Calculus Rules of Differentiation
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The Big Shortcut
Rules of Differentiation

 
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) = X2 F'(X) = 2X
F(X) = X3 F'(X) = 3X2
See the pattern????
If we let "n" stand for any exponent we might have,
we can say:
Original Function Derivative Function
F(X) = Xn F'(X) = nXn-1
 
This puppy is called the "Power Rule"
Here are some examples:
Power Rule Examples:
Original Function   Derivative Function
F(X) = X4   F'(X) = 4X3
F(X) = 4X2   F'(X) = 8X
F(X) = X0.5   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) = X1 then F'(X) = 1X0

 
and anything to the zero power equals ONE so.
 

F'(X) = 11 = 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 0again?
 

 
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) = X3 + 4X   F'(X) = 3X2 + 4
F(X) = X2 - 5   F'(X) = 2X - 0 = 2X
F(X) = X3 - X2 + X   F'(X) = 3X2 - 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) = X3 X2   F'(x) = X3 2X + 3X2 X2
    F'(x) = 2X4 + 3X4
    F'(x) = 5X4
 
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(X2 - 2)3

The derivative is:

F'(X) = 9(X2 - 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) = X2 - 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) = aXn  then  F'(x) = naXn-1dx

 
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(X3 - X)3   F'(X) = 9(X3 - X)2(3X2 - 1)
F(X) = 2(X2 - 1)-2   F'(X) = -4(X2 - 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.
 
F(X) G(X)

H(X)
 
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) X3 + 2

X2 - 3
 
F'(X) (X2 - 3)(3X2) - (X3 + 2)(2X)

(X2 - 3)2
F'(X) 3X4 - 9X2 - 2X4 - 4X

(X2 - 3)2
F'(X) X4 - 9X2 - 4X

(X2 - 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

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