Calculus Review – Derivatives
Objectives: At the end of this lesson, you should be able to:
1. Find the derivative of functions involving sums, products, quotients and composition of polynomial,
exponential and logarithmic structures.
2. Evaluate the derivative (find the slope) at any point in the domain of the function.
Background
You need to review the formulas commonly used in finding derivatives. I know you don’t like the technical
definition. It took me about a year to get it down really solid. However, there are times in business where you
must use the definition, because the problem set is discrete. Here’s the formula again:
f ' (x ) = lim
h !0
f (x + h ) " f (x )
h
Just as a reminder, it says
1. Calculate the functional value at a point in domain x and at a point h units distant from it. This creates
the change in output or rise.
2. Create the ratio of the result of step 1 and h. Now you have rise over run, or the average rate of
change across the h units of domain from x to (x + h).
3. If it is possible, shrink h to as close to zero as possible. That creates the functions limit at point x.
4. If all the math works, we have the derivative, also known as the instantaneous rate of change, the
slope of the tangent line, or the rate of change at the point x, and maybe another three or four
common names.
Let’s get blunt. If this is a mystery to you, you need to get help now!
From this powerful calculation we developed a host of the following rules:
The Power Rule
d n
x )= n " x n !1 .
(
dx
Where u = f (x ), v = g (x ) are both continuous functions, we get these results:
1. Product Rule
d
(u ! v ) = u '! v + u ! v '
dx
2. Quotient Rule
d # u $ u '! v " u ! v '
% &=
dx ' v (
v2
3. Chain Rule
dy dy du
=
.
dx du dx
and the Generalized Power Rule
d n
u )= n " u n !1 " u '
(
dx
© Arizona State University, Department of Mathematics and Statistics
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Calculus Review – Derivatives
We created special cases to simplify our work.
They all come out of the Chain Rule, which may explain why I’ve never really memorized them!
1. Exponential Rules:
d x
d u
d x
e )= e x and
e ) = eu ! u ' ,
a )= a x ! ln a ,
(
(
(
dx
dx
dx
d u
a )= a u ! ln a ! u '
(
dx
2. Logarithmic Rules:
d
1
d
u'
ln x ) = and
ln u ) =
(
(
dx
x
dx
u
Finally, if y = f (x ) , the derivative can be written as y ' = f ' (x ) =
dy df (x )
=
.
dx
dx
Discussion
We developed the process of finding the derivatives of non-linear functions because the slope of a curve
changes at each point on the graph.
As we said, the formula for finding the first derivative is f ' (x ) = lim
h !0
was developed from the original formula m =
f (x + h ) " f (x )
. Remember how it
h
y2 ! y1
for finding the slope of a line joining two points
x2 ! x1
(x1 , y1 ) and (x2 , y2 ) . All other formulas we developed later were based on this.
The only problem with the formula is that it has too many names !The derivative is defined to be the slope of
the tangent line at any point on the graph so long as it is a unique value.* It is also considered as the rate of
change of the variable y with respect to the variable x .
*
Recall that the simple function f (x) = x is beautifully continuous. However, it does not have a derivative
at zero, because the slope of the tangent line has no unique value there.
Example: Suppose we have the function, f (x ) = 2 x 2 ! 5 x + 4 .
From algebra, you know this is a convex quadratic function that is that opens up. The slope is negative on the
left of the vertex and positive on the right of the vertex. However, the slope is different at every point on the
graph. If we use the definition of the derivative above, we can find the appropriate formula that defines the
slope at any point on the graph.
2 (x + h ) " 5 (x + h ) + 4 " (2 x 2 " 5 x + 4 )
f (x + h ) " f (x )
f ' (x ) = lim
= lim
h !0
h !0
h
h
2
We can now simplify using basic algebra.
2 x 2 + 2 " 2 xh + 2h 2 # 5 x # 5h + 4 # 2 x 2 + 5 x # 4
f ' (x ) = lim
h !0
h
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Calculus Review – Derivatives
h (4 x + 2h " 5 )
4 xh + 2h 2 " 5h
f ' (x ) = lim
= lim
= lim (4 x + 2h " 5 ) = 4 x " 5
h !0
h !0
h !0
h
h
Hence, f ' (x ) = 4 x ! 5 is the formula that defines the derivative of the function f (x ) = 2 x 2 ! 5 x + 4 .
Let’s apply the rule at the point (!2, 22 ) is on the graph.
1. Make sure the point is on the graph! Since f (!2 ) = 2 (!2 ) ! 5 (!2 ) + 4 = 8 + 10 + 4 = 22 , the point is
on the graph.
2
2. Now, evaluate the derivative. At this point the derivative is f ' (!2 ) = 4 (!2 ) ! 5 = !8 ! 5 = !13 . We
can say that the rate of change of y with respect to x at (!2, 22 ) is -13.
f (x + h ) " f (x )
becomes tedious, dangerously
h !0
h
fraught with calculation error, and time-consuming for anything beyond a quadratic. We cannot even use it
directly for most exponential and logarithmic functions! We developed all the other formulas to save time
and reduce using more complex algebra.
As much as I love to do it, this limit process f ' (x ) = lim
Examples
Let’s look at some (many) examples. The only trick is to use the appropriate formulas.
1.
f (x ) = x 250
The definition would be tedious to use. The power rule
derivative more quickly. It is
2.
d n
x )= n " x n !1 allows us to find the
(
dx
d 250
(x )= 250 x 249 .
dx
f (x ) = 4 x5 ! 2 x3 + 7 x ! 9
We can use the power rule on each of the 4 terms in this function.
Remember that f (x ) = 4 x 5 ! 2 x 3 + 7 x ! 9 can be seen as the sum of four power functions.
So, f ' (x ) = 4 " 5 x 4 ! 2 " 3 x 2 + 7 "1x 0 ! 0 " 9 x !1 = 20 x 4 ! 6 x 2 + 7 as we take the derivatives term-byterm.
3.
f (x ) = (3 x 2 + 4 x ! 8 )(2 x3 ! 5 )
This can be written as f (x ) = u ! v where u = 3 x 2 + 4 x ! 8 and v = 2 x 3 ! 5 .
The formula that fits the function is the product rule
d
(u ! v ) = u '! v + u ! v ' .
dx
We can now find the derivative of each of these two new and simpler functions in the same way
as before and then substituting them into the formula to get the derivative f ' (x ) .
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Calculus Review – Derivatives
Let u ' = 6 x + 4 and v ' = 6 x 2
Now using the product rule,
d
f ' (x ) = (u ! v ) = u '! v + u ! v ' = (6 x + 4 ) 2 x3 " 5 + 3 x 2 + 4 x " 8 6 x 2 . Simplify using basic
dx
algebra.
(
4.
) (
)( )
6 x2 ! 5x + 1
f (x ) =
3x 2 ! 5
Write it as f (x ) =
u
where u = 6 x 2 ! 5 x + 1 and v = 3 x 2 ! 5 .
v
The formula that now fits is the quotient rule
d # u $ u '! v " u ! v '
.
% &=
dx ' v (
v2
Find u ' = 12 x ! 5 and v ' = 6 x .
The quotient rule gives
2
2
d # u $ u '" v ! u " v ' (12 x ! 5 )(3 x ! 5 )! (6 x ! 5 x + 1)(6 x )
f ' (x ) = % & =
=
which can be
2
dx ' v (
v2
(3x 2 ! 5)
simplified using basic algebra.
3
5. Suppose y = 5u + 4 and u = 2 x ! 1 .
12
(
)
The derivative of the composition y = f (x ) = 5 2 x ! 1
Use the chain rule:
We get y ' =
3
12
+ 4 is difficult to find.
dy dy du
=
.
dx du dx
dy
du
= 60u11 with u ' =
= 6x2
du
dx
()
Applying them, f ' x =
(
)
11
dy dy du
=
. = 60u11 ! 6x 2 = 60 2x 3 " 1 ! 6x 2
dx du dx
()
Simplifying leads to f ' x = 360x
2
( 2x ! 1)
3
11
.
If you don’t simplify, be in for a stunning problem when you go for the second or third
derivative!
© Arizona State University, Department of Mathematics and Statistics
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Calculus Review – Derivatives
6.
y = f (x ) = (x 2 + 2 x ! 1)
8
(
) =u
Look at it as y = f (x ) = x + 2 x ! 1
Now,
2
8
8
where u = x + 2 x ! 1 .
2
d n
u )= n " u n !1 " u ' (one instance of the chain rule) fits perfectly with u ' = 2 x + 2
(
dx
(
)
That gets us y ' = f ' (x ) = 8 x + 2 x ! 1 " (2 x + 2 )
2
7
7. If y = 5 x ! 2e + 1 then y ' = 5 ! 2 x " 2e = 10 x " 2e
2
8. Let y = e
x
2 x2 + 7 x
x
x
.
2
This can be seen as y = e where u = 2 x + 7 x .
u
Hence, u ' = 2 ! 2 x + 7 = 4 x + 7 . So,
9. Look at y = 9 x . It fits
10. y = 7
4 x 2 !3 x
y ' = e2 x
2
+7 x
! (4 x + 7 )
by the chain rule.
( )
d u
a = au ! ( ln a ) u " and y ! = 9 x " ln 9 .
dx
d u
a )= a u ! ln a ! u '
(
4 x 2 !3 x
" ln 7 " (8 x ! 3) by the chain rule.
fits dx
So, y ' = 7
4
!1"
x
x
& $ 6 # 2 x + 2e = $ 12 x + 2e
x
'x(
2
x
11. y = 4 ln x ! 6 x + 2e leads to y ' = 4 # %
(
)
12. y = ln 6 x + 5 x ! 11 fits
and y ' =
2
12 x + 5
.
6 x 2 + 5 x ! 11
d
u'
2
ln u ) =
(
where u = 6 x + 5 x ! 11 . Hence u ' = 12 x + 5
dx
u
Functions can be constructed where one or more of these rules do fit simultaneously. You need to
make a conscious effort to recognize the one or more of these rules that fit. This is necessary to be able
to find the derivative of complex functions.
If you can use these rules appropriately, you will have a solid background when finding derivatives of
functions having two or more independent variables. You should practice finding derivatives of a few
functions before going on.
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