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Mathematics 10 Online
OpenStudy (yb1996):

Suppose F(x) = f(g(x)) and g(3) = 6, g '(3) = 4, f '(3) = 2, and f '(6) = 7. Find F '(3).

OpenStudy (amistre64):

chain rule ....

OpenStudy (yb1996):

yes, I understand that I have to do chain rule, I just don't understand how I'm supposed to get the answer.

OpenStudy (amistre64):

define the chain rule for me

OpenStudy (yb1996):

the chain rule is f(gx))' = f'(gx)) g'(x)

OpenStudy (amistre64):

correct so lets just sub in the values they give you f'(g(3)) * g'(3)

OpenStudy (yb1996):

ok, so that would be f'(6) * 4 = 7 * 4 = 28 is that right?

OpenStudy (amistre64):

yep

OpenStudy (yb1996):

Ok, thank you. I also have another question. With chain rule how would I take the derivative of function like so: W = uov ???

OpenStudy (yb1996):

so, how would I find the derivative of W?

OpenStudy (amistre64):

same way: just the names have changed uov = u(v) derive: u'(v) * v'

OpenStudy (amistre64):

im assuming u o v is the usual notation for function composition

OpenStudy (yb1996):

so "o" implies multiplication?

OpenStudy (amistre64):

not conventionally no. 'o' is shorthand for 'composed of' fog means f, composed of g ... or simply f(g)

OpenStudy (amistre64):

f*g means multiplication, or even fg

OpenStudy (yb1996):

oh wait, uov means " u of v", is that right?

OpenStudy (amistre64):

correct

OpenStudy (yb1996):

ahhh, that makes things so much more clear. Thank you for the help!

OpenStudy (amistre64):

youre welcome

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