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Mathematics (50)

MATA32H3 (3)

Raymond Grinnell (2)

Chapter 11

Department

MathematicsCourse Code

MATA32H3Professor

Raymond GrinnellChapter

11This

**preview**shows page 1. to view the full**5 pages of the document.**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+2x−3−x2−5x+2

Following the 3 steps outlined above I should have:

−3

(

7

)

x6+2

(

−3

)

x−4−1

(

2

)

x1−5

(

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

)

x−4−1

(

2

)

x1−5

(

1

)

x0+0=−21 x6−6x−4−2x−5

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 x6−6

x4−2x−5

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(x5−1)

…. Yellow represents the 1st term/function, green represents the 2nd . Following

the 3 steps above we have:

f'

(

x

)

=3

(

2

)

x1

(

x5−1

)

+3x2

(

5x4−0

)

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

addition

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f'

(

x

)

=3

(

2

)

x1

(

x5−1

)

+3x2

(

5x4−0

)

=6x

(

x5−1

)

+3x2

(

5x4

)

=6x

(

x5−1

)

+14 x6=¿

6x6−6x+14 x6=20 x6−6x

Ex2b: Derivative of

f

(

x

)

=(−8x+3)(2x−2−4x+1)

.. following the product rule again term by term, we have…

f'=

(

−8x0+0

)(

−4x−3−4x0+0

)

=−8

(

−4x−3−4

)

=−8

(

−4

x3−4

)

∨

(

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

2x2−4x

f'

(

x

)

=

(

−24 x2+10x+0

) (

2x2−4x

)

−(−8x3+5x2+1)(4x−4)

(

2x2−4x

)

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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