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Chapter 11

MATA32H3 Chapter Notes - Chapter 11: Rhyne, Chain Rule, 32X


Department
Mathematics
Course Code
MATA32H3
Professor
Raymond Grinnell
Chapter
11

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Derivatives (the easy way of doing limits!)
Topic One: Basic concept of derivative/rate of change/marginal cost (all the same words and same process!): This
is the simplest case, where you just have one or a few terms in a row (by terms I mean added/subtracted together not
multiplied and divided as those are different cases). You simply 1) leave the number in front of the variable alone (i.e. the
coefficient) 2) but bring down the exponent beside it and multiply, 3) then subtract 1 from the exponent for each term.
Examples: Determine the derivative for each of the following.
EX1:
f
(
x
)
=3x7+2x3x25x+2
Following the 3 steps outlined above I should have:
3
(
7
)
x6+2
(
3
)
x41
(
2
)
x15
(
1
)
x0+0
You’ll notice that in the 2nd term, the negative exponent doesn’t scare me, I continue as outlined. In the 3rd term the
exponent becomes 1 which I don’t really need to write. In the 4th term the
x0
always equals 1 so get used to not writing
x to power zero but quickly replacing it with 1. And the final term, any constant (i.e. number by itself) instantly becomes
zero! So, our simplified answer should look like.
f'
(
x
)
=3
(
7
)
x6+2
(
3
)
x41
(
2
)
x15
(
1
)
x0+0=−21 x66x42x5
Note: Continue simplifying as best you can. The only further I can go in my solution above is to realize that negative
exponents mean they need to move floors (either up or down depending where they were originally) so that the exponent
becomes positive. We usually only like positive exponents in math.
So, my final answer would look like this.
f'
(
x
)
=21 x66
x42x5
Topic two: Product Rule: Product in math means multiply. That means we use this rule if and only if we have two terms
being multiplied together. This too has a method to follow! 1) You take the derivative of the first term/function and take the
derivative of that and keep the 2nd term as is (i.e. copy & paste) multiplied beside it. 2) Then you add the first term left
alone (copy & paste) but write the derivative of the 2nd term multiplied beside it. 3) Simplify and you’re done.
The general and gross math definition looks like:
derivative=f'
(
x
)
g
(
x
)
+f
(
x
)
g '(x)
Ex2a: Derivative of
f
(
x
)
=3x2(x51)
…. Yellow represents the 1st term/function, green represents the 2nd . Following
the 3 steps above we have:
f'
(
x
)
=3
(
2
)
x1
(
x51
)
+3x2
(
5x40
)
You’ll see that I took the derivative of the 1st term just as I did in example 1 but copy &pasted the 2nd term. In the 2nd half of
product rule, I left the first term alone/copy&pasted this time and took the derivative of the 2nd term as per example 1
above. Now I simplify as much as I can and we have:
Terms left as is originally
Derivative of 2nd term
Derivative of 1st term
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f'
(
x
)
=3
(
2
)
x1
(
x51
)
+3x2
(
5x40
)
=6x
(
x51
)
+3x2
(
5x4
)
=6x
(
x51
)
+14 x6=¿
6x66x+14 x6=20 x66x
Ex2b: Derivative of
f
(
x
)
=(8x+3)(2x24x+1)
.. following the product rule again term by term, we have…
f'=
(
8x0+0
)(
4x34x0+0
)
=8
(
4x34
)
=8
(
4
x34
)
(
32
x3+32
)
Topic Three: Quotient Rule: Used for terms being divided by each other. Similar the product but with subtraction and a
denominator.
The general math definition looks like:
derivative=f'
(
x
)
g
(
x
)
f
(
x
)
g ' (x)
g
(
x
)
2
You’ll notice the numerator of this thing is just product rule with a minus sign. The only thing we have to add is the
denominator which is just the 2nd term all squared! Simplify and done.
Ex3a: Determine derivative of
f
(
x
)
=3x4
2x
Following the idea that I just do product rule up top with minus and
square the 2nd (green) term, we have…
f'
(
x
)
=
[
3
(
4
)
x3
]
[
2x
]
[
3x4
]
[2x0]
(
2x
)
2
Now simplify carefully (I did the pink parts first):
f'
(
x
)
=
[
12 x3
]
[
2x
]
[
3x4
]
[2]
4x2
Finally multiply pairs of terms in numerator and collect any like terms:
f'
(
x
)
=24 +6x4
4x2=18 x4
4x2=9x2
2
Ex 3b: Derivative of
f
(
x
)
=8x3+5x2+1
2x24x
f'
(
x
)
=
(
24 x2+10x+0
) (
2x24x
)
(8x3+5x2+1)(4x4)
(
2x24x
)
2
Note: You will usually get at least half marks (if not most marks) just by setting it properly! Full marks for simplifying…Go
as far as you feel you comfortably can (time permitting of course).
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