# Here - BCIT Commons ```MATH 1441
Technical Mathematics for Biological Sciences
The Derivative
The derivative is the fundamental &quot;thing&quot; of differential calculus. It is nothing more than a formula for the
slope of a graph. Coming up with ways of determining this formula involves the determination of a
relatively straightforward limit (thought the algebra can be quite tedious).
Obtaining a formula for the slope of a graph may not sound very exciting or very useful. However, slopes of
graphs are really measures of rates of change. And most of the principles governing the way the world
works seem to involve rates of change. Thus, mathematical machinery for working with rates of change will
be fundamental tools in understanding much of physics, chemistry, and biology.
The Slope of A Graph
We have a clear and easy to use definition of slope for a graph that consists of a straight line:
slope 
rise
run
(1)
This ratio of rise/run is the same between any two points on a straight line, and so, we can easily calculate
the slope of the entire line by just using the coordinates of any two distinct points on that line in formula (1).
The slope measures how steep the line is: a large valued
positive slope indicates a line rising steeply from left to
y
right. A small valued positive slope indicates a line rising
only slowly from left to right. Negative slopes indicate lines
C
dropping as you move along them from left to right.
This notion of steepness is not difficult to visualize for
curved graphs. The sketch to the right shows such a graph
with four locations labeled A, B, C, and D. Near A, the
graph is rising slowly , near B it is rising more rapidly, near
C, it is not rising or falling much, and near D is appears to
be fallilng quite rapidly. It is natural to think of the slope of
this graph as being positive but small near A, positive and
larger near B, nearly zero near C, and negative and quite
large numerically near D. A big difference from the
situation with a straight line graph is that when the graph
is a curve, the slope is not a single number that applies to
the whole graph, but is a quantity that varies from region
to region, even from point to point, along the graph, being
positive in some regions, negative in others, with values
ranging from small numbers or even zero at some points,
to quite large numbers in other regions.
D
B
A
x
y
The idea behind the formula for the slope of a curved
graph starts by identifying the slope of the curve at a point
as equal to the slope of the so-called tangent line to the
curve at that point. The tangent line is a line that just
touches the curve at that precise point. Three such lines
x
are sketched on the graph to the right. You can see that
the tangent lines seem to have the same slope as the
curve at the points where they touch. The advantage of this is that we know how to calculate the slope of a
line, and so to get a formula for the slope of the curve from point to point, we just need to find a formula for
the slope of the sequence of tangent lines.
David W. Sabo (1999)
The Derivative
Page 1 of 8
The next difficulty we encounter is that we need two points on a line to be able to calculate its slope, but we
only know the coordinates of one point on the tangent line -- the coordinates of that single point where it
touches the curve. Solving this problem involves one of the relatively few new mathematical ideas in
calculus.
The -Method
We will illustrate the way a formula for the slope of a tangent line is obtained using a specific example, and
then we will state a general formula. This general formula embodies what is usually called the -method for
finding derivatives. It can be quite cumbersome, particularly for complicated formulas, so it is not used for
routine work in calculus. Nevertheless, it is
the starting point of all formulas for
derivatives.
y
y = x2
As an example, consider the graph of
y = x2, shown with a lot of other stuff in the
figure to the right. The focus is on the lower
point with x-coordinate denoted 'x'. The
dashed line through that point represents
the tangent line at that point, and so the
goal is to determine the slope of the dashed
line. The difficulty, as mentioned before, is
that we know the coordinates of only one
point on this line, which is insufficient
information to be able to compute its slope.
(x + x, y + y)
y
The -method uses the following stratagem.
Pick a second point on the curve some
distance away from the point of interest 
this is the point shown in the figure with
coordinates (x + x, y + y). Here the
symbols x and y are intended to indicate small quantities. We can now easily write down the slope of the
line between these two points (at least formally), since with the notation adopted, the &quot;rise&quot; of this line is just
y, and the &quot;run&quot; is just x:
x
x
x
x + x
slope 
rise y

run x
In fact, since the second point is the point on the curve with horizontal coordinate x + x, we can even
express y in terms of x:
y   x  x   x 2
2
So, we can express the slope of the line (shown in the figure as a solid line  sometimes called a secant
line for the curve) between the two points in terms of the quantity x, the horizontal distance between the
point of interest at (x, y), and the arbitrarily chosen second point at (x + x, y + y):
2
rise y  x  x   x
slope 


run x
x
2
The thing to notice is that if we picture shifting the upper point towards the lower point (that is, decrease the
value of x towards zero), the solid secant line becomes more and more like the dashed tangent line we
wish to determine, and in particular, the formula for the slope of the secant line given just above must give a
better and better estimate of the desired slope of the tangent line the closer x gets to zero. If we could get
a useful result out of this slope formula by substituting x = 0 into it, our problem would be solved, since
when x = 0, the two lines would have to be identical. Unfortunately, substituting x = 0 into this formula
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The Derivative
David W. Sabo (1999)
gives us the problematic result 0/0. This is not surprising in a way, because setting x = 0 is really the same
thing as trying to calculate the slope of a line by using the coordinates of the same point twice.
All is not lost however. Although we cannot set x exactly to zero, we can imagine its value decreasing to
as close to zero as we wish, and so, it seems reasonable to propose the following:
y
x 0 x
slope of the tangent line  lim
 lim
 x  x 
2
 x2
x
x 0
x 2  2x  x   x   x 2
2
 lim
x
x 0
 lim
x  2x  x 
x 0
x
 lim  2 x  x   2 x
x 0
The use of
lim
x  0
allows us to think of x becoming as close to zero as we wish, but not exactly zero. This
avoids the problem with the 0/0 result, but still permits approximating the slope of the tangent line by the
slope of a line which is for all practical purposes indistinguishable from the tangent line. The series of
algebraic manipulations just above is prompted by our experience with the examples of limits in the
preceding document that involved the 0/0 form: this form must arise because there is a common factor of x
in the numerator and denominator of the expression. We simplify the numerator, and factor it so that the
common factor can be cancelled, leaving us with evaluating the limit of 2x + x as x  0. But obviously,
the closer to zero x is, the more and more this expression looks like just 2x.
The conclusion is that the slope of the graph of y = x2 is given by 2x. This should seem fairly plausible. For
instance, at the vertex of the parabola at x = 0, the tangent line is horizontal, and this expression gives
2  0 = 0. If we consider a sequence of increasing positive values of x, we know that the parabola is
increasing more steeply to the right, and the expression 2x will give us increasing positive values. On the
other hand, the left-half of the parabola has a negative slope, and for negative values of x, 2x gives a
negative value.
So, what we have found is that the expression that gives the slope of the graph of y = x 2 is just 2x, and so
we would call this expression, 2x, the derivative of the function y = x2 with respect to x. Several different
notations are used to denote the derivative:
d
dy
[y ] 
 y   2x
dx
dx
We recommend the first form shown here, because it emphasizes that the determination of the derivative is
an operation that is done on the function y. You can think of the symbol d/dx as similar to the symbols  or
log, etc., in that they stand for a specific kind of operation that is done to some expression to which they are
applied. You should not regard the symbol d/dx or even dy/dx as a fraction, nor should you consider dx to
indicate &quot;d times x&quot;. Using primes to denote derivatives, as in y', is very popular because it minimizes the
amount of writing, but it can be confusing, since no mention is made in this notation of what is the intended
independent variable. In the present example, the only other variable is x, and so it is reasonable to
interpret the symbol y' as meaning the derivative of y with respect to x. However, when a calculation
involves several variables, the exact meaning of y' may be ambiguous.
David W. Sabo (1999)
The Derivative
Page 3 of 8
You need to study the example above very carefully to make sure you understand the mathematical and
geometric strategies it involves. The limiting process used to get the slope of the tangent line was initially
quite controversial, but mathematicians now understand the circumstances under which it is justified, and so
use this approach without reservation for defining derivatives as required in typical scientific and technical
applications.
We can now state the gist of the -method in some generality. For the graph of the function y = f(x), the
derivative, or the formula for the slope of the graph when y is plotted against x, is given by
f  x  x   f  x 
d
y   lim

x 0
dx
x
(2)
To implement this formula, you simply use the function definition f(x) to get the initial expression for the
quotient whose limit is required, manipulate the numerator until a x can be factored out in order to cancel
the x in the denominator, and then usually, the value of the desired limit will be obvious by inspection.
The -method is the fundamental formula of differential calculus. However, it is not used routinely to find
derivatives. Instead, what we will do in the next few documents in this series is use the -method formally to
derive recipes that relate the derivative of an expression to derivatives of its simpler parts. By applying
these recipes, the derivatives of even very complicated expressions can be obtained relatively easily without
going through all the work of the -method from scratch.
However, even though you will not use the -method routinely for determining derivatives, you do need to
understand the principles of the method, and how to use it for very simple expressions. To help you with
this, we conclude this document with a few more quick examples of the method.
Example 1: Use the -method to determine the derivative of y = x3 with respect to x.
Solution
From the basic definition
x  x   x 3

d
d
3
 y   dx  x   lim
x 0
dx
x
3
x 3  3 x 2  x  3 x  x    x   x 3
2
 lim
x 0
3
x
2
x 3 x 2  3 x  x    x  


 lim
x 0
x
2
 lim 3x 2  3x  x    x  

x 0 
y
y = x3
x
 3x 2
The details here are a lot like those for the earlier example involving y = x2,
so they should be fairly easy to follow. The initial goal is to simplify the
numerator to be able to remove a factor of x from it. In the second last expression, both terms that contain
x as a factor become negligible in the limit that x  0, leaving the single term result, 3x2.
You can see from the small sketch of the graph of y = x3 just above to the right that this formula corresponds
to the major features of the shape of the graph. The formula says that the slope of the graph is always a
positive number, which is true, since that graph is always varying upwards to the right. It also says, that the
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The Derivative
David W. Sabo (1999)
slope of the graph will be zero at x = 0, and that also is clear in the sketch. We also expect that the graph
will get steeper the farther x is from zero, and that also shows up in the graph.

Example 2: Use the -method to determine the derivative of
y x
with respect to x.
Solution
We have immediately that
x  x  x
d
d 
y
x   lim


x 0
dx
dx
x
and we seem to be stymied, because there is no obvious way to factor a x out of the numerator to cancel
the x in the denominator. The difficulty is that the only reference to x is buried inside one of the square
roots in this difference of two square roots. However, you may recall a technique from basic algebra that
can be used to turn a difference of two square roots into an expression with no square roots (used when it is
necessary to &quot;rationalize&quot; the denominator of a quotient). In this case, we can use the same approach, but
we will &quot;rationalize&quot; the numerator instead. Thus
x  x  x
d 
x   lim


x

0
dx
x
x  x  x
x  x  x

x
x  x  x
 lim
x 0
 lim
x 0
 lim
x 0
 lim
x 0
x

x

x  x  x

x  x  x

x
x  x  x
1
x  x  x


y

1
2 x
y = x
Thus,
d 
1
x  

dx
2 x
x
This formula should seem plausible in view of the shape of the graph of
y x
, shown above to the
right. The graph appears to go vertical right at x = 0, consistent with the infinite value of the slope at x = 0
given by the derivative. Then, as x increases from zero, the square root of x in the denominator of the
derivative will make the value of the derivative decrease gradually, and we see the original graph tending to
rise less steeply the farther to the right you go. The value of derivative here is always positive, and the slope
of the original graph is also always positive. Note that the fact that
David W. Sabo (1999)
The Derivative
y x
exists only for non-negative
Page 5 of 8
values of x corresponds to the derivative existing only for non-negative values of x as well (though in the
case of the derivative, even the value x = 0 is excluded).

Example 3: Use the -method to determine the derivative of
y
1
x
with respect to x.
Solution
We have immediately that
1
1

d
d 1
x  x x
 y      lim
dx
dx  x  x 0
x
1
x 1 x  x
  
x  x x x x  x
 lim
x 0
x
 lim
x 0
 lim
x 0
x   x  x 
x  x  x 
y
x
y = 1/x
 x
x  x  x  x 
x
1
1
 2
x 0 x  x  x 
x
 lim
Here, the -method leads immediately to a complex fraction. The first few steps above are required to
reduce the original complex fraction to a simple fraction in simplest form. By the time that is accomplished,
the sought-after factor of x in the numerator is obvious.
As expected, this derivative formula gives an undefined result when x = 0. Notice that the graph of the
original function is decreasing wherever it exists, and indeed, the derivative has only negative values. For
values of x very close to zero, the slope of the graph is very steeply downwards, and the value of the
derivative will be a numerically large negative number. For values of x well away from zero, the graph of the
original function is nearly horizontal, and the derivative gives a value very close to zero.

Example 4: Use the -method to determine the derivative of y = 1 with respect to x.
Solution
This is quite a simple function. Since the formula for y does not contain x at all, the value of y is the not
affected by the value of x. Thus, the -method gives
d
1 1
0
1  lim
 lim
 lim 0  0

x 0 x
x 0 x
x 0
dx
We know that the graph of y = 1 is just a horizontal line at y = 1, so it isn't surprising that the formula for the
slope of that graph is just the value 0.

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The Derivative
David W. Sabo (1999)
A Suggestive Pattern: #1
These last five examples presented so far in this document have all involved functions which are simple
powers of x. You can see a pattern in the results when they are written in a common power-of-x format:
d
 x3   3x 2
dx  
d
 x 2   2 x1
dx
d  12  1 12
x
 x
 2
dx 
d
 x0   0x0  0
dx  
d
 x 1    1 x 2
dx
You can see that in each case, the derivative is also a power of x, but with the exponent decreased by one
from the original function. Thus, the derivative of x3 contains an x2, the derivative of x2 contains an x1 or x,
the derivative of x1/2 contains x-1/2. Furthermore, the power of x in the derivative is multiplied by a numerical
factor which appears to be equal to the exponent in the original function. That is, the derivative of x 3
contains a factor 3, the derivative of x2 contains a factor 2, the derivative of x-1 contains a factor of -1, and so
on.
In fact, this apparent pattern is not a coincidence. With a bit of work, it is possible to demonstrate that in
general,
d
 x n   n  x n  1
dx
(3)
We will resist the temptation to attempt to prove this formula in part or completely, but it is valid for any realvalued exponent n. With this formula, we now never have to use the -method to find the derivative of any
power of x with respect to x.
Example 5:
Find the derivative of the function
yx
5
3
with respect to x.
Solution
Believe it or not, it is not that hard to determine this derivative using the full-blown -method as done in the
previous examples. However, even that amount of work is not necessary now that formula (3) is available.
The requested derivative here has the form of formula (3) with n = 5/3. Thus,
d  53  5 53  1 5 23
x
 x
 x
 3
dx 
3

A second important pattern is suggested by the following example.
David W. Sabo (1999)
The Derivative
Page 7 of 8
Example 6: Use the -method to determine the derivative of y = 3x2 + 7x -2 with respect to x.
Solution
Although this function is a bit more complicated than the ones in the previous examples, in that it has more
than one term, we can easily apply the standard procedure. The numerator of the quotient in the -method
formula is formed by subtracting the expression of the function above from the expression obtained when x
is replaced by x + x:
3  x  x 2  7  x  x   2  3 x 2  7 x  2
d

 
2
3 x  7 x  2  lim 

x

0
dx
x
3 x 2  6 x  x  3  x 2  7 x  7x  2   3 x 2  7 x  2 

 
 lim 
x 0
x
6 x  x   x   7x
2
 lim
x
x 0
 lim
y
x  6 x  x  7 
x 0
x
 lim 6x  x  7  6x  7
x 0
This formula seems to reflect some of the major features of the
graph of the function, sketched above to the right. In particular,
notice that the derivative will be zero when
6x  7  0

x
7
x     1.17
6
and we observe that the vertex of the parabola, the point where the tangent line would be horizontal is just to
the left of x = -1. To the left of this point, the slope of the parabola is negative, and the value of 6x + 7 is
negative for x &lt; -7/6. To the right of this point, the slope of the parabola is positive, and the value of 6x + 7
is positive for x &gt; -7/6. Thus, this simple formula forming the derivative seems to reflect the shape of the
graph of the original function.
Notice a couple of other things about this result. The '6x' term in the derivative appears to originate with the
3x2 term in the original function. We know that the derivative of x2 is 2x, and so it looks as if the derivative
of 3x2 is just three times the derivative of x2 or 6x. The derivative of x is just 1, and it appears from this then
that the derivative of 7x is just seven times that or 7. In fact, we know that the function y = 7x is a straight
line with slope 7, so it makes sense that the derivative of 7x is 7. Also, it appears that to get the derivative of
a sum and/or difference of two or more terms, you just form the corresponding sum and/or difference of the
derivatives of the individual terms. All of these conjectures are true, and we'll state them formally and
illustrate them more fully in the next document.

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The Derivative
David W. Sabo (1999)
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