$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 1.9: Partial Derivatives

### Definition of a Partial Derivative

Let $$f(x,y)$$ be a function of two variables. Then we define the partial derivatives as:

Definition: Partial Derivative

$f_x = \dfrac{\partial f}{\partial x} = \lim_{h\to{0}} \dfrac{f(x+h,y)-f(x,y)}{h}$

$f_y = \dfrac{\partial f}{\partial y} = \lim_{h\to{0}} \dfrac{f(x,y+h)-f(x,y)}{h}$

if these limits exist.

Algebraically, we can think of the partial derivative of a function with respect to $$x$$ as the derivative of the function with $$y$$ held constant. Geometrically, the derivative with respect to $$x$$ at a point $$P$$ represents the slope of the curve that passes through $$P$$ whose projection onto the $$xy$$ plane is a horizontal line (if you travel due East, how steep are you climbing?)

Example 1

Let

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

then

\begin{align} \dfrac{\partial f}{ \partial } &= \lim_{h\to{0}}\dfrac{(2(x+h)+3y) - (2x+3y)}{h} \\ &= \lim_{h\to{0}} \dfrac{2x+2h+3y-2x-3y}{h} \\ &= \lim_{h\to{0}} \dfrac{2h}{h} =2 . \end{align}

We also use the notation $$f_x$$  and $$f_y$$ for the partial derivatives with respect to $$x$$ and $$y$$ respectively.

Find $$f_y$$ for the function from the example above.

### Finding Partial Derivatives the Easy Way

Since a partial derivative with respect to $$x$$ is a derivative with the rest of the variables held constant, we can find the partial derivative by taking the regular derivative considering the rest of the variables as constants.

Example 2

Let

$f(x,y) = 3xy^2 - 2x^2y$

then

$f_x = 3y^2 - 4xy$

and

$f_y = 6xy - 2x^2.$

Exercises

Find both partial derivatives for

1. $$f(x,y) = xy \sin x$$

2. $$f(x,y) = \dfrac{ x + y}{ x - y}$$.

### Higher Order Partials

Just as with function of one variable, we can define second derivatives for functions of two variables. For functions of two variables, we have four types:  $$f_{xx}$$, $$f_{xy}$$, $$f_{yx}$$, and $$f_{yy}$$.

Example 3

Let

$f(x,y) = ye^x$

then

$f_x = ye^x$

and

$f_y=e^x.$

Now taking the partials of each of these we get:

$f_{xx}=ye^x \;\;\; f_{xy}=e^x \;\;\; \text{and} \;\;\; f_{yy}=0 .$

Notice that

$f_{x,y} = f_{yx}.$

Theorem

Let $$f(x,y)$$ be a function with continuous second order derivatives, then

$f_{xy} = f_{yx}.$

### Functions of More Than Two Variables

Suppose that

$f(x,y,z) = xy - 2yz$

is a function of three variables, then we can define the partial derivatives in much the same way as we defined the partial derivatives for three variables.

We have

$f_x=y \;\;\; f_y=x-2z \;\;\; \text{and} \;\;\; f_z=-2y .$

Example 4: The Heat Equation

Suppose that a building has a door open during a snowy day.  It can be shown that the equation

$H_t = c^2H_{xx}$

models this situation where $$H$$ is the heat of the room at the point $$x$$ feet away from the door at time $$t$$.  Show that

$H = e^{-t} \cos(\frac{x}{c})$

satisfies this differential equation.

Solution

We have

$H_t = -e^{-t} \cos (\dfrac{x}{c})$

$H_x = -\dfrac{1}{c} e^{-t} \sin(\frac{x}{c})$

$H_{xx} = -\dfrac{1}{c^2} e^{-t} \cos(\dfrac{x}{c}) .$

So that

$c^2 H_{xx}= -e^{-t} \cos (\dfrac{x}{c}) .$

And the result follows.

### Contributors

• Integrated by Justin Marshall.