1.9: Partial Derivatives
- Page ID
- 599
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
\[\begin{align*} \dfrac{\partial f}{ \partial } &= \lim_{h\to{0}}\dfrac{(2(x+h)+3y) - (2x+3y)}{h} \nonumber \\[4pt] &= \lim_{h\to{0}} \dfrac{2x+2h+3y-2x-3y}{h} \nonumber \\[4pt] &= \lim_{h\to{0}} \dfrac{2h}{h} =2 . \end{align*} \]
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 and Attributions
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.