CHAPTER 7

Derivatives. One and Several Variables

Derivatives

MATLAB implements a special set of commands enabling you to work with derivatives, which is particularly important in the world of computing and its applications. The derivative of a real function at a point measures the instantaneous rate of change of that function in a neighborhood of the point; i.e. how the dependent variable changes as a result of a small change in the independent variable. Geometrically, the derivative of a function at a point is the slope of the tangent to the function at that point. The origin of the idea derived from the attempt to draw the tangent line at a given point on a given curve.

A function f(x) defined in a neighborhood of a point a is said to be differentiable at a if the following limit exists:

The value of the limit, if it exists, is denoted by f'(a), and is called the derivative of the function f at the point a. If f is differentiable at all points of its domain, it is simply said to be differentiable. The continuity of a function is a necessary (but not sufficient) condition for differentiability, and any differentiable function is continuous.

The following table shows the basic commands that enable MATLAB to work with derivatives.

As a first example, we consider the function f(x) = x⁵-3x⁴-11x³+27x²+10x-24 and graph its derivative in the interval [-4,5].

>> x=-4:0.1:5;
>> f=x.^5-3*x.^4-11*x.^3+27*x.^2+10*x-24;
>> df=diff(f)./diff(x);
>> plot(x,f)

As a second example, we calculate the derivative of the function log(sin(2x)) and simplify the result.

>> pretty(simplify(diff('log(sin(2*x))','x')))

  2 cot(2 x)

As a third example, we calculate the first four derivatives of the following function:

f(x) = 1/x2

>> f='1/x^2'

f =
1/x^2

>> [diff(f),diff(f,2),diff(f,3),diff(f,4)]

ans =

[ -2/x^3, 6/x^4, -24/x^5, 120/x^6]

Partial Derivatives

The MATLAB commands for differentiation described above can also be used for partial differentiation.

As an example, given the function f(x,y) = sin(xy)+cos(xy²), we calculate:

∂f/∂x, ∂f/∂y, ∂2f/∂x2, ∂2f/∂y2, ∂2f/∂x∂y, ∂2f/∂y∂x and ∂4f/∂2x∂2y

>> syms x y
>> f=sin(x*y)+cos(x*y^2)

f =

sin(x*y)+cos(x*y^2)

>> diff(f,x)

ans =

cos(x*y)*y-sin(x*y^2)*y^2

>> diff(f,y)

ans =

cos(x*y)*x-2*sin(x*y^2)*x*y

>> diff(diff(f,x),x)

ans =

-sin(x*y)*y^2-cos(x*y^2)*y^4

>> diff(diff(f,y),y)

ans =

-sin(x*y)*x^2-4*cos(x*y^2)*x^2*y^2-2*sin(x*y^2)*x

>> diff(diff(f,x),y)

ans =

-sin(x*y)*x*y+cos(x*y)-2*cos(x*y^2)*x*y^3-2*sin(x*y^2)*y

>> diff(diff(f,y),x)

ans =

-sin(x*y)*x*y+cos(x*y)-2*cos(x*y^2)*x*y^3-2*sin(x*y^2)*y

>> diff(diff(diff(diff(f,x),x),y,y))

ans =

sin(x*y)*y^3*x-3*cos(x*y)*y^2+2*cos(x*y^2)*y^7*x+6*sin(x*y^2)*y^5

Applications of Differentiation. Tangents, Asymptotes, Extreme Points and Points of Inflection

A direct application of differentiation allows us to find the tangent to a function at a given point, horizontal, vertical and oblique asymptotes of a function, intervals of increase and concavity, maxima and minima and points of inflection.

With this information it is possible to give a complete study of curves and their representation.

If f is a function which is differentiable at x₀, then f ' (x₀) is the slope of the tangent line to the curve y = f(x) at the point (x₀, f (x₀)). The equation of the tangent will be .

The horizontal asymptotes of the curve are limit tangents, as , which are horizontal. They are defined by the equation .

The vertical asymptotes of the curve are limit tangents, as , which are vertical. They are defined by the equation , where x₀ is a value such that .

The oblique asymptotes to the curve at the point have the equation , where And .

If f is a function for which f ' (x₀) and f ' ' (x₀) both exist, then, if and , the function f   has a local maximum at the point(x₀, f (x₀)).

If f is a function for which f ' (x₀) and f ' ' (x₀) both exist, then, if and , the function f has a local minimum at the point(x0, f (x0)).

If f is a function for which f ' (x₀), f ' ' (x₀) and f ' ' ' (x₀) exist, then, if and and f ' ' ' (x₀) 0, the function f has a turning point at the point(x₀, f(x₀)).

If f is differentiable, then the values of x for which the function f is increasing are those for which f ' (x) is greater than zero.

If f is differentiable, then the values of x for which the function f is decreasing are those for which f ' (x) is less than zero.

If f is twice differentiable, then the values of x for which the function f is concave are those for which f ' ' (x) is greater than zero.

If f is twice differentiable, then the values of x for which the function f is convex are those for which f ' ' (x) is less than zero.

As an example, we perform a comprehensive study of the function:

calculating the asymptotes, maxima, minima, inflexion points, intervals of increase and decrease and intervals of concavity and convexity.

>> f='x^3/(x^2-1)'

f =

x^3/(x^2-1)

>> syms x, limit(x^3/(x^2-1),x,inf)

ans =

NaN

Therefore, there are no horizontal asymptotes. To see if there are vertical asymptotes, we consider the values of x that make the function infinite:

>> solve('x^2-1')

ans =

[ 1]
[-1]

The vertical asymptotes are the lines x = 1 and x = -1. Let us see if there are any oblique asymptotes:

>> limit(x^3/(x^2-1)/x,x,inf)

ans =

1

>> limit(x^3/(x^2-1)-x,x,inf)

ans =

0

The line y = x is an oblique asymptote. Now, the maxima and minima, inflection points and intervals of concavity and growth will be analyzed:

>> solve(diff(f))

ans =

[       0]
[       0]
[ 3^(1/2)]
[-3^(1/2)]

The first derivative vanishes at the points with abscissa x = 0, x =√ 3 and x = - √ 3. These points are candidates for maxima and minima. To determine whether they are maxima or minima, we find the value of the second derivative at these points:

>> [numeric(subs(diff(f,2),0)),numeric(subs(diff(f,2),sqrt(3))),
    numeric(subs(diff(f,2),-sqrt(3)))]

ans =

         0    2.5981   -2.5981

Therefore, at the point with abscissa x = - √3 there is a maximum and at the point with abscissa x = √ 3 there is a minimum. At x = 0 we know nothing:

>> [numeric(subs(f,sqrt(3))),numeric(subs(f,-sqrt(3)))]

ans =

    2.5981   -2.5981

Therefore, the maximum point is (- √ 3, -2.5981) and the minimum point is (√ 3, 2.5981).

We will now analyze the inflection points:

>> solve(diff(f,2))

ans =

[         0]
[ i*3^(1/2)]
[-i*3^(1/2)]

The only possible point of inflection occurs at x = 0, and since f (0) = 0, the possible turning point is (0, 0):

>> subs(diff(f,3),0)

ans =

-6

As the third derivative at x = 0 does not vanish, the origin is indeed a turning point:

>> pretty(simple(diff(f)))

                              2   2
                            x  (x  - 3)
                            ------------
                                2     2
                             (x  - 1)

The curve is increasing when y ' > 0, i.e., in the intervals (- ∞, - √ 3) and (√ 3, ∞).

The curve is decreasing when y ' < 0, that is, in the intervals (-√3,-1), (-1,0), (0,1) and (1, √3).

>> pretty(simple(diff(f,2)))

                                   2
                              x (x  + 3)
                           2 ------------
                                  2     3
                               (x  - 1)

The curve will be concave when y'' > 0, that is, in the intervals (-1,0) and (1, ∞).

The curve will be convex when y'' < 0, that is, in the intervals (0,1) and (- ∞, -1).

The curve has horizontal tangents at the three points at which the first derivative is zero. The equations of the horizontal tangents are y = 0, y = 2.5981 and y = -2.5981.

The curve has vertical tangents at the points that make the first derivative infinite. These points are x = 1 and x = -1. Therefore, the vertical tangents coincide with the two vertical asymptotes.

Next, we graph the curve together with its asymptotes:

>> fplot('[x^3/(x^2-1),x]',[-5,5,-5,5])

We can also represent the curve, its asymptotes and their horizontal and vertical tangents in the same graph.

>> fplot('[x^3/(x^2-1),x,2.5981,-2.5981]',[-5,5,-5,5])

Differentiation in Several Variables

The concept of differentiation for functions of one variable is generalizable to differentiation for functions of several variables. Below we consider partial derivatives for the case of two variable functions.

Given the function f: R² → R, the partial derivative of f with respect to the variable x at the point (a, b) is defined as follows:

In the same way, the partial derivative of f with respect to the variable y at the point (a, b) is defined in the following way:

Generally speaking, we can define the partial derivative with respect to any variable for a function of n variables.

Given the function f: Rⁿ → R, the partial derivative of f with respect to the variable x_i (i = 1,2,...,n) at the point (a₁,a₂,...,a_n) is defined as follows:

The function f is differentiable if all partial derivatives with respect to x_i  (i = 1,2,...,n) exist and are continuous.

Every differentiable function is continuous, and if a function is not continuous it cannot be differentiable.

The directional derivative of the function f with respect to the vector v=(v₁,v₂,...,v_n) is defined as the following scalar product:

is called the gradient vector of f.

The directional derivative of the function f with respect to the vector v =(dx₁,dx₂,...,dx_n) is called the total differential of f. Its value is:

Partial derivatives can be calculated using the MATLAB commands for differentiation that we already know.

As a first example, we study the differentiability and continuity of the function:

if and

To see if the function is differentiable, it is necessary to check whether it has continuous partial derivatives at every point. We consider any point other than the origin and calculate the partial derivative with respect to the variable x:

>> syms x y
>> pretty(simplify(diff((2*x*y)/(x^2+y^2)^(1/2),x)))

        3
     2 y
  ----------
           3
           -
           2
    2    2
  (x  + y )

Now, let's see if this partial derivative is continuous at the origin. When calculating the iterated limits at the origin, we observe that they do not coincide.

>> limit(limit(2*y^3/(x^2+y^2)^(3/2),x,0),y,0)

ans =

NaN

>> limit(limit(2*y^3/(x^2+y^2)^(3/2),y,0),x,0)

ans =

0

The limit of the partial derivative does not exist at (0, 0), so we conclude that the function is not differentiable.

However, the function is continuous, since the only problematic point is the origin, and the limit of the function is 0 = f (0, 0):

>> limit(limit((2*x*y)/(x^2+y^2)^(1/2),x,0),y,0)

ans =

0

>> limit(limit((2*x*y)/(x^2+y^2)^(1/2),y,0),x,0)

ans =

0

>> m = sym('m', 'positive')
>> limit((2*x*(m*x))/(x^2+(m*x)^2)^(1/2),x,0)

ans =

0

>> a = sym ('a', 'real'),
>> f =(2*x*y) /(x^2+y^2) ^(1/2);
>> limit(subs(f,{x,y},{r*cos(a),r*sin(a)}),r,0)

ans =

0

The iterated limits and the directional limits are all zero, and by changing the function to polar coordinates, the limit at the origin turns out to be zero, which coincides with the value of the function at the origin. Thus this is an example of a non-differentiable continuous function. A graphical representation helps us to interpret the result.

As a second example, we consider the function:

We verify the equation:

>> syms x y z
>> f=1/(x^2+y^2+z^2)^(1/2)

f =

1/(x^2 + y^2 + z^2)^(1/2)

>> diff(f,x,2)+diff(f,y,2)+diff(f,z,2)

ans =

(3*x^2)/(x^2 + y^2 + z^2)^(5/2) - 3/(x^2 + y^2 + z^2)^(3/2) + (3*y^2)/(x^2 + y^2 + z^2)^(5/2) + (3*z^2)/(x^2 + y^2 + z^2)^(5/2)

>> simplify(diff(f,x,2)+diff(f,y,2)+diff(f,z,2))

ans =

0

As a third example, we calculate the directional derivative of the function:

at the point (2,1,1) in the direction of the vector v= (1,1,0). We also find the gradient vector of f.

Recall that the directional derivative of the function f in the direction of the vector v= (v₁,v₂,...,v_n) is defined as the following dot product:

is called the gradient vector of f.

First, we calculate the gradient of the function f.

>> syms x y z
>> f=1/(x^2+y^2+z^2)^(1/2)

f =

1/(x^2 + y^2 + z^2)^(1/2)

>> Gradient_f=simplify([diff(f,x),diff(f,y),diff(f,z)])

Gradient_f =

[ -x/(x^2 + y^2 + z^2)^(3/2), -y/(x^2 + y^2 + z^2)^(3/2), -z/(x^2 + y^2 + z^2)^(3/2)]

We then calculate the gradient vector at the point (2,1,1).

>> Gradient_f_p = subs(Gradient_f,{x,y,z},{2,1,1})

Gradient_f_p =

   -0.1361   -0.0680   -0.0680

Finally, we calculate the directional derivative.

>> Directional_derivative_p = dot(Gradient_f_p, [1,1,0])

Directional_derivative_p =

-0.2041

Extreme Points in Several Variables

MATLAB allows you to easily calculate maxima and minima of functions of several variables.

A function f: Rⁿ→R, which maps the point (x₁, x₂,..., x_n)ÎR to f(x₁,x₂,...,x_n)ÎR, has an extreme point at (a₁, a₂, ..., a_n) if the gradient vector is zero at (a₁, a₂, ..., a_n).

By setting all the first order partial derivatives equal to zero and solving the resulting system, we can find the possible maxima and minima.

To determine the nature of the extreme point, it is necessary to construct the Hessian matrix, which is defined as follows:

First, suppose that the determinant of H is non-zero at the point(a₁, a₂, ..., a_n). In this case, we say that the point is non-degenerate and, in addition, we can determine the nature of the extreme point via the following conditions:

If the Hessian matrix at the point (a₁, a₂, ..., a_n)  is positive definite, then the function has a minimum at that point.

If the Hessian matrix at the point (a₁, a₂, ..., a_n) is negative definite, then the function has a maximum at that point.

In any other case, the function has a saddle point at(a₁, a₂, ..., a_n).

If the determinant of H is zero at the point(a₁, a₂, ..., a_n), we say that the point is degenerate.

As an example, we find and classify the extreme points of the function:

We start by finding the possible extreme points. To do so, we set the partial derivatives with respect to all the variables (i.e. the components of the gradient vector of f ) equal to zero and solve the resulting system in three variables:

>> syms x y z
>> f=x^2+y^2+z^2+x*y

f =

x^2 + x*y + y^2 + z^2

>> [x y z] = solve(diff(f,x), diff(f,y), diff(f,z), x, y, z)

x =

0

y =

0

z =

0

The single extreme point is the origin (0,0,0). We will analyze what kind of extreme point it is. To do this, we calculate the Hessian matrix and express it as a function of x, y and z:

>> clear all
>> syms x y z
>> f=x^2+y^2+z^2+x*y

f =

x^2 + x*y + y^2 + z^2

>> diff(f,x)

ans =

2*x + y

>> H=simplify([diff(f,x,2),diff(diff(f,x),y),diff(diff(f,x),z);
               diff(diff(f,y),x),diff(f,y,2),diff(diff(f,y),z);
                    diff(diff(f,z),x),diff(diff(f,z),y),diff(f,z,2)])

H =

[ 2, 1, 0]
[ 1, 2, 0]
[ 0, 0, 2]

>> det(H)

ans =

6

We have seen that the Hessian matrix is constant (i.e. it does not depend on the point at which it is applied), therefore its value at the origin is already found. The determinant is non-zero, so there are no degenerate extreme points.

>> eig(H)

ans =

 1
 2
 3

We see that the Hessian matrix at the origin is positive definite, because its eigenvalues are all positive. We then conclude that the origin is a minimum of the function.

MATLAB additionally incorporates specific commands for the optimization and search for zeros of functions of several variables. The following table shows the most important ones.

As a first example, we minimize the function cos(x) in the interval (3,4).

>> x = fminbnd(inline('cos(x)'),3,4)

x =

    3.1416

In the following example we conduct the same minimization with a tolerance of 8 decimal places and find both the value of x that minimizes the cosine in the range given and the minimum value of the cosine function in that interval, presenting information relating to all iterations of the process.

>> [x, fval, f] = fminbnd (@cos, 3, 4, optimset('TolX',1e-8,...)) (( 'Display', 'iter'));

 Func-count     x          f(x)         Procedure
    1        3.38197    -0.971249        initial
    2        3.61803    -0.888633        golden
    3        3.23607    -0.995541        golden
    4        3.13571    -0.999983        parabolic
    5         3.1413           -1        parabolic
    6        3.14159           -1        parabolic
    7        3.14159           -1        parabolic
    8        3.14159           -1        parabolic
    9        3.14159           -1        parabolic

Optimization terminated successfully:
the current x satisfies the termination criteria using OPTIONS.TolX of 1.000000e-008

In the following example, taking as initial values (-1.2, 1), we minimize and find the target value of the function of two variables:

>> [x,fval] = fminsearch(inline('100*(x(2)-x(1)^2)^2+...
   (((1-x (1)) ^ 2'), [- 1.2, 1])

x =

   1.0000 1.0000

FVal =

     8. 1777e-010

The following example computes a zero of the sine function near 3 and a zero of the cosine function between 1 and 2.

>> x = fzero(@sin,3)

x =

   3.1416

>> x = fzero(@cos,[1 2])

x =

   1.5708

Conditional minima and maxima. The method of “Lagrange multipliers”

Suppose we want to optimize (i.e. maximize or minimize) the function f(x₁,x₂,...,x_n), called the objective function, but subject to certain restrictions given by the equations:

g1(x1,x2,...,xn)=0
g2(x1,x2,...,xn)=0
..........................
gk(x1,x2,...,xn)=0

This is the setting in which the Lagrangian is introduced. The Lagrangian is a linear combination of the objective function and the constraints, and has the following form:

The extreme points are found by solving the system obtained by setting the components of the gradient vector of L to zero, that is, ∇ L(x₁,x₂,...,x_n,λ) =(0,0,...,0). Which translates into:

By setting the partial derivatives to zero and solving the resulting system, we obtain the values of x₁, x₂,..., x_n, λ₁, λ₂,...,λ_k corresponding to possible maxima and minima.

To determine the nature of the points (x₁, x₂,..., x_n) found above, the following bordered Hessian matrix is used:

The nature of the extreme points can be determined by studying the set of bordered Hessian matrices:

For a single restriction g₁, if H1 < 0, H2 < 0, H3 < 0,..., H < 0, then the extreme point is a minimum.

For a single restriction g₁, if H1 > 0, H2 < 0, H3 > 0, H4 < 0, H5 > 0, ... then the extreme point is a maximum.

For a collection of restrictions g_i(x₁,..., x_n) (i = 1, 2,..., k) the lower right 0 will be a block of zeros and the conditions for a minimum will all have sign (-1)^k, while the conditions for a maximum will have alternating signs with H1 having sign (-1)^k+1. When considering several restrictions at the same time, it is easier to determine the nature of the extreme point by simple inspection.”

As an example we find and classify the extreme points of the function:

subject to the restriction:

First we find the Lagrangian L, which is a linear combination of the objective function and the constraints:

>> syms x y z L p
>> f = x + z

f =

x + z

>> g = x ^ 2 + y ^ 2 + z ^ 2-1

g =

x ^ 2 + y ^ 2 + z ^ 2 - 1

>> L = f + p * g

L =

x + z + p *(x^2 + y^2 + z^2-1)

Then, the possible extreme points are obtained by solving the system obtained by setting the components of the gradient vector of L equal to zero, that is, Ñ L(x₁,x₂,...,x_n,λ) =(0,0,...,0). Which translates into:

>> [x, y, z, p] = solve (diff(L,x), diff(L,y), diff(L,z), diff(L,p), x, y, z, p)

x =

 -2^(1/2)/2
  2^(1/2)/2

y =

  2^(1/2)/2
 -2^(1/2)/2

z =

 0
 0

p =

  2^(1/2)/2
 -2^(1/2)/2

By matching all the partial derivatives to zero and solving the resulting system, we find the values of x₁, x₂,..., x_n, λ₁, λ₂,...,λ_k corresponding to possible maxima and minima.

We already see that the possible extreme points are:

(-√2/2, √2/2, 0) and (√2/2, -√2/2, 0)

Now, let us determine the nature of these extreme points. To this end, we substitute them into the objective function.

>> clear all
>> syms x y z
>> f=x+z

f =

x + z

>> subs(f, {x,y,z},{-sqrt(2)/2,sqrt(2)/2,0})

ans =

   -0.7071

>> subs(f, {x,y,z},{sqrt(2)/2,-sqrt(2)/2,0})

ans =

    0.7071

Thus, at the point (-√2/2, √2/2, 0) the function has a maximum, and at the point (√2/2, -√2/2, 0) the function has a minimum.”

Vector Differential Calculus

Here we shall introduce four classical theorems of differential calculus in several variables: the chain rule or composite function theorem, the implicit function theorem, the inverse function theorem and the change of variables theorem.

Consider a function : R^m→  Rⁿ:

(x₁, x₂,..., x_m) → [F₁(x₁, x₂,..., x_m),...,F_n(x₁, x₂,..., x_m)]

The vector function is said to be differentiable at the point a = (a₁,...,a_m) if each of the component functions F₁, F₂,..., F_n is differentiable.

The Jacobian matrix of the above function is defined as:

The Jacobian of a vector function is an extension of the concept of a partial derivative for a single-component function. MATLAB has the command jacobian which enables you to calculate the Jacobian matrix of a function.

As a first example, we calculate the Jacobian of the vector function mapping (x,y,z) to (x * y * z, y, x + z).

>> syms x y z
>> jacobian([x*y*z; y; x+z],[x y z])

ans =

[y * z, x * z, x * y]
[    0,     1,     0]
[    1,     0,     1]

As a second example, we calculate the Jacobian of the vector function at the point (0, -π / 2, 0).

>> syms x y z

>> J = jacobian ([exp(x), cos(y), sin(z)], [x, y, z])

J =

[exp(x), 0, 0]
[0,-sin(y), 0]
[0, 0, cos(z)]

>> subs(J,{x,y,z},{0,-pi/2,0})

ans =

    1 0 0
    0 1 0
    0 0 1

Thus the Jacobian turns out to be the identity matrix.

The Composite Function Theorem

The chain rule or composite function theorem allows you to differentiate compositions of vector functions. The chain rule is one of the most familiar rules of differential calculus. It is often first introduced in the case of single variable real functions, and is then generalized to vector functions. It says the following:

Suppose we have two vector functions

where U and V are open and consider the composite function .

If is differentiable at and is differentiable at , then is differentiable at and we have the following:

MATLAB will directly apply the chain rule when instructed to differentiate composite functions.

Let us take for example and . If we calculate the Jacobian of g at (0, 0) as follows.  

>> syms x y u
>> f = x ^ 2 + y

f =

x ^ 2 + y

>> h = [sin(3*u), cos(8*u)]

h =

[sin(3*u), cos(8*u)]

>> g = compose (h, f)

g =

[sin(3*x^2 + 3*y), cos(8*x^2 + 8*y)]

>> J = jacobian(g,[x,y])

J =

[6 * x * cos(3*x^2 + 3*y), 3 * cos(3*x^2 + 3*y)]
[- 16 * x * sin(8*x^2 + 8*y), - 8 * sin(8*x^2 + 8*y)]

>> H = subs(J,{x,y},{0,0}) 

H =

0  3
0  0

The Implicit Function Theorem

Consider the vector function : A Ì R^{n + m}→ R^m  where A is an open subset of R^{n + m}

If F_i (i = 1, 2,..., m) are differentiable with continuous derivatives up to order r and the Jacobian matrix J = ∂ (F₁,..., F_m) / ∂ (y₁,..., y_m) has non-zero determinant at a point such that , then there is an open UÌRⁿ containing and an open VÌ R^m containing to and a single-valued function : U→ V such that ∀ x ÎU and is differentiable of order r with continuous derivatives.  

This theorem guarantees the existence of certain derivatives of implicit functions. MATLAB allows differentiation of implicit functions and offers the results in those cases where the hypotheses of the theorem are met.

As an example we will show that near the point (x, y, u, v) = (1,1,1,1) the following system has a unique solution:

where u and v are functions of x and y (u = u(x, y), v = v(x, y)).

First, we check if the hypotheses of the implicit function theorem are met at the point (1,1,1,1).

The functions are differentiable and have continuous derivatives. We need to show that the corresponding Jacobian determinant is non-zero at the point (1,1,1,1).

>> clear all
>> syms x y u v
>> f = x * y + y * v * u ^ 2-2

f =

v * y * u ^ 2 + x * y - 2

>> g = x * u ^ 3 + y ^ 2 * v ^ 4-2

g =

x * u ^ 3 + v ^ 4 * y ^ 2 - 2

>> J = simplify (jacobian([f,g],[u,v]))

J =

[2 * u * v * y, u ^ 2 * y]
[3 * u ^ 2 * x, 4 * v ^ 3 * y ^ 2]

>> D = det (subs(J,{x,y,u,v},{1,1,1,1}))

D =

     5

The Inverse Function Theorem

Consider the vector function : U Ì Rⁿ→  Rⁿ where U is an open subset of Rⁿ

(x₁, x₂,..., x_n) → [f₁(x₁, x₂,..., x_n),...,f_n(x₁, x₂,..., x_n)]

and assume it is differentiable with continuous derivative.

If there is an such that |J| = |∂(f₁,...,f_n) / ∂(x₁,...,x_n)| ≠ 0 at x₀, then there is an open set A containing and an open set B containing such that and has an inverse function that is differentiable with continuous derivative. In addition we have:

and if J = ∂ (f1,..., fn) / ∂ (x₁,..., x_n) then |J^-1| = 1 / |J|.

MATLAB automatically performs the calculations related to the inverse function theorem, provided that the assumptions are met.

As an example, we consider the vector function (u(x, y), v(x, y)), where:

We will find the conditions under which the vector function (x(u,v), y(u,v)) is invertible, with x = x (u, v) and y = y(u,v), and find the derivative and the Jacobian of the inverse transformation. We will also find its value at the point (π/4,-π/4).

The conditions that must be met are those described in the hypotheses of the inverse function theorem. The functions are differentiable with continuous derivatives, except perhaps at x= 0. Now let us consider the Jacobian of the direct transformation ∂ (u(x, y), v(x,y)) /∂(x, y):  

>> syms x y
>> J = simple ((jacobian ([(x^4+y^4)/x, sin(x) + cos(y)], [x, y])))

J =

[3 * x ^ 2-1/x ^ 2 * y ^ 4, 4 * y ^ 3/x]
[cos(x),-sin(y)]

>> pretty(det(J))

                               4          4     3
                     3 sin(y) x - sin(y) y + 4 y cos (x) x
                   - ---------------------------------------
                                        2
                                       x

Therefore, at those points where this expression is non-zero, we can solve for x and y in terms of u and v. In addition, it must be true that x π 0.

We calculate the derivative of the inverse function. Its value is the inverse of the initial Jacobian matrix and its determinant is the reciprocal of the determinant of the initial Jacobian matrix:

>> I=simple(inv(J));
>> pretty(simple(det(I)))

                                        2
                                       x
                   - ---------------------------------------
                               4          4     3
                     3 sin(y) x - sin(y) y + 4 y cos (x) x

Observe that the determinant of the Jacobian of the inverse vector function is indeed the reciprocal of the determinant of the Jacobian of the original function.  

We now find the value of the inverse at the point (π/4,-π/4):

>> numeric(subs(subs(determ(I),pi/4,'x'),-pi/4,'y'))

ans =

   0.38210611216717

>> numeric(subs(subs(symdiv(1,determ(J)),pi/4,'x'),-pi/4,'y'))

ans =

   0.38210611216717

Again these results confirm that the determinant of the Jacobian of the inverse function is the reciprocal of the determinant of the Jacobian of the function.

The Change of Variables Theorem

The change of variable theorem is another key tool in multivariable differential analysis. Its applications extend to any problem in which it is necessary to transform variables.

Suppose we have a function f(x,y) that depends on the variables x and y, and that meets all the conditions of differentiation and continuity necessary for the inverse function theorem to hold. We introduce new variables u and v, relating to the above, regarding them as functions u = u(x,y) and v = v(x,y), so that u and v also fulfill the necessary conditions of differentiation and continuity (described by the inverse function theorem) to be able to express x and y as functions of u and v: x=x (u,v) and y=y(u,v).

Under the above conditions, it is possible to express the initial function f as a function of the new variables u and v using the expression:

f(u,v) = f (x(u,v), y(u,v))|J| where J is the Jacobian ∂ (x (u, v), y(u,v)) /∂(u, v).

The theorem generalizes to vector functions of n components.

As an example we consider the function and the transformation u = u(x,y) = x + y, v = v(x,y) = x to finally find f(u,v).

We calculate the inverse transformation and its Jacobian to apply the change of variables theorem:

>> syms x y u v
>> [x, y] = solve('u=x+y,v=x','x','y')

x =

v

y =

u-v

>> jacobian([v,u-v],[u,v])

ans =

[0,   1]
[1, - 1]

>> f = exp(x-y);
>> pretty (simple (subs(f,{x,y},{v,u-v}) * abs (det (jacobian ()))
   ((([v, u-v], [u, v])))

                                 exp(2 v-u)

The requested function is f(u,v) = e^2v-u.

Series Expansions in Several Variables

The familiar concept of a power series representation of a function of one variable can be generalized to several variables. Taylor’s theorem for several variables theorem reads as follows:

Let , (x₁,...,x_n) → f(x₁,...,x_n), be differentiable k times with continuous partial derivatives.

The Taylor series expansion of order k of at the point is as follows:

Here

R = remainder.

Normally, the series are given up to order 2.

As an example we find the Taylor series up to order 2 of the following function at the point (1,0):

>> pretty(simplify(subs(f,{x,y},{1,0})+subs(diff(f,x),{x,y},{1,0})*(x-1)
+subs(diff(f,y),{x,y},{1,0})*(y)+1/2*(subs(diff(f,x,2),{x,y},{1,0})* (x-1)^2+subs(diff(f,x,y),{x,y},{1,0})*(x-1)*(y)+ subs(diff(f,y,2),{x,y},{1,0})* (y)^2)))

              2
         2   y
  (x - 1)   --- + 1
             2

Curl, Divergence and the Laplacian

The most common concepts used in the study of vector fields are directly treatable by MATLAB and are summarized below.

Definition of gradient: If h = f(x,y,z), then the gradient of f, which is denoted by D f(x, y, z), is the vector:

Definition of a scalar potential of a vector field: A vector field is called conservative if there is a differentiable function f such that . The function f is known as a scalar potential function for .

Definition of the curl of a vector field: The curl of a vector field is the following:

Definition of a vector potential of a vector field: A vector field F is a vector potential of another vector field G if F = curl (G).

Definition of the divergence of a vector field: The divergence of the vector field is the following:

Definition of the Laplacian: The Laplacian is the differential operator defined by:

As a first example, we calculate gradient and Laplacian of the function:

>> gradient=simplify([diff(f,x), diff(f,y), diff(f,z)])

gradient =

[x /(-x^2-y^2-z^2 + 1) ^(3/2), y /(-x^2-y^2-z^2 + 1) ^(3/2), z /(-x^2-y^2-z^2 + 1) ^(3/2)]

>> pretty (gradient)

  +-                                                                     -+
  |            x                      y                      z            |
  |  ---------------------, ---------------------, ---------------------  |
  |                      3                      3                      3  |
  |                      -                      -                      -  |
  |                      2                      2                      2  |
  |      2    2    2            2    2    2            2    2    2        |
  |  (- x  - y  - z  + 1)   (- x  - y  - z  + 1)   (- x  - y  - z  + 1)   |
  +-                                                                     -+

>> Laplacian = simplify ([diff(f,x,2) + diff(f,y,2) + diff(f,z,2)])

Laplacian = 

3 /(-x^2-y^2-z^2 + 1) ^(5/2)

>> pretty (Laplacian)

            3
  ---------------------
                      5
                      -
                      2
      2    2    2
  (- x  - y  - z  + 1)

As a second example, we calculate the curl and the divergence of the vector field:

>> M = atan (x/y)

M =

atan (x/y)

>> N = log (sqrt(x^2+y^2))

N =

log ((x^2 + y^2) ^(1/2))

>> P = 1

P =
     1

>> Curl = simplify ([diff(P,y)-diff(N,z), diff(P,x)-diff(M,z), diff(N,x)-diff(M,y)])

Curl =

[0, 0, (2 * x) /(x^2 + y^2)]

>> pretty (Curl)

  +-               -+
  |          2 x    |
  |  0, 0, -------  |
  |         2   2   |
  |        x + y    |
  +-               -+

>> Divergence = simplify (diff(M,x) + diff(N,y) + diff(P,z))

Divergence =

(2 * y) /(x^2 + y^2)

>> pretty (divergence)

    2 y
  -------
   2   2
  x + y

Rectangular, Spherical and Cylindrical Coordinates

MATLAB allows you to easily convert cylindrical and spherical coordinates to rectangular, cylindrical to spherical coordinates, and their inverse transformations. As the cylindrical and spherical coordinates, we have the following:

In a cylindrical coordinate system, a point P in the space is represented by a triplet (r, θ, z), where:

• r is the distance from the origin (O) to the projection P' of P in the XY plane
• θ is the angle between the X axis and the segment OP'
• z is the distance PP'

In a spherical coordinate system, a point P in the space is represented by a triplet (ρ, θ, φ), where:

• ρ is the distance from P to the origin
• θ is the same angle as the one used in cylindrical coordinates
• φ is the angle between the positive Z axis and the segment OP

The following conversion equations are easily found:

Cylindrical to rectangular:

Rectangular to cylindrical:

Spherical to rectangular:

Rectangular to spherical:

As a first example we express the surfaces with equations given by xz = 1 and x² + y² + z² = 1 in spherical coordinates.

>> clear all
>> syms x y z r t a
>> f = x * z-1

f =

x * z - 1

>> equation = simplify (subs (f, {x, y, z}, {r * sin(a) * cos(t), r * sin(a) * sin(t), r * cos(a)}))

equation =

r ^ 2 * cos(a) * sin(a) * cos(t) - 1

>> pretty (equation)

   2
  r cos(a) sin(a) cos(t) - 1

g =

x ^ 2 + y ^ 2 + z ^ 2 - 1

>> equation1 = simplify (subs (g, {x, y, z}, {r * sin(a) * cos(t), r * sin(a) * sin(t), r * cos(a)}))

equation1 =

r ^ 2 - 1

>> pretty (equation1)

   2
  r -1

However, MATLAB provides commands that allow you to transform between different coordinate systems.

Below are the basic MATLAB commands which can be used for coordinate transformation.

The following example transforms the point (π1, 2) in polar coordinates to Cartesian coordinates.

>> [X, Y, Z] = pol2cart(pi,1,2)

X =

    -1

Y =

  1. 2246e-016

Z =

     2

Next we transform the point (1,1,1) in Cartesian coordinates to spherical and cylindrical coordinates.

>> [X, Y, Z] = cart2sph(1,1,1)

X =

    0.7854

Y =

    0.6155

Z =

    1.7321

>> [X, Y, Z] = cart2pol(1,1,1)

X =

    0.7854

Y =

    1,4142

Z =

     1

The following example transforms the point (2,π/4) in polar coordinates into Cartesian coordinates.

>> [X, Y] = pol2cart(2,pi/4)

X =

   -0.3268

Y =

    0.7142