Derivatives

## 9   Derivatives

### 9.1   Difference quotients

Concept: Definition of a derivative

Let f(x) be a function defined in a neighborhood of the point x=x0. The derivative of f(x) at x=x0 is defined by the limit
f (x0)=
 lim x ® x0
(f(x)-f(x0))/(x-x0),
where the quantity (f(x)-f(x0))/(x-x0), whose limit is being taken, is called the difference quotient since it is the ratio of the change in f to the change in x.

Since x is close to x0 during the limit process described above, it is convenient to let x=x0+h and think of h as being small and approaching to zero. Note that h can be positive or negative corresponding to the approach from the right or from the left, respectively. Thus we have the definitions (you may be reminded of the secant line example )

difquo = (f(x0+h)-f(x0))/h
and
f (x0)=
 lim h ® 0
(f(x0+h)-f(x0))/h.

Therefore, given a small value of h, difquo, as defined above, is an approximation of the derivative of f(x) at x0. The smaller the value of h is, the better the approximation. Furthermore, since x0 can be any point in the domain of f(x), we write
dy/dx = f (x)=
 lim h ® 0
(f(x+h)-f(x))/h
to emphasize that f(x) is not just a number but a function of x which is defined at all points where the limit exists. If f(x) exists at all points in an interval, f(x) is said to be differentiable in that interval.

So, to find the approximate derivative of a differentiable function f(x) over an interval, we may use the difquo function which is
difquo(x)= (f(x+h)-f(x)) /h,
using a small value of h such as h=0.1 or 0.01.

Example: Graphing the difference quotient:

Let us create a plot of the derivative of f(x)=cos x over the interval [0,2p] based on the numerical evaluation of its difference quotient. We shall use h=0.01 in evaluating the difference quotient. The following MATLAB commands will plot both the function and its approximate derivative together.
>> h=0.01;
>> x=0:0.001:2*pi;                      % step size should be smaller
% than h
>> difquo=(cos(x+h)-cos(x))./h;         % difference quotient
>> plot(x,cos(x),'r',x,difquo,'b'),grid % plot cos in red and difquo in blue
>> axis([0 2*pi -1 1])                  % a better frame for the
%  graph


Figure 21: plot of cos(x) and its difference quotient

Here is another example which shows that we can reduce typing quite a bit, and hence possible errors.

Example: Baseball with wind correction
If there is constant acceleration, a baseball thrown from a height of x0 feet with an initial upward velocity of v0 feet/sec will fall according to the formula (0 £ t £ T).
y(t) = x0 + v0 t - 16 t2.
Here T is the time that the ball hits the ground.

If we assume a slightly more realistic model, then the acceleration will depend upon the velocity of the baseball. A model for this yields the new formula for 0 £ t £ T.
 w(t) = x0 - 16 t + ( (16 + v0)/2 ) ( 1 - e-2t ) .

Let's suppose we want to follow the path of a baseball thrown from a fourth story window during the Yankees World Series Championship parade. Let x0 = 48, and v0 = 32.

• Plot both functions y(t) and w(t) so that you can determine from the graph the respective values of T.
To find an estimate on the domain to plot the the two functions we note that the baseball under y(t) will hit the ground when the polynomial y = -16 t2 + 32 t + 48.
>> roots([-16 32 48])           % find the zeroes
ans = 3, -1                     % actually a column vector

So T=3 for y(t) when the ball hits the ground. We expect the ball with wind-resistance to fall more slowly, so let's plot the functions over the interval [0,5].

Figure 22: plot of ball falling under 2 different models

>> t = linspace(0,5); x0=48;v0=32;
>> y = -16*t.^2 + v0*t + x0;    % remember the .^
>> w = x0 - 16*t + (16+v0)/2 * ( 1 - exp(-2*t));
>> plot(t,y,t,w);
>> T = 4.5;                     % an approximation

We see that 4.5 is about correct for T for w(t).

• What is the average velocity of the baseball during [0,T].
This is given by (w(T) - w(0))/T. To compute this without typing too much, let's use the up-arrow keys to give w. Notice w is defined for t = linspace(0,5) but we'll just redefine it to be T.
>> t = 0;
>> w0 = x0 - 16*t + (16+v0)/2 * ( 1 - exp(-2*t)); % use up arrow key,
% and edit
>> t = T;
>> wT = x0 - 16*t + (16+v0)/2 * ( 1 - exp(-2*t)); % use up arrow key
% and edit
>> avg = (wT - w0)/T            % pretty easy if done smartly

• Let h= 0.1, plot the difference quotient of w(t) over the interval [0,T].
We want to again use our definition of w(t) to minimize typing. Notice, we need to add an h in the right place.
>> t = linspace(0,T); h = 0.1;
>> w = x0 - 16*t + (16+v0)/2 * ( 1 - exp(-2*t)); % from up arrow
>> wh = x0 - 16*(t+h) + (16+v0)/2 * ( 1 - exp(-2*(t+h))); % from up
% arrow.
>> difquo = (wh - w)/h;
>> plot(t,difquo)

• What is the time t when the ball is at its highest point? What is the height?
Using Calculus, we know this is when the derivative is 0, from the graph of the difference quotient this is t=1.

• The difference quotient is a good approximation to the instantaneous velocity. From the graph, find the instantaneous velocity when the ball hits the ground?
We see from the graph that it has a horizontal asymptote of -16. This says that the velocity stops decreasing after a certain amount. This is called a terminal velocity. From the graph it appears that T is big enough so that the ball is very near the terminal velocity, so we'll say the instantaneous velocity is -16 when the ball strikes the ground/

Concept: Implicit differentiation
Sometimes you are confronted with finding rates of change but you don't have a function relating x and y but rather an equation. That is y is implicitly defined in terms of x. The trick in mathematics is to take d/dx of both sides, use the chain rule properly and then solve for dy/dx in terms of x and y.

Using MATLAB to find numeric estimates requires a slightly different tack. Here is the main idea.
• Turn you equation into a problem involving level curves of a function: F(x,y) = c for some constant c.
• Differentiate using the chain rule:
d
dx
F(x,y) = (Fx(x,y),Fy(x,y)) · (1,
dy
dx
) = 0
Then solve to get
dy
dx
= -
Fx(x,y)
Fy(x,y)
where Fx is the partial derivative in the x variable. We can use MATLAB to find this just as before because partial derivatives are just like regular derivatives of a function of a single variable.
Here is an example. Let x2 -y6 = 1. Find dy/dx at the point x = 1/2 and y = (1 - (1/2)2)(1/6). Theoretically, we get
d
dx
(x2 - y6) = 2x - 6y5
dy
dx
=
d
dx
1 = 0
or
dy
dx
=
2x
6y5

Using the difference quotient to estimate Fx and Fy we get instead the following. Suppose F(x,y) = x2 - 6y5 is defined in an m-file.
>> y0 = (1 - 1/4)^(1/6)
y0 = 0.95318
>> x0 = 1/2
x0 = 0.50000
>> h = .01
h = 0.010000
>> -(F(x0+h,y0) - F(x0,y0)) / (F(x0,y0+h) - F(x0,y0))
ans = -0.20839                  % this is the approximate value
>> (-2*x0)/(6*y0^5)
ans = -0.21182                  % this is the exact answer

The answer is not too far off and could be improved by using a smaller value of h. Notice that we used the secant line approximation for Fx and Fy
Fx(x,y) »
F(x+h,y) - F(x,y)
h
, Fy(x,y) »
F(x,y+h) - F(x,y)
h