
# 1.9: Partial Derivatives

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### 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 $$\PageIndex{1}$$

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 \\[5pt] &= \lim_{h\to{0}} \dfrac{2x+2h+3y-2x-3y}{h} \nonumber \\[5pt] &= \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.

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

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

### 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

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

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

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

satisfies this differential equation.

Solution

We have

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

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

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

So that

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

And the result follows.