



OK,
now you've seen the official math way that these problems are
formally done. 





The deal is, almost nobody uses
it.






IT TAKES TOO LONG!






And as the problems get
nastier, this would just get worse.






There HAS to be an easier way.






There is...






In the
examples we did on the last page, we had: 





Original
Function 

Derivative
Function 
F(X)
= X^{2} 

F'(X)
= 2X 
F(X)
= X^{3} 

F'(X)
= 3X^{2} 






See the
pattern???? 


If we let "n" stand for
any exponent we might have,



we can say:



Original
Function 

Derivative
Function 
F(X)
= X^{n} 

F'(X)
= nX^{n1} 






This puppy is called the
"Power Rule"



Here are some
examples: 


Power
Rule Examples: 
Original
Function 

Derivative
Function 
F(X)
= X^{4} 

F'(X)
= 4X^{3} 
F(X)
= 4X^{2} 

F'(X)
= 8X 
F(X)
= X^{0.5} 

F'(X)
= .5X^{.5} 
F(X)
= X^{2} 

F'(X)
= 2X^{3} 
F(X)
= X 

F'(X)
= 1 






That last one
is a bit tricky. The deal is, X is the same as X ^{1} so... 





If F(X) = X^{1}
then F'(X) = 1´X^{0}






and anything to
the zero power equals ONE so. 





F'(X) =
1´1 = 1 











OK, what about
something like: 


F(X) = 3 


One's like this
have a simple rule. 


The derivative
of any constant is zero. 


So:



Constant
Rule Examples 
Original
Function 

Derivative
Function 
F(X)
= 5 

F'(X)
= 0 
F(X)
= 10 

F'(X)
= 0 
F(X)
= p 

F'(X)
= 0 







The formal rule
is: 





If
F(X) = C then F'(X) = 0 


where
C is any constant 





For extra
credit, can you show that the rule for constants we just did 


is
really just the Power Rule? 


hint:
What was X ^{0}again? 











Next rule: 


If F(X) has
more than one term added together or subtracted from each other. 


Just find the
derivatives of each term by itself. 


These are
usually called the sum and difference rules in formal math type
talk. 


So: 


Sum
and Difference Rule Examples 
Original
Function 

Derivative
Function 
F(X)
= X^{3} + 4X 

F'(X)
= 3X^{2} + 4 
F(X)
= X^{2}  5 

F'(X)
= 2X  0 = 2X 
F(X)
= X^{3}  X^{2} + X 

F'(X)
= 3X^{2}  2X + 1 






The general
rule for this is: 





If F(X) =
G(X)
± H(X) then
F'(X) =
G'(X)
± H'(X)












If F(X) is the
product of two functions, 


either multiply it out and use the rules
we already have, 


or use this
rule ... 





If F(X) =
G(X)
´
H(X) then
F'(X) =
G(X)
´
H'(X) +
G'(X)
´
H(X) 


this is
called the product rule 





You generally
need this one when you have trig functions, 


or log functions or some
other weird thing. 


We haven't
talked at all about those yet, so we can't do one until later. 





But let's
use this on a simple function that we could have multiplied out 


and
used another rule just to see how
it goes. 








Product
Rule Example 
Original
Function 

Derivative
Function 
F(x)
= X^{3} ´
X^{2} 

F'(x)
= X^{3} ´
2X + 3X^{2} ´
X^{2} 


F'(x)
= 2X^{4} + 3X^{4} 


F'(x)
= 5X^{4} 






If we had
multiplied the X ^{3} and X ^{2} right off, we would
have gotten X ^{5}. 


We would then
have used the power rule and gotten the same answer. 





This does bring
up a point. 


Probably the
hardest thing about the last example 


was knowing how to add and
multiply exponent terms. 


That part is
Algebra, not Calculus. 





That's what I said at the beginning. 


Most of the
time, the Calculus is easier than the Algebra. 











If we have a
function like: 


F(X) = 3(X^{2}
 2)^{3} 


The derivative
is: 


F'(X) = 9(X^{2}
 2)^{2}(2X) 





What is going
on here is that we have two functions 


where one is INSIDE the other. 





The inside
function is: 


G(X) = X^{2}
 2 


The outside
function is: 


H(X) = 3(stuff)^{3} 


Put them
together, and you get: 


F(X) =
H(G(x)) 





The rule here,
is that the derivative of a nested double function like this 


is the
derivative of the outside times the
derivative of the inside. 





If F(X) =
H(G(X))
then F'(X) =
H'(G(X))
x G'(X) 


This
puppy is called the power chain rule 


Sometimes
it's just called the chain rule 





The really
important part to notice here 


is that when we did the derivative of
the outside part. 


The function
inside it DID NOT CHANGE! 





Scroll back and
look at the example. 


The inside part
was X ^{2}  2 


When we found
the derivative of the outside part, 


it was still sitting there on
the inside. 


That term was
then multiplied by 2X. 


2X is the
derivative of X ^{2}  2 





The first rule
we talked about was called the power rule. 


Really, it was
just a special case of this rule. 


In that case, the
"inside" function was just X.



We could have
multiplied by the derivative of the inside function there too. 


But it wouldn't have changed
much.






Why? 


You can figure
it out. 


Here's a Hint:
what's the derivative of X? 





The full official Power rule is: 





If F(x) = aX^{n}
then F'(x) = naX^{n1}dx 





The dx thing at
the end means the derivative of x 


(or whatever nasty inside function
we have) 





Power
Chain Rule Example 
Original
Function 

Derivative
Function 
F(X)
= 3(X^{3}  X)^{3} 

F'(X)
= 9(X^{3}  X)^{2}(3X^{2}  1) 
F(X)
= 2(X^{2}  1)^{2} 

F'(X)
= 4(X^{2}  1)^{3}(2X) 












The last rule
we need to talk about is a rule for taking the derivative 


of two
functions where one is divided by the
other. 











We actually
don't need a special rule for division. 


We can
just write the denominator part with an exponent of 1: 





F(X) =
G(X)(H(X))^{1} 





And then solve
this using the rules we have up to now. 


We would have
to use the product rule on the two multiplied terms 


But since one
of the terms is an outside/inside thing (the H(X)) 


We would need
the power chain rule when we take the derivative of that. 





But if you
really want to use a special division rule here it is: 





If
F(X) = 
G(X) 
Then
F'(X) = 
H(X)
´
G'(X) 
H'(X)
´
G(X) 


H(X) 
(H(X))^{2} 






Division is
really just a fraction. 


A fancy math
name for a fraction is a quotient (pronounced quo  shunt) 


So this one is
called the quotient rule. 





This rule is
more involved and harder to remember than the others. 


It's easy to
remember part of it: 





something
 something 

denominator
^{2} 






The hard part
is remembering exactly how the top part goes. 


You can
probably remember that it's just like the product rule 


but with a
minus sign. 





One term times
the derivative of the other 


Minus the other term times the
derivative of the first.



But because of
the minus sign, it's important which terms go first. 





The way I
remember it is by saying to myself: 


This is a
denominator type problem, so we start with the denominator 


NOT the
derivative of the denominator. 


Using a single
quote (') as a
code for a derivative, we have: 





denominator
x numerator'  denominator' x numerator 

denominator
^{2} 






Let's try
one: 


Quotient
Rule Example 
Original
Function 

Derivative
Function 
F(X)
= 
X^{3}
+ 2 

X^{2}
 3 


















F'(X)
= 
(X^{2}
 3)(3X^{2})  (X^{3} + 2)(2X) 

(X^{2}
 3)^{2} 


F'(X)
= 
3X^{4}
 9X^{2}  2X^{4}  4X 

(X^{2}
 3)^{2} 


F'(X)
= 
X^{4}
 9X^{2}  4X 

(X^{2}
 3)^{2} 







You noticed
that we didn't multiply out the denominator in the last example. 


The question of
whether or not to multiply out stuff in the answer 


comes up with all
kinds of problems. 





To me, there
are two times when you should multiply stuff out. 


1  When it
lets you do something worthwhile, like simplifying something 


2  When your
teacher says to 





Two reasons above often don't both happen on the same problem. 





That's all the
rules for now. 


Using these
rules, and a few tricks for dealing with logs and trig
functions 


we will get
through most of Calculus 1 





copyright 2005 Bruce Kirkpatrick 
