace $y$ with $c$ or $k$, which are two variables that are commonly used to represent constant values. Using this trick and replacing $y$ with $k$, we have

$f(x, y) = kx + kx^2$

Now, we can find the partial derivative $\frac{\partial{f}}{\partial{x}}$ using the derivative rules. Remember to change $k$ back to $y$ when you have your final answer.

$\frac{\partial{(kx+kx^2)}}{\partial x}=k+2kx$
$\frac{\partial{(xy + x^2y)}}{\partial{x}} = y + 2yx$

We can do the same to find $\frac{\partial{f}}{\partial{y}}$, the partial derivative with respect to $y$. The first thing to do is treat $x$ as a constant. Remember that the square of any constant is simply another constant.

$f(x, y) = ky + k^2y$

Now, we can find the partial derivative $\frac{\partial{f}}{\partial{y}}$ using the derivative rules. Remember to change $k$ back to $x$ when you have your final answer.

$\frac{\partial{(ky + k^2y)}}{\partial{y}} = k + k^2$
$\frac{\partial{(xy + x^2y)}}{\partial{y}} = x + x^2$

## Partial Derivatives Examples

Let’s take a look at some more partial derivative examples.

### Example 1

Find the partial derivatives of $f(r, h) = \pi r^2h$.

### Solution:

This function represents the volume of a cylinder. When we find the partial derivative $\frac{\partial{(\pi r^2h)}}{\partial{r}}$, we find the rate of change of the cylinder’s volume as only the radius changes.

When we find the partial derivative $\frac{\partial{(\pi r^2h)}}{\partial{h}}$, we find the rate of change of the cylinder’s volume as only the height changes. So,

$\frac{\partial{(\pi r^2h)}}{\partial{r}} = 2\pi rh$
$\frac{\partial{(\pi r^2h)}}{\partial{h}} = \pi r^2$

### Example 2

Find the partial derivatives of $f(x, y, z) = xy^3 - zx + z$.

### Solution:

How do partial derivatives work in more than two variables? Just the same! For a function with three variables, we change only one variable and treat the other two as constants. So,

$\frac{\partial{(xy^3 - zx + z)}}{\partial{x}} = y^3 - z$
$\frac{\partial{(xy^3 - zx + z)}}{\partial{y}} = 3xy^2$
$\frac{\partial{(xy^3 - zx + z)}}{\partial{z}} = 1 - x$

### Example 3

Find the partial derivatives of $f(x, y) = x^2\sin{(y)} - y^2\cos{(x)}$.

### Solution:

We can use the trigonometry derivative rules for this problem. Remember that $\sin{(y)}$ acts as a constant when we calculate $\frac{\partial{(f(x, y))}}{\partial{x}}$, and $\cos{(x)}$ acts as a constant when we calculate $\frac{\partial{(f(x, y))}}{\partial{y}}$. So,

$\frac{\partial{(x^2\sin{(y)} - y^2\cos{(x)})}}{\partial{x}} = 2x\sin{(y)} + y^2\sin{(x)}$
$\frac{\partial{(x^2\sin{(y)} - y^2\cos{(x)})}}{\partial{y}} = x^2\cos{(y)} - 2y\cos{(x)}$

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