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=
x_{0}. The
derivative
of
f(
x) at
x=
x_{0} is defined by
the limit
f 

(x_{0})= 

(f(x)f(x_{0}))/(xx_{0}),

where the quantity (
f(
x)
f(
x_{0}))/(
x
x_{0}), 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
x_{0} during the limit process described
above, it is convenient to let
x=
x_{0}+
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(x_{0}+h)f(x_{0}))/h
and
f 

(x_{0})= 

(f(x_{0}+h)f(x_{0}))/h.

Therefore, given a small value of h, difquo, as defined above, is an
approximation of the derivative of
f(
x) at
x_{0}. The smaller the
value of h is, the better the approximation. Furthermore, since
x_{0}
can be any point in the domain of
f(
x), we write
dy/dx = f 

(x)= 

(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 [0,2
p] 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
x_{0} feet with an initial upward velocity of
v_{0} feet/sec will
fall according to the formula (0
£ t £ T).
y(t) = x_{0} + v_{0} t  16 t^{2}.
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) = x_{0}  16 t + 
( 
(16 + v_{0})/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
x_{0} = 48, and
v_{0} = 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 t^{2} + 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 windresistance 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).
Now answer questions about the ball with windresistance.
 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 uparrow 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:


F(x,y) = (F_{x}(x,y),F_{y}(x,y)) · (1, 

)
= 0

Then solve to get
where F_{x} 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
x^{2} 
y^{6} = 1. Find
dy/
dx at the point
x
= 1/2 and
y = (1  (1/2)
^{2})
^{(}1/6). Theoretically, we get


(x^{2}  y^{6}) = 2x  6y^{5} 

=


1 = 0

or
Using the difference quotient to estimate
F_{x} and
F_{y} we get
instead the following. Suppose
F(
x,
y) =
x^{2}  6
y^{5} is defined in
an mfile.
>> 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
F_{x} and
F_{y}
F_{x}(x,y) » 

,
F_{y}(x,y) » 

9.2 Symbolic derivatives