- #1
Feodalherren
- 604
- 6
I think the topic sums it up pretty well. I have no idea what I'm supposed to be doing here.
Product Rule for Derivatives: What Happens When Two Straight Lines Meet?
- Thread starter Feodalherren
- Start date
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- Tags
- Derivative
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Homework Help Overview
The discussion revolves around the application of the product rule for derivatives, particularly in the context of piecewise linear functions. Participants are exploring how to differentiate a combination of two functions that meet at specific points.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning
Approaches and Questions Raised
- Participants discuss the differentiation of piecewise functions and the implications of the product rule. There are attempts to evaluate derivatives at specific points, with some questioning the geometric meaning of the derivative and the values of the functions involved.
Discussion Status
The discussion includes various attempts to clarify the differentiation process and the values of the functions at certain points. Some participants have provided calculations, while others are seeking to understand the reasoning behind the results. There is an ongoing exploration of the geometric interpretation of the derivatives.
Contextual Notes
Participants are working under the assumption that the functions are piecewise linear and are examining intervals where these functions are defined. There is an acknowledgment of the need to differentiate functions that may change behavior at specific points.
I think the topic sums it up pretty well. I have no idea what I'm supposed to be doing here.
- #2
Feodalherren
- 604
- 6
Maybe I figured it out now.
Is a)
u'(1) = f(1) g'(1) + g(1) f'(1)
= 0
?
Is a)
u'(1) = f(1) g'(1) + g(1) f'(1)
= 0
?
Last edited:
- #3
SqueeSpleen
- 138
- 5
They are piecewise functions.
So u(x) and v(x) will be piecewise functions aswell.
The intervals are (-∞,0), (0,2) and (2,∞)
(I assumed they don't change after the graph ends)
So u(x) and v(x) will be piecewise functions aswell.
The intervals are (-∞,0), (0,2) and (2,∞)
(I assumed they don't change after the graph ends)
- #4
Feodalherren
- 604
- 6
That's not the question though, they want me to differentiate a combination of the two.
- #5
SqueeSpleen
- 138
- 5
I would do it in the following steps:
1) Find the functions f(x) and g(x), they're piecewise and linear.
2) Find the functions u(x) and v(x), they're also piece wise.
3) Derivate the functions.
4) Evaluate the u'(x) in 1 and v(x) in 5.
1) Find the functions f(x) and g(x), they're piecewise and linear.
2) Find the functions u(x) and v(x), they're also piece wise.
3) Derivate the functions.
4) Evaluate the u'(x) in 1 and v(x) in 5.
- #6
voko
- 6,053
- 391
Feodalherren said:
The beginning is good, but why zero in the end?
What is the geometric meaning of the derivative?
- #7
CompuChip
Science Advisor
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Feodalherren said:
That is correct. What are the values of f(1), f'(1), g(1) and g'(1)?
- #8
Feodalherren
- 604
- 6
The meaning is instantaneous rate of change.
Lets see here
(2)(-1)+(1)(2) = 0
Still get it to zero.
f(1) = 2
g'(1) = -1
g(1) = 1
f'(1) = 2
Lets see here
(2)(-1)+(1)(2) = 0
Still get it to zero.
f(1) = 2
g'(1) = -1
g(1) = 1
f'(1) = 2
- #9
HallsofIvy
Science Advisor
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For x between 0 and 2, the graph of f is the straight line between (0, 0) and (2, 4): f(x)= 2x. For x between 0 and 2, the graph of y is the straight line between (0, 2) and (2, 0): g(x)= 2- x. Their product is fg(x)= 2x(2- x)= 4x- 2x^2. (fg)'= 4- 2x and at x= 1, that is, indeed, 0.
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