Partial DerivativesPartial Derivatives

GoalsGoals Define Define partial derivativespartial derivatives Learn Learn notationnotation and and rules for calculatingrules for calculating

partial derivativespartial derivatives InterpretInterpret partial derivatives partial derivatives Discuss Discuss higher derivativeshigher derivatives Apply to Apply to partial differential equationspartial differential equations

IntroductionIntroduction

If If ff is a function of two variables is a function of two variables xx and and yy, , suppose we let only suppose we let only xx vary while keeping vary while keeping yy fixed, say fixed, say yy = = bb, where , where bb is a constant. is a constant.

Then we are really considering a function Then we are really considering a function of a single variable of a single variable xx, namely, , namely, gg((xx) = ) = ff((xx, , bb).).

If If gg has a derivative at has a derivative at aa, then we call it , then we call it the the partial derivative of f with respect to partial derivative of f with respect to x at x at ((aa, , bb) and denote it by ) and denote it by ffxx((aa, , bb).).

Introduction (cont’d)Introduction (cont’d)

ThusThus

The definition of a derivative givesThe definition of a derivative gives

Introduction (cont’d)Introduction (cont’d)

Similarly, the Similarly, the partial derivative of f partial derivative of f with respect to y at with respect to y at ((aa, , bb), denoted ), denoted by by ffyy((aa, , bb), is obtained by holding ), is obtained by holding xx = =

aa and finding the ordinary derivative and finding the ordinary derivative at at bb of the function of the function GG((yy) = ) = ff((aa, , yy):):

DefinitionDefinition

If we now let the point (If we now let the point (aa, , bb) vary, ) vary, ffxx

and and ffyy become functions of two become functions of two

variables:variables:

NotationsNotations

There are many alternate notations There are many alternate notations for partial derivatives:for partial derivatives:

Finding Partial Finding Partial DerivativesDerivatives

The partial derivative with respect to x is just the ordinary derivative of the function g of a single variable that we get by keeping y fixed:

ExampleExample

If If ff((xx, , yy) = ) = xx33 + + xx22yy33 – 2 – 2yy22, find , find ffxx(2, 1) (2, 1)

and and ffyy(2, 1).(2, 1).

SolutionSolution Holding Holding yy constant and constant and differentiating with respect to differentiating with respect to xx, we get, we get

ffxx((xx, , yy) = 3) = 3xx22 + 2 + 2xyxy33

and soand so

ffxx(2, 1) = 3 · 2(2, 1) = 3 · 222 + 2 · 2 · 1 + 2 · 2 · 133 = 16 = 16

Solution (cont’d)Solution (cont’d)

Holding Holding xx constant and constant and differentiating with respect to differentiating with respect to yy, we , we getget

ffxx((xx, , yy) = 3) = 3xx22yy22 – 4 – 4yy

and soand so

ffxx(2, 1) = 3 · 2(2, 1) = 3 · 222 · 1 · 122 – 4 · 1 = 8 – 4 · 1 = 8

InterpretationsInterpretations

To give a geometric interpretation of To give a geometric interpretation of partial derivatives, we recall that the partial derivatives, we recall that the equation equation zz = = ff((xx, , yy) represents a surface ) represents a surface SS..

If If ff((aa, , bb) = ) = cc, then the point , then the point PP((aa, , bb, , cc) ) lies on lies on SS. By fixing . By fixing yy = = bb, we are , we are restricting our attention to the curve restricting our attention to the curve CC11

in which the vertical plane in which the vertical plane yy = = bb intersects intersects SS..

Interpretations (cont’d)Interpretations (cont’d)

Likewise, the vertical plane Likewise, the vertical plane xx = = aa intersects intersects SS in a curve in a curve CC22..

Both of the curves Both of the curves CC11 and and CC22 pass pass

through the point through the point PP.. This is illustrated on the next slide:This is illustrated on the next slide:

Interpretations (cont’d)Interpretations (cont’d)

Interpretations (cont’d)Interpretations (cont’d)

Notice that…Notice that… CC11 is the graph of the function is the graph of the function gg((xx) = ) = ff((xx, , bb), so ), so

the slope of its tangent the slope of its tangent TT11 at at PP is is gg′(′(aa) = ) = ffxx((aa, ,

bb);); CC22 is the graph of the function is the graph of the function GG((yy) = ) = ff((aa, , yy), so ), so

the slope of its tangent the slope of its tangent TT22 at at PP is is GG′(′(bb) = ) = ffyy((aa, ,

bb).).

Thus Thus ffxx and and ffyy are the are the slopes of the tangent slopes of the tangent

lineslines at at PP((aa, , bb, , cc) to ) to CC11 and and CC22..

Interpretations (cont’d)Interpretations (cont’d)

Partial derivatives can also be Partial derivatives can also be interpreted as interpreted as rates of changerates of change..

If If zz = = ff((xx, , yy), then…), then… ∂∂zz/∂/∂xx represents the rate of change of represents the rate of change of zz

with respect to with respect to xx when when yy is fixed. is fixed. Similarly, ∂Similarly, ∂zz/∂/∂yy represents the rate of represents the rate of

change of change of zz with respect to with respect to yy when when xx is is fixed.fixed.

ExampleExample

If If ff((xx, , yy) = 4 – ) = 4 – xx22 – 2 – 2yy22, find , find ffxx(1, 1) (1, 1)

andandffyy(1, 1).(1, 1).

Interpret these numbers as slopes.Interpret these numbers as slopes. SolutionSolution We have We have

Solution (cont’d)Solution (cont’d)

The graph of The graph of ff is the paraboloid is the paraboloidzz = 4 – = 4 – xx22 – 2 – 2yy22 and the vertical plane and the vertical plane yy = 1 = 1 intersects it in the parabola intersects it in the parabola zz = 2 – = 2 – xx22, , yy = 1. = 1.

The slope of the tangent line to this parabola The slope of the tangent line to this parabola at the point (1, 1, 1) is at the point (1, 1, 1) is ffxx(1, 1) = –2.(1, 1) = –2.

Similarly, the plane Similarly, the plane xx = 1 intersects the = 1 intersects the graph of graph of ff in the parabola in the parabola zz = 3 – 2 = 3 – 2yy22, , xx = 1. = 1.

Solution (cont’d)Solution (cont’d)

The slope of the tangent line to this The slope of the tangent line to this parabola at the point (1, 1, 1) is parabola at the point (1, 1, 1) is ffyy(1, (1,

1) = –4:1) = –4:

ExampleExample

IfIf

SolutionSolution Using the Chain Rule for Using the Chain Rule for functions of one variable, we havefunctions of one variable, we have

. and calculate ,1

sin,yf

xf

yx

yxf

ExampleExample

Find ∂Find ∂zz/∂/∂xx and ∂ and ∂zz/∂/∂yy if if zz is defined is defined implicitly as a function of implicitly as a function of xx and and yy by by

xx33 + + yy33 + + zz33 + 6 + 6xyzxyz = 1 = 1 SolutionSolution To find ∂ To find ∂zz/∂/∂xx, we , we

differentiate implicitly with respect differentiate implicitly with respect to to xx, being careful to treat , being careful to treat yy as a as a constant:constant:

Solution (cont’d)Solution (cont’d)

Solving this equation for ∂Solving this equation for ∂zz/∂/∂xx, we , we obtainobtain

Similarly, implicit differentiation Similarly, implicit differentiation with respect to with respect to yy gives gives

More Than Two VariablesMore Than Two Variables Partial derivatives can also be defined Partial derivatives can also be defined

for functions of three or more for functions of three or more variables, for examplevariables, for example

If If ww = = ff((xx, , yy, , zz), then ), then ffxx = ∂ = ∂ww/∂/∂xx is the is the

rate of change of rate of change of ww with respect to with respect to xx when when yy and and zz are held fixed. are held fixed.

More Than Two Variables More Than Two Variables (cont’d)(cont’d)

But we can’t interpret But we can’t interpret ffxx

geometrically because the graph of geometrically because the graph of ff lies in four-dimensional space.lies in four-dimensional space.

In general, if In general, if uu = = ff((xx11, , xx2 2 ,…, ,…, xxnn), then), then

and we also writeand we also write

ExampleExample

Find Find ffxx ,, f fy y , and , and ffzz if if ff((xx, , yy, , zz) = ) = eexyxy ln ln zz..

SolutionSolution Holding Holding yy and and zz constant and constant and differentiating with respect to differentiating with respect to xx, we , we havehave

ffxx = = yeyexyxy ln ln zz

SimilarlySimilarly,,

ffyy = = xexexyxy ln ln zz and and ffzz = = eexyxy//zz

Higher DerivativesHigher Derivatives

If If ff is a function of two variables, then is a function of two variables, then its partial derivatives its partial derivatives ffxx andand f fyy are also are also

functions of two variables, so we can functions of two variables, so we can consider their partial derivativesconsider their partial derivatives

((ffxx))xx ,, ((ffxx))yy ,, ((ffyy))x x , and (, and (ffyy))y y ,,

which are called the which are called the second partial second partial derivativesderivatives of of ff..

Higher Derivatives Higher Derivatives (cont’d)(cont’d)

If If zz = = ff((xx, , yy), we use the following ), we use the following notation:notation:

ExampleExample Find the second partial derivatives ofFind the second partial derivatives of

ff((xx, , yy) = ) = xx33 + + xx22yy33 – 2 – 2yy22

SolutionSolution Earlier we found that Earlier we found that

ffxx((xx, , yy) = 3) = 3xx22 + 2 + 2xyxy3 3 ffyy((xx, , yy) = 3) = 3xx22yy2 2 – –

44yy ThereforeTherefore

Mixed Partial DerivativesMixed Partial Derivatives

Note that Note that ffxyxy = = ffyxyx in the preceding in the preceding

example, which is not just a example, which is not just a coincidence.coincidence.

It turns out that It turns out that ffxyxy = = ffyxyx for most for most

functions that one meets in practice:functions that one meets in practice:

Mixed Partial Derivatives Mixed Partial Derivatives (cont’d)(cont’d)

Partial derivatives of order 3 or higher Partial derivatives of order 3 or higher can also be defined. For instance,can also be defined. For instance,

and using Clairaut’s Theorem we can and using Clairaut’s Theorem we can show that show that ffxyyxyy = = ffyxyyxy = = ffyyxyyx if these if these

functions are continuous.functions are continuous.

ExampleExample

Calculate Calculate ffxxyzxxyz if if ff((xx,, y y, , zz) = sin(3) = sin(3xx + +

yzyz).). SolutionSolution

Partial Differential Partial Differential EquationsEquations

Partial derivatives occur in Partial derivatives occur in partial partial differential equationsdifferential equations that express that express certain physical laws.certain physical laws.

For instance, the partial differential For instance, the partial differential equationequation

is called is called Laplace’s equationLaplace’s equation..

02

2

2

2

yu

xu

Partial Differential Eqns. Partial Differential Eqns. (cont’d)(cont’d)

Solutions of this equation are called Solutions of this equation are called harmonic functionsharmonic functions and play a role and play a role in problems of heat conduction, fluid in problems of heat conduction, fluid flow, and electric potential.flow, and electric potential.

For example, we can show that the For example, we can show that the function function uu((xx, , yy) = ) = eexx sin  sin yy is a is a solution of Laplace’s equation:solution of Laplace’s equation:

Partial Differential Eqns. Partial Differential Eqns. (cont’d)(cont’d)

Therefore, Therefore, uu satisfies Laplace’s satisfies Laplace’s equation.equation.

Partial Differential Eqns. Partial Differential Eqns. (cont’d)(cont’d)

The The wave equationwave equation

describes the motion of a waveform, describes the motion of a waveform, which could be an ocean wave, which could be an ocean wave, sound wave, light wave, or wave sound wave, light wave, or wave traveling along a string.traveling along a string.

2

22

2

2

xu

atu

Partial Differential Eqns. Partial Differential Eqns. (cont’d)(cont’d)

For instance, if For instance, if uu((xx, , tt) represents the ) represents the disaplacement of a violin string at time disaplacement of a violin string at time tt and at a distance and at a distance xx from one end of from one end of the string, then the string, then uu((xx, , tt) satisfies the ) satisfies the wave equation. (See the next slide.)wave equation. (See the next slide.)

Here the constant Here the constant aa depends on the depends on the density of the string and on the tension density of the string and on the tension in the string.in the string.

Partial Differential Eqns. Partial Differential Eqns. (cont’d)(cont’d)

ExampleExample

Verify that the function Verify that the function uu((xx, , tt) = sin() = sin(xx – – atat) satisfies the wave equation.) satisfies the wave equation.

Solution Solution Calculation givesCalculation gives

So So uu satisfies the wave equation. satisfies the wave equation.

ReviewReview

Definition of partial derivativeDefinition of partial derivative Notations for partial derivativesNotations for partial derivatives Finding partial derivativesFinding partial derivatives Interpretations of partial derivativesInterpretations of partial derivatives Function of more than two variablesFunction of more than two variables Higher derivativesHigher derivatives Partial differential equationsPartial differential equations