The Derivative -- Definition

If someone asks you how fast you were driving, the usual answer would be something like: 90 km/hr or "90 kilometres per hour". (90 k.p.h. really means which really means that, on average , 90 kilometers were covered in one hour (or would have been, if the trip were long enough). i.e. when people report their speed, they are reporting their average speed over an interval of time.

Definition : The average velocity of an object travelling in a straight line over an interval of time = (if the motion is all in one direction, then the total displacement is the same as the distance covered. When we speak about velocity as opposed to speed , we allow that velocity has a direction component to it. In the strictly one-dimensional case that we will study, direction can only be '+' or '-' (i.e. up or down, left or right, etc.). In this case, speed is the absolute value or magnitude of the velocity: .

Big e.g.

Problem : A ball is dropped from the top of the CN Tower (450 m above the ground!) on a day after which an alien invading race has just ripped Earth's atmosphere away (i.e. air resistance can be neglected). The height of the ball t seconds after release is given by . Find the velocity of the ball after exactly 5 seconds.

> restart:f:=t->450-4.9*t^2;

The average velocity of the ball over the time interval [t,5] is:

> v_avg:=(f(t)-f(5))/(t-5);

For instance, to find the average velocity of the ball over the interval beween t = 4s and t = 5s we plug in t=4 into v_avg and get -44.1 m/s which means that during the interval of the fourth second, the ball travelled 44.1 meters in the downward direction.

> subs(t=4,v_avg);

Let's see what happens to the average velocity over smaller and smaller time intervals that include t = 5s by plugging in values of t into v_avg that are closer and closer to 5:

From the table, it seems that the smaller the time interval that contains t = 5 seconds, the closer the average velocity for the ball is to -49 m/s. It seems natural to call -49 m/s the velocity of the ball when t is exactly 5 seconds. What we are really doing is defining the velocity to be the limit of the average velocities as the time intervals around the time in question shrink to zero. Now, find the limit in question and compare to your "guess:"

= = -49

Definition : If the position at time t of a particle moving in a straight line is given by then the (instantaneous) velocity of the particle at time a is v( a ) = f'( a ) = = where stands for "the change in t " and is the corresponding change in y = f(t).

e.g.

Suppose a ball is thrown upward with a velocity of 40 f/s and has the positon equation .

a) Find its (instantaneous) velocity at time 2 seconds.

b) Find its maximum height.

c) Find its speed when it hits the ground.

> s:=t->40*t-16*t^2;

a)

> v_avg:=(s(t)-s(2))/(t-2);

> Limit(v_avg,t=2)=limit(v_avg,t=2);

==> It's velocity after 2 seconds is -24 f/s (why '-' ?).

Note:

> simplify(v_avg);

b) For (b) we first need its velocity at any time a:

> v_avg2:=(s(t)-s(a))/(t-a);

> Limit(v_avg2,t=a);

> velocity:=value(%);

==> at time a , we have v(a) = -32a +40 and the ball is at max. height when v(a) = 0!

> solve(velocity=0);

==> Maximum height after 5/4 seconds.

c) It hits ground when the height is zero (s(t) =0):

> solve(s(t)=0);

Its velocity at impact is therefore:

> subs(a=5/2,velocity);

==> The speed on impact is 40 f/s.

General Rates of Change.

Definition : If then the average rate of change of f with respect to x (wrt x) over the interval [a,x] is and the ( instantaneous) rate of change of y wrt x is f'( a ) = = .

f'( a ) is called the derivative of f when x = a .

Big Fact : f'(a) = = .

e.g. Find the rate of change of f(x) = when x = -1 using both limit definitions.

Note : The rate of change when x = -1 is none other than f'(-1).

> restart:

> f:=x->x^2-2*x;

Note: in both definitions, we take a = -1.

 1)

> (f(x)-f(-1))/(x+1);

> anwser1:=Limit((x^2-2*x-3)/(x+1),x=-1)=limit((x^2-2*x-3)/(x+1),x=-1);

2)

> (f(-1+h)-f(-1))/h;simplify(%);

> answer2:=Limit((f(-1+h)-f(-1))/h,h=0)=limit((f(-1+h)-f(-1))/h,h=0);

Note: We get the same answer with both approaches and are forced to conclude that f'(-1) = -4 = the rate of change of f with respect to x when x = -1.

e.g. Find g'(a) if both ways.

1)

> g:=x->1/x;

> (g(x)-g(a))/(x-a);

> answer1:=limit(%,x=a);

==> g'(a) = .

2)

> (g(a+h)-g(a))/h;

> answer2:=limit(%,h=0);

==> g'(a) = .

The Slope of the Tangent Line to at ( a, f ( a )).

> restart:with(plots):f:=x->4-x^2;

Consider the secant lines to y = f(x) that pass through (1,f(1)) and (1+h, f(1+h)) for "small" values of h. We can use use point slope formula:

The following will make m, the slope of the secant line to f, a function of h and f:

> m:=h->(f(1+h)-f(1))/h;

> m(h);

> simplify(m(h));

This will make the secant line through (1,f(1)) and (1+h,f(1+h)) a function of f and h.

> line:=h->f(1)+m(h)*(x-1);

> simplify(line(h));

> m(1);line(1);

> plot([f(x),line(1.5),line(1),line(1/2),line(1/4)],x=0..3,y=-2.5..5,colour=[black,red,navy,magenta,sienna],thickness=2);

Notice that as h gets smaller, the secant line through (1,f(1)) and (1+h,f(1+h)) gets closer and closer to a line that just touches the graph at (1,f(1)).

If you are viewing this in Maple you can go back and experiment with different choices for f and different values of h. Try it!!!

Definition : The tangent line to y = f(x) at (a,f(a)) is the line through (a,f(a)) with slope m = = = f'( a ) if this limit exists . It's equation is therefore y - f(a) = f'(a)(x - a) .

To finish off the above example then:

> fprime:=Limit(m(h),h=0)=limit(m(h),h=0);

You can then quickly see that the tangent line is then y = -2x + 5.

> plot({f(x),-2*x+5},x=0..3,y=-2.5..5,colour=[black,red],thickness=2);

e.g. Find the slope of y = f(x) at (a,f(a)) if f(x)=x^3. Use this to find the tangent line to f at (2,8).

> f:=x->x^3;

> m:=h->(f(a+h)-f(a))/h;

> m(h);

> simplify(%);

> slope:=Limit(m(h),h=0)=limit(m(h),h=0);

Note: "slope" is f'(a) = the slope to y = f(x) at (a,f(a)). At (2,8) the slope is m1 = f'(2):

> m1:=subs(a=2,slope);

It follows that the tangent line there is y - 8 = 12(x - 2) or y = 12x - 16. To check, we can look at a plot of the two:

> plot([x^3,12*x-16],x=-1..4,y=-2..18,colour=[red,black],thickness=2);

Looks good to me!

More Definitions and E.G.'s

Definition : The derivative of f at a is f'( a) = = if the limit exists. If the number a is replaced by the independent variable x then we have the function f', called the derivative of f .

A function f is called differentiable at a if f'( a ) exists and is differentiable on the open interval I if it is differentiable at every number a I.

e.g. Find the derivative of .

> restart:f:=x->sqrt(x-1);

> expr1:=(f(x+h)-f(x))/h;expr2:=(f(x)-f(a))/(x-a);

> limit(expr1,h=0);limit(expr2,x=a);

> plot([f(x),1/2*1/(sqrt(x-1))],x=0..10,y=0..6,colour=[black,red],numpoints=1000,thickness=2);

Compare the domains of f and f' and make a generalization.

Show that is not differentiable.

> restart:expr:=(abs(x+h)-abs(x))/h;

> Limit(expr,h=0)=limit(expr,h=0);

"expr2" below will become h'(0). Why?

> expr2:=subs(x=0,expr);

> simplify(%);

> Limit(expr2,h=0,left)=limit(expr2,h=0,left);Limit(expr2,h=0,right)=limit(expr2,h=0,right);Limit(expr2,h=0)=limit(expr2,h=0);

> plot(abs(x),x,colour=red,thickness=2);

For a slightly different approach you might consider the following.

> assume(x<0);

> limit(expr,h=0);

> assume(x>0);limit(expr,h=0);

Now, draw a plot of y = , y = -1 and y = 1 all on the same axes and comment intelligently.

Notations for the Derivative

Functional Notations: f' and y'

If y = f( x ) is a function then f'( x ) stands for the derivative function of f with respect to x . y' is shorthand for f'( x ) .

e.g. If y = = then y' = f'( x ) = .

It is wrong to write f'( ) = . Because this is functional notation , f'( ) means plug in into the f' function. Therefore, if f(x) = then f'( ) would mean which is not f'(x) !!

Operator Notations:

If y = f( x ) then the symbol is read "the derivative of y with respect to x" and is an instruction to differentiate whatever y is: it is telling you to perform the operation of differentiation on y.

Therefore, if then = = . i.e. means differentiate it does not mean plug into the derivative funcion as f'( ) would mean.

The following are all equivalent operator notations for the differentiation process:

= =

Differentiability ==> Continuity

Theorem : If f is differentiable at a then f is continuous at a.

proof : See the blackboard! (Or study page 169 in Stewart.)