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Mathematics 4 Online
OpenStudy (anonymous):

Use the given information to find f'(2). MEDAL WILL BE REWARDED! g(2)=3 g'(2)=-2 h(2)=-1 h'(2)=4 f(x)=g(x)h(x)

OpenStudy (anonymous):

@hartnn

hartnn (hartnn):

use product rule

OpenStudy (anonymous):

How do I incorporate these into this equation? Am I changing the equation first?

hartnn (hartnn):

f'(x) = ... ? then you just put x =2

OpenStudy (anonymous):

okay so hold on give me one sec...

OpenStudy (anonymous):

f(2)=3(2)-1(2)??

hartnn (hartnn):

why 2 ? \(\large f'= g'h +h'g \) right ?

OpenStudy (anonymous):

it was x so I thought I was subsitituing f(2)?

hartnn (hartnn):

\(\large f'(x)= g'(x)h(x) +h'(x)g(x)\) \(\large f'(2)= g'(2)h(2) +h'(2)g(2)\) makes sense ?

OpenStudy (anonymous):

oh okay, so I plug in h and g now?

hartnn (hartnn):

yes, all values are given...

OpenStudy (anonymous):

okay let me calculate

OpenStudy (anonymous):

okay so before I go any further I have f'(2)=(-2)(2)(-1)(2)+(4)(2)(3)(2)

hartnn (hartnn):

what ? \(\large f'(2)= g'(2)h(2) +h'(2)g(2) \\ \large f'(2)=-2 (-1)+4(3)\)

OpenStudy (anonymous):

oh okay....so my final answer would be f'(2)=2+12 so f'(2)=14?

hartnn (hartnn):

yes.

OpenStudy (anonymous):

Awesome! thank you, this one really was confusing!

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