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Is this work right?...
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Looks fine to me!First a table is given (im gunna try to reconstruct it; the dashes represent spaces)
X ---U(x)--- V(x)--- U'(x)---- V'(x)
2----3--------4------(-1)-------2
3----2--------1--------3------(-1)
4----1--------3--------0-------(-2)
U(x) and V(x) are defined and differentiable for all real numbers x. the following data is known about U,V, and their derivatives.
Define f(x)=[U(x)]3+2[v(x)]2
Find f '(2)
_________________________________________________________________
My work below
I was able to use the chain rule to simply it. And so far I got
f '(x)=3[U(x)]2(U'(x)) + 4[V(x)][V'(x)]
This is where I'm not really to sure what to do.
I'm thinking you plug in the values from the table in here but I think I might be putting in the wrong values. Do I only use the values at x=2 because its asking for f '(2)?
f '(2)=3[3]2(-1) + 4[4](2)
f '(2)= -27+32
f '(2)=5
Is this work right?
Yes.First a table is given (im gunna try to reconstruct it; the dashes represent spaces)
X ---U(x)--- V(x)--- U'(x)---- V'(x)
2----3--------4------(-1)-------2
3----2--------1--------3------(-1)
4----1--------3--------0-------(-2)
U(x) and V(x) are defined and differentiable for all real numbers x. the following data is known about U,V, and their derivatives.
Define f(x)=[U(x)]3+2[v(x)]2
Find f '(2)
_________________________________________________________________
My work below
I was able to use the chain rule to simply it. And so far I got
f '(x)=3[U(x)]2(U'(x)) + 4[V(x)][V'(x)]
This is where I'm not really to sure what to do.
I'm thinking you plug in the values from the table in here but I think I might be putting in the wrong values. Do I only use the values at x=2 because its asking for f '(2)?
Yes. Good job!f '(2)=3[3]2(-1) + 4[4](2)
f '(2)= -27+32
f '(2)=5
Is this work right?
What do _you_ think? You know the answer; you just need to be more confident.Alright thanks guys!
So for example if it said f '(3) I would only use the x values at 3 which are [2,1,3,-1]?