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

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Start by thinking of the quotient rule, it's (f'(x)g(x)-f(x)g'(x))/(g(x))^2
Then you just replace the derivitives using the chainrule
\[\frac{ f'(x)*(x-3)^{5} - (3x-2)^{3}g'(x) }{(x-3)^{10} }\]

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Other answers:

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?
You have to do the chainrule when replacing the derivitives in the function i wrote above
I think it's easier to start of writing it as above
okay so I did the chain rule and got F'(g(x))= 9(3x-2)^2 (top function deriv) and g'(x)= 5(x-3)^4 (denominator deriv) so do I plug in those values or do I have to take the chain rule further?
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.
You've done with derivative u' and v' Plug them into the formula ( u'v - uv' ) / v²
Thank you!!

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