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start with finding expressions for u'(x) and v'(x)

u'(x)=f'(g(x))g'(x)
v'(x)=g'(f(x))f'(x)

yeah i dont understand this. am i just looking at 3 on the graph?

looks like 4

looks like g(3)=10 to me
each square is 2 vertically, and 1 horizontally as far as I can tell

g(3)=8 I mean, sorry

I'm not really sure about the increments on the graph though, it's a bit vague...

8, how so?

also remember that f'(4) represents the slope of f(x) at x=4

f(4) is 2

when x is 4, f is 2 right?

ohh

y2-y1/x2-x1?

rise over run right? 4/3? no?

at x=3 it looks to me like g has a slope of -1

how did you figure that out?

here's a sketch of g|dw:1329536578080:dw|

ok i see now i see

here's the part from x=2 to x=4

|dw:1329536693848:dw|in your graph you can see it goes one over and one down on that region

(I'm not gonna draw the little squares in...)

i see that thanks

what is g(3) ?

4
slope is -1

right, g(3)=4
and what is f'(g(3)) ?

f is also 4 right?
so that makes 4 x 4?

no, we want f'(g(3))=f'(4)
because g(3)=3, right ?
what is f'(4) ?

the slope of at 4 is 3

you are thinking about the idea correctly, but it looks like 2 to me

ok the gradients on this graph is crap

4/2

right, looks that way
so what is u'(3)=f'(g(3))g'(3) ?

(2)(4) (4)

there should be no 4 in the answer
what is f'(g(3)) ?

(we just figured it out above)

yes, and what is g'(3) ?

-1

so f'(g(3))g'(3) is what?

(2) (-1)

u'(3)=f'(g(3))g'(3)=-2
yep :D

thanks for the help

so now you get to try for v'(3) !
good luck :)

I got zero

yep!