# 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

- \(f(x,y) = xy \sin x \)
- \( f(x,y) = \dfrac{ x + y}{ x - y}\).

### 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.

### Contributors

- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.