1.9: Partial Derivatives
- Page ID
- 599
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Definition of a Partial Derivative
Let \(f(x,y)\) be a function of two variables. Then we define the partial derivatives as:
\[ f_x = \dfrac{\partial f}{\partial x} = \lim_{h\to{0}} \dfrac{f(x+h,y)-f(x,y)}{h} \nonumber \]
\[ f_y = \dfrac{\partial f}{\partial y} = \lim_{h\to{0}} \dfrac{f(x,y+h)-f(x,y)}{h} \nonumber \]
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?)
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.
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.
Let
\[ f(x,y) = 3xy^2 - 2x^2y \nonumber \]
then
\[ f_x = 3y^2 - 4xy \nonumber \]
and
\[ f_y = 6xy - 2x^2. \nonumber \]
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 \(f_{yy}\).
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
\[ f_{x,y} = f_{yx}.\nonumber \]
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
\[f_x=y \;\;\; f_y=x-2z \;\;\; \text{and} \;\;\; f_z=-2y . \nonumber \]
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.