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marytheshade
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I need to find the derivative using the chain and quotient rules of F(x)= ((3x2)^3)/((x3)^5). I just don't know when to stop doing chain rule and start the quotient
 2 years ago
 2 years ago
marytheshade Group Title
I need to find the derivative using the chain and quotient rules of F(x)= ((3x2)^3)/((x3)^5). I just don't know when to stop doing chain rule and start the quotient
 2 years ago
 2 years ago

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frx Group TitleBest ResponseYou've already chosen the best response.3
Start by thinking of the quotient rule, it's (f'(x)g(x)f(x)g'(x))/(g(x))^2
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.3
Then you just replace the derivitives using the chainrule
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.3
\[\frac{ f'(x)*(x3)^{5}  (3x2)^{3}g'(x) }{(x3)^{10} }\]
 2 years ago

marytheshade Group TitleBest ResponseYou've already chosen the best response.0
don't you have to do the chain rule first? is it possible to derive the F(x) and g(x) equations without the chain rule?
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.3
You have to do the chainrule when replacing the derivitives in the function i wrote above
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.3
I think it's easier to start of writing it as above
 2 years ago

marytheshade Group TitleBest ResponseYou've already chosen the best response.0
okay so I did the chain rule and got F'(g(x))= 9(3x2)^2 (top function deriv) and g'(x)= 5(x3)^4 (denominator deriv) so do I plug in those values or do I have to take the chain rule further?
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.3
You just put them in as you counted them, the derivitive of f(x) is f'(x) and it's what you're looking for according to the quotientrule.
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
You've done with derivative u' and v' Plug them into the formula ( u'v  uv' ) / v²
 2 years ago

marytheshade Group TitleBest ResponseYou've already chosen the best response.0
Thank you!!
 2 years ago
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