Definition

The partial derivative of a function of more than one variable is its derivative with respect to one of those variables, holding the others constant. The partial derivative (sometimes "the partial") gives the slope of a function along one of its dimensions.

The notation is similar to the Leibniz notation for derivatives, but with a different-looking "d":

$$\frac{\partial f(x, y)}{\partial x}$$

A quick example

Consider the two-variable function $f(x, y) = 4x^3 + 3x^2y - 2xy^2 + y^3$.

(a) Find the partial derivative of $f$ with respect to $x$.

Here we treat the variable $y$ as a constant and find $\frac{df}{dx}$

$$\frac{\partial f(x, y)}{\partial x} = 12x^2 + 6xy - 2y^2$$

Pro tip: getting the hang of it

Derivation

Consider a function of two variables, $f(x, y)$, and let $(x, y)$ be a point in the domain of $f(x, y)$. Then

$$ \begin{align} \frac{\partial f(x,y)}{\partial x} = \lim_{h \rightarrow 0} \frac{f(x + h, y) - f(x, y)}{h} \\[8pt] \frac{\partial f(x,y)}{\partial y} = \lim_{k \rightarrow 0} \frac{f(x, y + k) - f(x, y)}{k} \end{align}$$

as long as that limit exists. This is just how we developed the derivative for a function of a single variable.

In more general terms, the partial derivatives of a function of an $n$-dimensional vector $\vec a$

We define $f(\vec a) = f(a_1, a_2, \dots , a_n)$, where $\vec a$ is an $n$-dimensional vector. Then we define the partial derivative of $f$ with respect to the $i^{th}$ independent variable, $x_i$ as

$$ \begin{align}\frac{\partial f(\vec a)}{\partial x_i} &= \lim_{h \rightarrow 0} \frac{f(a_1, a_2, \dots, a_{i-1}, a_i + h, a_{i+1}, \dots , a_n) - f(a_1, a_2, \dots, a_i, \dots, a_n)}{h} \\[5pt] &= \lim_{h \rightarrow 0} \frac{f(\vec a + h\vec{e}_i)- f(\vec a)}{h} \end{align}$$

In the last expression, the vector $\vec {e}_i$ is the unit vector in the direction of our variable $x_i$. for example, for a three-variable equation in which our function $f$ is a function of the 3D vector $\vec a = (x_1, x_2, x_3)$, and we're taking the partial derivative with respect to $x_3$, $\vec{e}_3 = (0, 0, 1)$. It's just a fancy way of selecting the variable for which we're finding the slope of this function.

Example 1

Find and interpret the partial derivatives of $z = 2x^2-3y^2 + 4$.

The graph of this function is shown below. It has a saddle point at $(0, 0, 4)$

Example

Calculate the gradient of the 3D function $f(x, y) = 1 - \frac{1}{2} x^2 - y^2$.

Solution: The part of the function near $(x, y) = (0, 0)$ is graphed below. The gradient is

$$ \begin{align} \nabla f(x, y) &= \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \\[5pt] &= (-x, -2y) \end{align}$$

In the graph below, the unit vectors for the direction of the gradients at $(0, -\frac{1}{2})$ and $(\frac{1}{2}, -\frac{1}{2})$ are shown.

Here's a top view of that paraboloid showing the same unit vectors marking those gradients vectors. You're looking down into that "box," along the z-axis.

Practice problems

Consider the multidimensional functions below. For 3D functions (functions of $x$ and $y$), findl $\frac{\partial f(x,y)}{\partial x}$ and $\frac{\partial f(x,y)}{\partial y}$. For 4D functions (functions of $x$, $y$ and $z$) find the partial derivative with respect to $z$ as well.

1. $f(x, y) = x^4 y^3 - 7 x^2 y - y^4 + 5x$

   Solution

   $$ \begin{align} \frac{\partial f}{\partial x} &= 4x^3 y^3 - 14 xy + 5 \\[5pt] \frac{\partial f}{\partial y} &= 3x^4 y^2 - 7x^2 - 4y^3 \end{align}$$




2. $f(x, y) = (x^2 + y^3)^9 + 2ln(x)$

   Solution

   $$ \begin{align} \frac{\partial f}{\partial x} &= 9(x^2 + y^3)^8 (2x) + \frac{2}{x} \\[5pt] &= 18(x^2 + y^3)^8 + \frac{2}{x} \\[5pt] \frac{\partial f}{\partial y} &= 9(x^2 + y^3)^8 (3y^2) \\[5pt] &= 27 y^2 (x^2 + y^3)^8 \end{align}$$




3. $f(x, y) = x e^{xy}$

   Solution

   $$ \begin{align} \frac{\partial f}{\partial x} &= e^{xy} + x e^{xy}(y) = e^{xy}(1 + xy) \\[5pt] \frac{\partial f}{\partial y} &= xe^{xy} (x) = x^2 e^{xy} \end{align}$$




4. $f(x, y, z) = \frac{x}{x+y+z}$

   Solution

   $$ \begin{align} \frac{\partial f}{\partial x} &= \frac{(x+y+z)(1) - x(1)}{(x+y+z)^2} \\[5pt] &= \frac{y + z}{(x+y+z)^2} \\[5pt] \frac{\partial f}{\partial y} &= \frac{0 - x(1)}{(x+y+z)^2} - \frac{-x}{(x+y+z)^2}\\[5pt] \frac{\partial f}{\partial z} &= \frac{0 - x(1)}{(x+y+z)^2} - \frac{-x}{(x+y+z)^2} \end{align}$$

   Here we've used the quotient rule for all derivatives

Second partials

We can also find partial second derivatives of multivariable or vector functions. The notation of the second partial derivatives of $f(x, y)$ with respect to $x$ and $y$ are

$$\frac{\partial^2 f}{\partial x^2} \phantom{000} \text{ and } \phantom{000} \frac{\partial^2 f}{\partial y^2}$$

Some other notations are reviewed in the box below.

Example

Here's an example of some partial second derivatives of the function $f(x, y) = x^3 y^2 - y^4 x$.

$$ \begin{align} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x} (3x^2 y^2 - y^4) \\[5pt] &= 6x y^2 \\[5pt] \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} (2y x^3 - 4y^3) \\[5pt] &= 2x^3 - 12 y^2 x \end{align}$$

The

The Gradient, $\nabla$

The gradient of a multidimensional function is a vector function that gives the direction of the fastest increase in slope (steepest slope). It is found by taking the partial derivative of the function with respect to each component of the vector.

The gradient is given the symbol $\nabla$, which is called "nabla", but is often referred to as "del". So when someone says "calculate del $f$," they mean "calculate the gradient of the function $f$."

$$\nabla f(x_1, x_2, \dots x_n) = \left( \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \dots, \frac{\partial f}{\partial x_n} \right)$$

Here's a quick example. Let's take the 3D function

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

and find the gradient:

$$ \begin{align} \nabla f(x, y) &= (2xy + 2y, \; x^2 + 2x) \\[5pt] &= \langle 2y(x + 1), \; x(x + 2) \rangle \end{align}$$

That set of vectors gives the direction of the steepest slope anywhere on $f(x, y)$.