Transcript Section 11.3 Partial Derivatives

Section 11.3
Partial Derivatives

Goals

Define partial derivatives

Learn notation and rules for calculating
partial derivatives

Interpret partial derivatives

Discuss higher derivatives

Apply to partial differential equations
Introduction

If f is a function of two variables x and y,
suppose we let only x vary while keeping
y fixed, say y = b, where b is a constant.

Then we are really considering a function
of a single variable x, namely, g(x) = f(x, b).

If g has a derivative at a, then we call it the
partial derivative of f with respect to x at (a, b)
and denote it by fx(a, b).
Introduction (cont’d)

Thus

The definition of a derivative gives
Introduction (cont’d)
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Similarly, the partial derivative of f with
respect to y at (a, b), denoted by fy(a, b), is
obtained by holding x = a and finding the
ordinary derivative at b of the function
G(y) = f(a, y):
Definition

If we now let the point (a, b) vary, fx and fy
become functions of two variables:
Notations

There are many alternate notations for
partial derivatives:
Finding Partial Derivatives
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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:
Example
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If f(x, y) = x3 + x2y3 – 2y2, find fx(2, 1) and
fy(2, 1).

Solution Holding y constant and
differentiating with respect to x, we get
fx(x, y) = 3x2 + 2xy3
and so
fx(2, 1) = 3 · 22 + 2 · 2 · 13 = 16
Solution (cont’d)

Holding x constant and differentiating
with respect to y, we get
fx(x, y) = 3x2y2 – 4y
and so
fx(2, 1) = 3 · 22 · 12 – 4 · 1 = 8
Interpretations
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To give a geometric interpretation of
partial derivatives, we recall that the
equation z = f(x, y) represents a surface S.
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If f(a, b) = c, then the point P(a, b, c) lies on
S. By fixing y = b, we are restricting our
attention to the curve C1 in which the
vertical plane y = b intersects S.
Interpretations (cont’d)

Likewise, the vertical plane x = a intersects
S in a curve C2.
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Both of the curves C1 and C2 pass through
the point P.
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This is illustrated on the next slide:
Interpretations (cont’d)
Interpretations (cont’d)


Notice that…
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C1 is the graph of the function g(x) = f(x, b), so
the slope of its tangent T1 at P is g′(a) = fx(a, b);
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C2 is the graph of the function G(y) = f(a, y), so
the slope of its tangent T2 at P is G′(b) = fy(a, b).
Thus fx and fy are the slopes of the tangent
lines at P(a, b, c) to C1 and C2.
Interpretations (cont’d)
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Partial derivatives can also be interpreted
as rates of change.
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If z = f(x, y), then…
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∂z/∂x represents the rate of change of z with
respect to x when y is fixed.
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Similarly, ∂z/∂y represents the rate of change
of z with respect to y when x is fixed.
Example
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If f(x, y) = 4 – x2 – 2y2, find fx(1, 1) and
fy(1, 1).
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Interpret these numbers as slopes.
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Solution We have
Solution (cont’d)
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The graph of f is the paraboloid
z = 4 – x2 – 2y2 and the vertical plane y = 1
intersects it in the parabola z = 2 – x2, y = 1.
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The slope of the tangent line to this
parabola at the point (1, 1, 1) is fx(1, 1) = –2.
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Similarly, the plane x = 1 intersects the
graph of f in the parabola z = 3 – 2y2, x = 1.
Solution (cont’d)
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The slope of the tangent line to this
parabola at the point (1, 1, 1) is fy(1, 1) = –4:
Example
 x 
f
f
 If f x , y   sin 
 1  y  , calculate x and y .


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Solution Using the Chain Rule for
functions of one variable, we have
Example
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Find ∂z/∂x and ∂z/∂y if z is defined
implicitly as a function of x and y by
x3 + y3 + z3 + 6xyz = 1
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Solution To find ∂z/∂x, we differentiate
implicitly with respect to x, being careful
to treat y as a constant:
Solution (cont’d)
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Solving this equation for ∂z/∂x, we obtain
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Similarly, implicit differentiation with
respect to y gives
More Than Two Variables
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Partial derivatives can also be defined for
functions of three or more variables, for
example
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If w = f(x, y, z), then fx = ∂w/∂x is the rate of
change of w with respect to x when y and z
are held fixed.
More Than Two Variables (cont’d)
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But we can’t interpret fx geometrically
because the graph of f lies in fourdimensional space.
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In general, if u = f(x1, x2 ,…, xn), then
and we also write
Example
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Find fx , fy , and fz if f(x, y, z) = exy ln z.
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Solution Holding y and z constant and
differentiating with respect to x, we have
fx = yexy ln z
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Similarly,
fy = xexy ln z
and
fz = exy/z
Higher Derivatives
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If f is a function of two variables, then its
partial derivatives fx and fy are also
functions of two variables, so we can
consider their partial derivatives
(fx)x , (fx)y , (fy)x , and (fy)y ,
which are called the second partial
derivatives of f.
Higher Derivatives (cont’d)
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If z = f(x, y), we use the following notation:
Example
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Find the second partial derivatives of
f(x, y) = x3 + x2y3 – 2y2
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Solution Earlier we found that
fx(x, y) = 3x2 + 2xy3
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Therefore
fy(x, y) = 3x2y2 – 4y
Mixed Partial Derivatives
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Note that fxy = fyx in the preceding
example, which is not just a coincidence.
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It turns out that fxy = fyx for most functions
that one meets in practice:
Mixed Partial Derivatives (cont’d)
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Partial derivatives of order 3 or higher can
also be defined. For instance,
and using Clairaut’s Theorem we can
show that fxyy = fyxy = fyyx if these functions
are continuous.
Example
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Calculate fxxyz if f(x, y, z) = sin(3x + yz).
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Solution
Partial Differential Equations
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Partial derivatives occur in partial
differential equations that express certain
physical laws.
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For instance, the partial differential
equation
 u
2
x
2
 u
2

y
2
0
is called Laplace’s equation.
Partial Differential Eqns. (cont’d)

Solutions of this equation are called
harmonic functions and play a role in
problems of heat conduction, fluid flow,
and electric potential.

For example, we can show that the
function u(x, y) = ex sin y is a solution of
Laplace’s equation:
Partial Differential Eqns. (cont’d)
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Therefore, u satisfies Laplace’s equation.
Partial Differential Eqns. (cont’d)

The wave equation
 u
2
t
2
 u
2
a
2
x
2
describes the motion of a waveform,
which could be an ocean wave, sound
wave, light wave, or wave traveling along
a string.
Partial Differential Eqns. (cont’d)

For instance, if u(x, t) represents the
disaplacement of a violin string at time t
and at a distance x from one end of the
string, then u(x, t) satisfies the wave
equation. (See the next slide.)

Here the constant a depends on the
density of the string and on the tension in
the string.
Partial Differential Eqns. (cont’d)
Example

Verify that the function u(x, t) = sin(x – at)
satisfies the wave equation.
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Solution Calculation gives
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So u satisfies the wave equation.
Review
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Definition of partial derivative
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Notations for partial derivatives
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Finding partial derivatives
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Interpretations of partial derivatives
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Function of more than two variables
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Higher derivatives
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Partial differential equations