Cal Cab Proof of Product Rule
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Transcript of Cal Cab Proof of Product Rule
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(The Movie)
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Now that we know the basics of differentiation wemight want to differentiate something more difficult,
like the product of two functions:
3 23 7 5 2xp x x e x x
f x g x p x
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:
'
p x f x g xfind
p x
We want to find a rule for findingderivatives of the product of twofunctions:
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0
' limh
p x h p xp xh
0
' limh
p x
So we want to evaluate the definition of aderivative:
f x g xGiven:
And therefore:
f x h g x h
And so the derivative is:
p x
p x h
f x h g x h
f x g x
h
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0
' limh
p x f x h g x h f x g x
h
f x h g x
We want to simplify the derivative expression toevaluate the limit.
0
First add and subtract an identical expression.This has the effect of adding zero and doesnt
change the expression.
f x h g x
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0
' limh
p x f x h
f x h f x h
g x h
hh
f x h
f x h
g x
f x h g x f x h g x
f x h
g x
g x g x
g x
f x h
g x
g x
f x
g x
f x
Time to simplify
First split the fraction, right between the twoexpressions we added and subtracted.
h h
Factor out the f(x+h) from the left expression,and g(x) from the right expression
Now use limit laws to express each limitseparately:
'p x
0
limh
f x h
0limh
g x h g x
h
0limh
g x
0limh
f x h f x
h
Split here!
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'p x
0
limh
f x h
0limh
g x h g x
h
0limh
g x
0limh
f x h f x
h
Time to evaluate the 4 limits!
f x
Plug in zero for h.
0
Its f(x)!
This is the definition of a derivative of
g(x)!
Its g(x)!
'g x
This doesnt depend on h, so the limit
doesnt affect itIts g(x)!
g x
This is the definition of a derivative of
f(x)!
'f x
Its f(x)!
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p x f x g x
' ' 'p x f x g x g x f x
3 23 7 5 2xp x x e x x
f x
g x
2' 9 5 xf x x e
' 1 4 g x x
'p x f x 'g x 'f x g x 33 7 5 xx e 1 4x 22x x 29 5 xx e
If a function is a product of two functions:
Then the derivative of the function is the ProductRule:
The example given earlier was:
The derivative is:
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p x f x g x
' ' 'p x f x g x g x f x
If a function is a product of two functions:
Then the derivative of the function is the Product
Rule:
The Product Rule
