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8 Differential Calculus
1. Derivative of a Function . . . 8-1 2. Elementary Derivative Operations . . . 8-1 3. Critical Points . . . 8-2 4. Derivatives of Parametric Equations . . . 8-3 5. Partial Differentiation . . . 8-4 6. Implicit Differentiation . . . 8-4 7. Tangent Plane Function . . . 8-5 8. Gradient Vector . . . 8-5 9. Directional Derivative . . . 8-6 10. Normal Line Vector . . . 8-6 11. Divergence of a Vector Field . . . 8-7 12. Curl of a Vector Field . . . 8-7 13. Taylor’s Formula . . . 8-7 14. Maclaurin Power Approximations . . . 8-8
1. DERIVATIVE OF A FUNCTION
In most cases, it is possible to transform a continuous function, fðx1; x2; x3; . . .Þ, of one or more independent variables into a derivative function.1 In simple cases, the derivative can be interpreted as the slope (tangent or rate of change) of the curve described by the original function. Since the slope of the curve depends on x, the derivative function will also depend on x. The deriva-tive, f0ðxÞ, of a function f ðxÞ is defined mathematically by Eq. 8.1. However, limit theory is seldom needed to actually calculate derivatives.
f0ðxÞ ¼ lim
Dx!0
Df ðxÞ
Dx 8:1
The derivative of a function fðxÞ, also known as the first derivative, is written in various ways, including
f0ðxÞ;dfðxÞ dx ;df
dx; Df ðxÞ; DxfðxÞ; _fðxÞ; sf ðsÞ A second derivative may exist if the derivative operation is performed on the first derivative—that is, a derivative is taken of a derivative function. This is written as
f00ðxÞ;d2fðxÞ dx2 ;d2f
dx2; D2fðxÞ; D2xfðxÞ; €fðxÞ; s2fðsÞ
A regular (analytic or holomorphic) function possesses a derivative. A point at which a function’s derivative is undefined is called a singular point, as Fig. 8.1 illustrates.
2. ELEMENTARY DERIVATIVE OPERATIONS Equation 8.2 through Eq. 8.5 summarize the elemen-tary derivative operations on polynomials and expo-nentials. Equation 8.2 and Eq. 8.3 are particularly useful. (a, n, and k represent constants. fðxÞ and g(x) are functions of x.)
Dk ¼ 0 8:2
Dxn¼ nxn1 8:3
D ln x ¼1
x 8:4
Deax ¼ aeax 8:5
Equation 8.6 through Eq. 8.17 summarize the elemen-tary derivative operations on transcendental (trigono-metric) functions.
D sin x ¼ cos x 8:6
D cos x ¼ sin x 8:7
D tan x ¼ sec2x 8:8
D cot x ¼ csc2x 8:9
D sec x ¼ sec x tan x 8:10
D csc x ¼ csc x cot x 8:11 D arcsin x ¼ 1
ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2
p 8:12
D arccos x ¼ D arcsin x 8:13
1A function, fðxÞ, of one independent variable, x, is used in this section to simplify the discussion. Although the derivative is taken with respect to x, the independent variable can be anything.
Figure 8.1 Derivatives and Singular Points f (x)
xp xs x
regular point
singular point slope ⫽ f ⬘(xp)
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Equation 8.24 through Eq. 8.29 summarize the elemen-tary derivative operations on functions and combinations of functions.
The derivative function found from Eq. 8.3 determines the slope.
f0ðxÞ ¼ 3x2 2 The slope at x = 3 is
f0ð3Þ ¼ ð3Þð3Þ2 2 ¼ 25
Example 8.2
What are the derivatives of the following functions?
(a) fðxÞ ¼ 5 ffiffiffiffiffi
Derivatives are used to locate the local critical points of functions of one variable—that is, extreme points (also known as maximum and minimum points) as well as the inflection points (points of contraflexure). The plurals extrema, maxima, and minima are used without the word “points.” These points are illustrated in Fig. 8.2.
There is usually an inflection point between two adja-cent local extrema.
Figure 8.2 Extreme and Inflection Points global maximum
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The first derivative is calculated to determine the loca-tions of possible critical points. The second derivative is calculated to determine whether a paricular point is a local maximum, minimum, or inflection point, according to the following conditions. With this method, no dis-tinction is made between local and global extrema.
Therefore, the extrema should be compared with the function values at the endpoints of the interval, as illus-trated in Ex. 8.3.2Generally, f0ðxÞ 6¼ 0 at an inflection point.
f0ðxcÞ ¼ 0 at any extreme point; xc 8:30 f00ðxcÞ < 0 at a maximum point 8:31 f00ðxcÞ > 0 at a minimum point 8:32
f00ðxcÞ ¼ 0 at an inflection point 8:33
Example 8.3
Find the global extrema of the function fðxÞ on the interval½2; þ2.
fðxÞ ¼ x3þ x2 x þ 1
Solution
The first derivative is
f0ðxÞ ¼ 3x2þ 2x 1
Since the first derivative is zero at extreme points, set f0ðxÞ equal to zero and solve for the roots of the quad-ratic equation.
3x2þ 2x 1 ¼ ð3x 1Þðx þ 1Þ ¼ 0
The roots are x1¼1=3, x2¼ 1. These are the locations of the two local extrema.
The second derivative is
f00ðxÞ ¼ 6x þ 2 Substituting x1and x2into f00(x),
f00ðx1Þ ¼ ð6Þ 13
þ 2 ¼ 4
f00ðx2Þ ¼ ð6Þð1Þ þ 2 ¼ 4
Therefore, x1is a local minimum point (because f00ðx1Þ is positive), and x2 is a local maximum point (because f00ðx2Þ is negative). The inflection point between these two extrema is found by setting f00ðxÞ equal to zero.
f00ðxÞ ¼ 6x þ 2 ¼ 0 or x ¼ 13
Since the question asked for the global extreme points, it is necessary to compare the values of fðxÞ at the local extrema with the values at the endpoints.
fð2Þ ¼ 1 fð1Þ ¼ 2 f 13
¼ 22=27 fð2Þ ¼ 11
Therefore, the actual global extrema are the endpoints.
4. DERIVATIVES OF PARAMETRIC EQUATIONS
The derivative of a function fðx1; x2; . . . ; xnÞ can be calculated from the derivatives of the parametric equa-tions f1ðsÞ; f2ðsÞ; . . . ; fnðsÞ. The derivative will be expressed in terms of the parameter, s, unless the deriv-atives of the parametric equations can be expressed explicitly in terms of the independent variables.
Example 8.4
A circle is expressed parametrically by the equations x¼ 5 cos
y¼ 5 sin
Express the derivative dy/dx (a) as a function of the parameter and (b) as a function of x and y.
Solution
(a) Taking the derivative of each parametric equation with respect to,
dx
d¼ 5 sin dy
d¼ 5 cos Then,
dy dx¼
dy d dx d
¼ 5 cos
5 sin ¼ cot
2It is also necessary to check the values of the function at singular points (i.e., points where the derivative does not exist).
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(b) The derivatives of the parametric equations are closely related to the original parametric equations.
dx
Derivatives can be taken with respect to only one inde-pendent variable at a time. For example, f0ðxÞ is the derivative of fðxÞ and is taken with respect to the inde-pendent variable x. If a function, fðx1; x2; x3; . . .Þ, has more than one independent variable, a partial derivative can be found, but only with respect to one of the inde-pendent variables. All other variables are treated as constants. Symbols for a partial derivative of f taken with respect to variable x are∂f/∂x and fx(x, y).
The geometric interpretation of a partial derivative∂f/∂x is the slope of a line tangent to the surface (a sphere, ellipsoid, etc.) described by the function when all variables except x are held constant. In three-dimensional space with a function described by z = f (x, y), the partial derivative∂f/∂x (equivalent to ∂z/∂x) is the slope of the line tangent to the surface in a plane of constant y.
Similarly, the partial derivative ∂f/∂y (equivalent to
∂z/∂y) is the slope of the line tangent to the surface in a plane of constant x.
Example 8.5
What is the partial derivative ∂z/∂x of the following function?
z¼ 3x2 6y2þ xy þ 5y 9 Solution
The partial derivative with respect to x is found by considering all variables other than x to be constants.
@z@x¼ 6x 0 þ y þ 0 0 ¼ 6x þ y
Example 8.6
A surface has the equation x2+ y2+ z2 9 = 0. What is the slope of a line that lies in a plane of constant y and is tangent to the surface atðx; y; zÞ ¼ ð1; 2; 2Þ?3
Solution
Solve for the dependent variable. Then, consider vari-able y to be a constant.
z¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
plane of constant y
slope = −1 2
6. IMPLICIT DIFFERENTIATION
When a relationship between n variables cannot be manipulated to yield an explicit function of n 1 inde-pendent variables, that relationship implicitly defines the nth variable. Finding the derivative of the implicit vari-able with respect to any other independent varivari-able is known as implicit differentiation.
An implicit derivative is the quotient of two partial derivatives. The two partial derivatives are chosen so that dividing one by the other eliminates a common differential. For example, if z cannot be explicitly extracted from fðx; y; zÞ ¼ 0, the partial derivatives
∂z/∂x and ∂z/∂y can still be found as follows.
@x@z¼ @f
3Although only implied, it is required that the point actually be on the surface (i.e., it must satisfy the equation f (x, y, z) = 0).
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Example 8.7
Find the derivative dy/dx of
fðx; yÞ ¼ x2þ xy þ y3 Solution
Implicit differentiation is required because x cannot be extracted from fðx; yÞ.
@f
@x¼ 2x þ y
@f
@y¼ x þ 3y2
dy dx¼ @f
@f@x
@y
¼ð2x þ yÞ xþ 3y2
Example 8.8
Solve Ex. 8.6 using implicit differentiation.
Solution
fðx; y; zÞ ¼ x2þ y2þ z2 9 ¼ 0
@f@x¼ 2x
@f
@z¼ 2z
@z@x¼ @f
@f@x
@z
¼ 2x 2z ¼ x
z
At the pointð1; 2; 2Þ,
@x@z¼ 1 2
7. TANGENT PLANE FUNCTION
Partial derivatives can be used to find the equation of a plane tangent to a three-dimensional surface defined by fðx; y; zÞ ¼ 0 at some point, P0.
Tðx0; y0; z0Þ ¼ ðx x0Þ@f ðx; y; zÞ
@x
P0
þ ðy y0Þ@f ðx; y; zÞ
@y
P0
þ ðz z0Þ@f ðx; y; zÞ
@z
P0
¼ 0 8:36
The coefficients of x, y, and z are the same as the coefficients of i, j, and k of the normal vector at point P0. (See Sec. 8.10.)
Example 8.9
What is the equation of the plane that is tangent to the surface defined by fðx; y; zÞ ¼ 4x2þ y2 16z ¼ 0 at the pointð2; 4; 2Þ?
Solution
Calculate the partial derivatives and substitute the coordinates of the point.
@f ðx; y; zÞ
@x
P0
¼ 8xjð2; 4; 2Þ¼ ð8Þð2Þ ¼ 16
@f ðx; y; zÞ
@y
P0
¼ 2yjð2; 4; 2Þ¼ ð2Þð4Þ ¼ 8
@f ðx; y; zÞ
@z
P0
¼ 16jð2; 4; 2Þ¼ 16
Tð2; 4; 2Þ ¼ ð16Þðx 2Þ þ ð8Þðy 4Þ ð16Þðz 2Þ
¼ 2x þ y 2z 4 Substitute into Eq. 8.36.
2xþ y 2z 4 ¼ 0
8. GRADIENT VECTOR
The slope of a function is the change in one variable with respect to a distance in a chosen direction. Usually, the direction is parallel to a coordinate axis. However, the maximum slope at a point on a surface may not be in a direction parallel to one of the coordinate axes.
The gradient vector function rf ðx; y; zÞ (pronounced
“del f ”) gives the maximum rate of change of the func-tion fðx; y; zÞ.
rf ðx; y; zÞ ¼@f ðx; y; zÞ
@x i þ@f ðx; y; zÞ
@y j
þ@f ðx; y; zÞ
@z k 8:37
Example 8.10
A two-dimensional function is defined as fðx; yÞ ¼ 2x2 y2þ 3x y
(a) What is the gradient vector for this function?
(b) What is the direction of the line passing through the pointð1; 2Þ that has a maximum slope? (c) What is the maximum slope at the pointð1; 2Þ?
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Solution
(a) It is necessary to calculate two partial derivatives in order to use Eq. 8.37.
@f ðx; yÞ
@x ¼ 4x þ 3
@f ðx; yÞ
@y ¼ 2y 1 rf ðx; yÞ ¼ ð4x þ 3Þi þ ð2y 1Þj
(b) Find the direction of the line passing throughð1; 2Þ with maximum slope by inserting x = 1 and y =2 into the gradient vector function.
V ¼
(c) The magnitude of the slope is jVj ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð7Þ2þ ð3Þ2
q ¼ 7:62
9. DIRECTIONAL DERIVATIVE
Unlike the gradient vector (covered in Sec. 8.8), which calculates the maximum rate of change of a function, the directional derivative, indicated by rufðx; y; zÞ, Dufðx; y; zÞ, or fu0ðx; y; zÞ, gives the rate of change in the direction of a given vector, u or U. The subscript u implies that the direction vector is a unit vector, but it does not need to be, as only the direction cosines are calculated from it.
The direction cosines are given by Eq. 8.40 and Eq. 8.41.
cos ¼Ux The partial derivatives are
@f ðx; yÞ
@x ¼ 6x þ y
@f ðx; yÞ
@y ¼ x 4y
The directional derivative is given by Eq. 8.38.
rufðx; yÞ ¼ 4
10. NORMAL LINE VECTOR
Partial derivatives can be used to find the vector normal to a three-dimensional surface defined by fðx; y; zÞ = 0 at some point P0. The coefficients of i, j, and k are the same as the coefficients of x, y, and z calculated for the equation of the tangent plane at point P0. (See Sec. 8.7.)
N ¼@f ðx; y; zÞ
What is the vector normal to the surface of fðx; y; zÞ ¼ 4x2þ y2 16z ¼ 0 at the point ð2; 4; 2Þ?
Solution
The equation of the tangent plane at this point was calculated in Ex. 8.9 to be
Tð2; 4; 2Þ ¼ 2x þ y 2z 4 ¼ 0
A vector that is normal to the tangent plane through this point is
N ¼ 2i þ j 2k
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11. DIVERGENCE OF A VECTOR FIELD The divergence, div F, of a vector field Fðx; y; zÞ is a scalar function defined by Eq. 8.44 through Eq. 8.46.4 The divergence of F can be interpreted as the accumu-lation of flux (i.e., a flowing substance) in a small region (i.e., at a point). One of the uses of the divergence is to determine whether flow (represented in direction and magnitude by F) is compressible. Flow is incompressible if div F = 0, since the substance is not accumulating.
F ¼ Pðx; y; zÞi þ Qðx; y; zÞj þ Rðx; y; zÞk 8:44
div F ¼@P
@x þ @Q
@y þ @R
@z 8:45
It may be easier to calculate divergence from Eq. 8.46.
div F ¼ r F 8:46
The vector del operator,r, is defined as r ¼ @
@xi þ @
@yj þ @
@zk 8:47
If there is no divergence, then the dot product calculated in Eq. 8.46 is zero.
Example 8.13
Calculate the divergence of the following vector function.
12. CURL OF A VECTOR FIELD
The curl, curl F, of a vector field Fðx; y; zÞ is a vector field defined by Eq. 8.49 and Eq. 8.50. The curl F can be interpreted as the vorticity per unit area of flux (i.e., a flowing substance) in a small region (i.e., at a point).
One of the uses of the curl is to determine whether flow (represented in direction and magnitude by F) is rota-tional. Flow is irrotational if curl F = 0.
F ¼ Pðx; y; zÞi þ Qðx; y; zÞj þ Rðx; y; zÞk 8:48
It may be easier to calculate the curl from Eq. 8.50. (The vector del operator,r, was defined in Eq. 8.47.)
curl F ¼ r F
If the velocity vector is V, then the vorticity is
! ¼ r V ¼ !xi þ !yj þ !zk 8:51
The circulation is the line integral of the velocity, V, along a closed curve.
¼I
V ds ¼I
! dA 8:52
Example 8.14
Calculate the curl of the following vector function.
Fðx; y; zÞ ¼ 3x2i þ 7exyj Expand the determinant across the top row.
i @
Taylor’s formula (series) can be used to expand a func-tion around a point (i.e., approximate the funcfunc-tion at one point based on the function’s value at another point). The approximation consists of a series, each term composed of a derivative of the original function and a polynomial. Using Taylor’s formula requires that the original function be continuous in the interval [a, b]
4A bold letter, F, is used to indicate that the vector is a function of x, y, and z.
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and have the required number of derivatives. To expand a function, fðxÞ, around a point, a, in order to obtain fðbÞ, Taylor’s formula is
fðbÞ ¼ f ðaÞ þf0ðaÞ
1! ðb aÞ þf00ðaÞ
2! ðb aÞ2 þ þfnðaÞ
n! ðb aÞnþ RnðbÞ 8:53
In Eq. 8.53, the expression fn designates the nth deriv-ative of the function fðxÞ. To be a useful approximation, two requirements must be met: (1) point a must be relatively close to point b, and (2) the function and its derivatives must be known or easy to calculate. The last term, Rn(b), is the uncalculated remainder after n deriv-atives. It is the difference between the exact and approx-imate values. By using enough terms, the remainder can be made arbitrarily small. That is, Rn(b) approaches zero as n approaches infinity.
It can be shown that the remainder term can be calcu-lated from Eq. 8.54, where c is some number in the interval [a, b]. With certain functions, the constant c can be completely determined. In most cases, however, it is possible only to calculate an upper bound on the remainder from Eq. 8.55. Mnis the maximum (positive) value of fnþ1ðxÞ on the interval [a, b].
RnðbÞ ¼ fnþ1ðcÞ
ðn þ 1Þ!ðb aÞnþ1 8:54 jRnðbÞj Mnjðb aÞnþ1j
ðn þ 1Þ! 8:55
14. MACLAURIN POWER APPROXIMATIONS If a = 0 in the Taylor series, Eq. 8.53 is known as the Maclaurin series. The Maclaurin series can be used to approximate functions at some value of x between 0 and 1. The following common approximations may be referred to as Maclaurin series, Taylor series, or power series approximations.
sin x x x3 3!þx5
5!x7 7!þ þ ð1Þn x2nþ1
ð2n þ 1Þ! 8:56
cos x 1 x2 2!þx4
4!x6
6!þ þ ð1Þn x2n
ð2nÞ! 8:57 sinh x x þx3
3!þx5 5!þx7
7!þ þ x2nþ1
ð2n þ 1Þ! 8:58 cosh x 1 þx2
2!þx4 4!þx6
6!þ þ x2n
ð2nÞ! 8:59
ex 1 þ x þx2 2!þx3
3!þ þxn
n! 8:60
lnð1 þ xÞ x x2 2 þx3
3 x4
4 þ þ ð1Þnþ1xn n
8:61
1
1 x 1 þ x þ x2þ x3þ þ xn 8:62
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