Product Rule for Derivatives: What Happens When Two Straight Lines Meet?

[Feodalherren]

I think the topic sums it up pretty well. I have no idea what I'm supposed to be doing here.

 

[Feodalherren]

Maybe I figured it out now.

Is a)

u'(1) = f(1) g'(1) + g(1) f'(1)

= 0

?

 

[SqueeSpleen]

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)

 

[Feodalherren]

That's not the question though, they want me to differentiate a combination of the two.

 

[SqueeSpleen]

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.

 

[voko]

Maybe I figured it out now.

Is a)

u'(1) = f(1) g'(1) + g(1) f'(1)

= 0

?

The beginning is good, but why zero in the end?

What is the geometric meaning of the derivative?

 

[CompuChip]

u'(1) = f(1) g'(1) + g(1) f'(1)

That is correct. What are the values of f(1), f'(1), g(1) and g'(1)?

 

[Feodalherren]

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

 

[HallsofIvy]

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.