finding deriv from graph...

G

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Hi Guys,
Got another one for you. Prob is :Consider the graph below. If f(x) is the red sketch and g(x) is the blue sketch, let
u(x) = f(g(x)), v(x) = g(f(x)) and w(x) = g(g(x)). Then find:

a. u ‘(1) b. v ‘(1) c. w ‘(1)

I've stared at this for about 20 minutes now and I think i'm having a brain cramp. Can someone point me in the right direction here? I think where I'm having problems is translating the graph into the original eqn. I'm not sure, im just not getting anywhere. Thanks in advance!
Chris




 
Use algebra to find the line equations: On the interval from 0 to 2, f(x) = 6 - 3x and g(x) = 2x.

Then f'(x) = -3 and g'(x) = 2 (on those intervals). Use this information to find u'(1), either by plug-n-chug with the chain-rule formula, or else by composing the functions and then differentiating and evaluating. Same instructions for v'(1) and w'(1).

Eliz.
 
Eliz,
Thank you for your reply... Im sorry but I am still confused on how you go about plugging this thing in to use the chain rule. From what I can gather,
u(x) = f(g(x)), and we know from the chain rule that F'(x) = f '(g(x)) g '(x). So in this problem for a) does that mean that u(x) is the equiv to F'(x)? If it does, how do we plug these in? I'm sorry, I just am not seeing it. Thanks again.
Chris
 
Taking Eliz's
f(x) = 6-3x and
g(x) = 2x
then
u(x) =
f(g(x)) =
f(2x) =
6-3(2x)) =
6-6x
 
Gene & Eliz,
I think I am getting a handle on this now... Let's see if you agree with what I have now.

We know u'(x) = f'(g(x))(g'(x))

We want to find u'(1)

We first find g(1), we know g(x) = 2x = 2(1) =2

Now we know u'(1) = f'(2)(g'(1))

f'(2) is the slope of f at x=2.
However, if you look at the graph don't we see that f is not differentiable at x=2, since the graph comes to a point at x=2? Doesn't this means that u'(1) has no solution?




The next part

v'(x) = g'(f(x))(f'(x)) at x=1

First we find f(1) = 3

so v'(x) = g'(3)(f'(1))

g'(3) is = to the slope of g at x=3. This value is -1/3.

f'(1) is = to the slope of f at x=1. This value is -3.

Using v'(x) = g'(3)(f'(1))
We can calculuate v'(x) = -1/3 * -3 = 1?





Now for part c)
w(x) = g(g(x))
so
w'(x) = g'(g(x))(g'(x)) at x=1


First we find g(1) = 2

So w'(1) = g'(2)(g'(1))

Once again, g'(2) does not exist because the graph of g has a point at the value of x=2. Like the first question, this, too, does not exist...

Both graphs of f and g are continuous at x=2, but not differentiable.
 
I ended up with
u(x)=6-6x which makes
u'(x)=-6

Eliz showed f'(x) = -3
so f'(anything [0 to 2] = -3
and g'(x) = 2
so f'(g(x))*g'(x) = -3*2 = -6
for x=[0 to 2]

Pheew, they agree. That's a relief. I hope Eliz is right about the others being the same. My printer is down and the scrolling is cramping my brain too.
 
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