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A “derivative” is just a fancy name that describes how something is changing with respect to something else. What follows will be a brief summary and insight into this world of ever changing quantities called derivatives.
1.
A Physics 100 Tutorial
2.
Why Do We Need Derivatives?
In physics, and life too, things are constantly
“changing.” Specifically, what we’ll be interested with
in the physics context is how physical quantities
change. For example, how does an object’s velocity
change over time, or how does the force acting on an
object change over a distance traveled. Such
changes are described mathematically by derivatives.
A “derivative” is just a fancy name that describes
how something is changing with respect to something
else. What follows will be a brief summary and
insight into this world of ever changing quantities
called derivatives.
3.
Getting Started
Let’s start simple. Consider the function 5
y(x) = 3 shown in the figure to the right. 4
y(x) = 3
If you were asked “How does this
function change with x?” or equivalently, y 3
“How does y change as a function of x? ,”
you would say, “It doesn’t change. It is a 2
constant value of 3 everywhere.” And 1
your response would be correct, but how
would we describe your response 0
mathematically? x1 x2 x
4.
Mathematics of “Change”
Well, we define the change in the function y(x) with respect to
the variable x, Δy/Δx, to be
y Y (x 2 ) y (x 1 )
.
x x 2 x1 Understanding the symbols…
In English, this equation will tell us how the function y(x) The symbol “Δ,” called
delta, represents the
changes over the distance between x1 and x2. Since y always change in a variable.
Translated, Δy reads:
equals 3 in our previous example, then “The change in y,” which
equals:
y 3 3 “y final minus y initial”
0. or mathematically:
x x2 x1 “Δy = yfinal – yinitial.”
Therefore, the function y(x) does not change over the interval
x1 and x2. This mathematics will be how we describe the change
in any function with respect to a given variable.
5.
Example of a Straight Line
Now consider the function y(x) = 3x + 1 as
drawn in the figure to the right. Again, how
would y(x) change with x? Let’s look at the The slope of the line
interval between x = 2 and x = 3: 20
equals Δy/Δx
y(x) = 3x + 1
Using our definition for the change in y(x)
with respect to x from the previous slide (hit
15
the left arrow key if you need to back to the
previous slide), we get: 10
Y
Δy=3
y y (x 3) y (x 2) [3 * (3) 1] [3 * (2) 1] 10 7 5
Δx=1
3
x 3 2 3 2 1
0
0 1 2 3 4 5 6
If we look at this graphically in Figure 2, X
we see that it is just the slope of the
line!!! If we look at any interval of x, we
would find that y(x) would change by the Δy/Δx = 3
same amount, 3, over that interval because The function y changes by 3 units
this function is just a straight line! Try it between x=2 and x=3.
and see!!!
6.
More than Straight Lines
On the previous slide, you learned that
the slope of a straight line gives you the
change in the function y(x) over the
change in x. That concept will become the
building block for everything that follows.
y
Now let us consider a more complicated
function other than a straight line.
Consider the graph to the right. How
would you describe how the function
y(x) changes at any given value of x?
Well, from what you just learned, the x
change in the function y(x) with
respect to x should be given by the
slope of a straight line. But what is the
straight line???
7.
Graphs that Curve
Let us just get started and choose a
Slope of the tangent line at
point X on the graph. The question to point X gives the change in
be answered is “How does y(x) change at y(x) with respect to x.:
Slope=Δy/Δx.
the point X?
y
To give the answer away, the change in
y(x) with respect to x at point X is
given by the slope of the tangent line
at point X !!!
The question now is “How do we
determine the slope of the tangent line
at point X?
X x
Understanding the terminology…
The word “tangent line” describes a line that intersects or touches a curve at only one point. At any given point on a
smooth curve, there is only one unique tangent line and therefore there is only one value for the slope of that tangent
line at that point.
8.
Determining the Slope of the
Tangent Line
Because h is smaller, the point X+h is closer to the
original point X than before. Therefore, the slope of
To determine the slope of the tangent this second line is closer in value to the slope of the
tangent line than what the slope of the first line was.
line, let us draw a different line that
intersects the curve at both point X and
point X + h.
What is the slope of this line? From
before, the slope of the line will be Δy/Δx:
y
Δy=y(X+h)–y(X)
Δy y(X h) y(X) y(X h) y(X)
Δx (X h) X h
Δx=(X+h)–X=h
This line is still not the tangent line at point
X, but we can make it look more like the
tangent line if we make the h a smaller value:
X X+hX+h x
If you calculate the slope for this second
line, it will have the same form as the above
equation, except now h is a smaller value and
therefore y(X+h) will be a different value.
9.
So What is a Derivative Anyway?
Tangent Line
Now, if we keep making h smaller and
smaller, then the line that passes through
the points X and X+h will start looking
more and more like the line tangent to the y
curve at point X. Eventually, as h goes to
zero, then the line that goes through X
and X+h will become the tangent line!!!
h goesh to
h gets zero!
even
gets smaller
smaller
Let animation run!!!
So, let’s write this out in mathematics…
X X+hX+hX+h x
y ( X h ) y (X )
Slope of tangent line = ( as h goes to zero)
h
dy
We give the symbol to represent the slope of the tangent line.
dx
dy
This symbol, , is what we call the derivative of y with respect to x.
dx
Therefore, the term “derivative” just represents how the function y(x) instantaneously
changes with respect to the variable x. As h goes to zero, Δy/Δx becomes dy/dx.
10.
Recap So Far
Let us recap what you have learned so
• The slope of a straight line tells you how the function y(x) changes as the
variable x changes:
y y (xfinal ) y (xinitial )
slope
x xfinal xinitial
• The “derivative” of y(x) at point X is the slope of the tangent line to the
curve of y(x) at point X.
• The derivative, dy/dx, is defined mathematically by the following equation:
dy y ( x h ) y (x )
( as h goes to zero)
dx h
• The derivative, dy/dx, is the instantaneous change of the function y(x).
• As h goes to zero, Δy/Δx becomes dy/dx.
11.
Using the Definition for Derivatives
Let us now apply our newly derived formula Tangent line at x = 5 has a slope of 10.
to calculate the derivative of y(x) = x2. Therefore, the function y(x) has an
instantaneous slope of 10 units at x=5.
dy y ( x h ) y (x )
( as h goes to zero) y = x2
dx h
((x h )2 ) (x 2 )
( as h goes to zero)
h
(x 2 2xh h 2 ) x 2
( as h goes to zero)
h
2xh h 2
( as h goes to zero)
h
2x h ( as h goes to zero)
2x -15 -10 -5 0 5 10 15
dy
And therefore, 2x
dx
Let us use this result to determine the derivative at x = 5. Since the derivative of
y(x)=x2 equals 2x, then the derivative at x = 5 is 2*5 = 10. Therefore, the slope of the
tangent line that passes through x = 5 has a slope of 10!
12.
Graphing the Derivative
In our previous example, we used the
y = x2
definition for the derivative to find the
derivative of the function y = x2. When we
did this, we found the derivative to be a
y
function itself: dy/dx = 2x. This is just a
straight line as plotted to the bottom
-15 -10 -5 0 5 10 15
x
Let us see how the two graphs are related. The slope of the tangent line at x = -4 is -8.
The
The
slope
slope
of of
thethe
tangent
tangent
lineline x =x-9
at at = 0isis-18.
0.
• You know that the derivative of a
function is just the slope of that
25
dy/dx = 2x 20
function. For example, look at the 15
graph of y = x2, for negative values of 10
x, the slope of the tangent line should 5
d y/d x
be negative. Looking at the graph of -15 -10 -5
0
0 5 10 15
dy/dx, when x is negative, dy/dx is also
-5
-8
-10
negative! -15
-18
• When the slope of the tangent line -20
equals zero, then the value of the -25
x
derivative will equal zero!
13.
Another Example
“Repetition is good for the soul” as one 60
teacher used to say, therefore, let us y=4x3–15x2+20 40
now do another example. Consider the 20
formula y =4x3-15x2+20. This function 0
is graphed to the right. Calculating the
-3 -2 -1 0 1 2 3 4 5
y
-20
derivative, we find: -40
-60
-80
dy y (x h ) y (x )
( as h goes to zero )
x
Definition of derivative
dx h
[4(x h )3 15(x h )2 20] [4x 3 15x 2 20]
( as h goes to zero )
Substituted in the expression for y(x)
h
[4(x 3 3x 2h 3xh 2 h 3 ) 15(x 2 2xh h 2 ) 20] [4x 3 15x 2 20]
( as h goes to zero)
h
12x 2h 12xh 2 h 3 30xh 15h 2
( as h goes to zeroTerms
) that survived after some terms canceled
h
12x 2 12xh h 2 30x 15h ( as h goes to zeroDivided ) each term by h
12x 2 30These
x terms survived after h went to zero
14.
Example Continued
We have found that the derivative of
60
y=4x –15x +20
3 2 The slope of the tangent lines
should be zero at these points.
y(x) = 4x3–15x2+20 to be: 40
dy/dx = 12x2-30x. The graph of the 20
original function is plotted in the top 0
right while the graph of its derivative is
-3 -2 -1 0 1 2 3 4 5
y
-20
plotted in the bottom right.
-40
Between x = 0 and x = 2.5, y(x)
Let us compare the two graphs -60
In this region,
shoulddy/dx
haveshould be positive.
a negative slope.
and some of their features: -80
• The original function y(x) in the x
region between x = -2 and x = 0 Indeed, dy/dx
Indeed, between x=0
Indeed, at x has
and x=2.5,
= 0 positive
dy/dx
values
and x = 2.5,
is negative
between
dy/dx equalsx=-2
inx=0.
and
zero.
value.
should have a positive slope. 120
dy/dx = 12x2-30x
• At x = 0 and at x = 2.5, y(x) has 100
critical points (points where the 80
slope of the tangent line equals 60
d y /d x
zero) and therefore its derivative 40
should equal zero at those points. 20
• Between x = 0 and 2.5, y(x) is -3 -2 -1
0
0 1 2 3 4 5
decreasing in value which implies -20
that its derivative is negative in -40
this region. x
15.
The Shortcut…
You have seen so far two examples on calculating derivatives and their graphical
representation and meaning. Certainly, the definition for the derivative can be used
each time when one needs to be determined, but there exists a shortcut when it
comes to functions of the form: y(x) = Axn, where “A” is just a numerical constant and
“n” is an integer, positive or negative. Plugging this expression into the definition for
the derivative, you will find that:
dy/dx = nAxn-1
Using this shortcut to calculate the derivative of y(x) = x2, we get:
dy/dx = 2*x2-1 = 2x.
This is exactly what we got when we used the definition of the derivative several
slides ago. In our second example, we found that the derivative of y =4x3-15x2+20 to
dy/dx = 12x2 – 30x
Does this shortcut work here? The answer is YES!!!
16.
Some Simple Rules of Differentiation
The subject of derivatives is a huge branch of mathematics in of itself and
cannot possibly be contained here in this one tutorial. Hopefully, though,
you now have some knowledge and appreciation for what derivatives are.
The following are commonly known formulas for derivatives.
What are these
symbols?
d d
1) [c ] 0 2) [x ] 1
dx dx
“f” and “g” are
functions of x:
f(x) and g(x).
d
[cx n ] ncx n 1 d “c” represents a
dx 4) [cf ] cf ' constant
dx numerical value
and therefore is
not a function of
d d
5) [f g ] f 'g ' 6) [f n ] nf n 1f ' x.
dx dx “n” represents an
integer number,
positive or
d d
7) [f * g ] f * g ' g * f ' 8) [sin x ] cos x negative.
dx dx f’ is shorthand
for df/dx.
Likewise, g’ is
d d
shorthand for
9) [cos x ] sin x 10) [tan x ] sec2 x dg/dx.
dx dx
17.
Problems to Solve
Use the knowledge you just learned to try to solve the following questions. If you need
assistance, feel free to ask any of the Physics 100 instructors.
1) Find dx / dt : x (t ) 30 10t 4.9t 2
2) Find dy/dθ : y θ 15 sinθ 3 θ 5
3) Based upon the graph shown, where should the derivative of this function be
positive, negative, and zero?
3
2
1
0
y -4 -3 -2 -1 0 1 2 3 4 5
-1
-2
-3
-4
x
18.
Answers to Problems
1) dx / dt 10 9.8t
2) dy/dθ 15cosθ 15θ 4
3
3) The derivative is:
a) negative between -3 and -2,
2
0 and 2, and between 1
3 and 4
0
b) positive between -2 and 0, y -4 -3 -2 -1 0 1 2 3 4 5
and between 2 and 3 -1
c) zero at points -2, 0, 2, and 3.
-2
-3
-4
x
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