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Mathematics LibreTexts

1.9: Partial Derivatives

  • Page ID
    599
  • [ "article:topic", "Partial Differentiation", "heat equation", "authorname:green", "Higher Order Partial Differentiation" ]

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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.

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