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

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

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

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 $\PageIndex{1}$

Let

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

then

$$\begin{align*} \dfrac{\partial f}{ \partial } &= \lim_{h\to{0}}\dfrac{(2(x+h)+3y) - (2x+3y)}{h} \nonumber \\[4pt] &= \lim_{h\to{0}} \dfrac{2x+2h+3y-2x-3y}{h} \nonumber \\[4pt] &= \lim_{h\to{0}} \dfrac{2h}{h} =2 . \end{align*} \nonumber$$

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

Exercise $\PageIndex{1}$

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 $\PageIndex{2}$

Let

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

then

$$f_x = 3y^2 - 4xy \nonumber$$

and

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

Exercises $\PageIndex{2}$

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 $\PageIndex{3}$

Let

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

then

$$f_x = ye^x \nonumber$$

and

$$f_y=e^x. \nonumber$$

Now taking the partials of each of these we get:

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

Notice that

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

Theorem

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

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

Functions of More Than Two Variables

Suppose that

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

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

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

Example $\PageIndex{4}$: The Heat Equation

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

$$H_t = c^2H_{xx} \nonumber$$

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

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

satisfies this differential equation.

Solution

We have

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

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

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

So that

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

And the result follows.