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marytheshade

  • 3 years ago

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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  1. frx
    • 3 years ago
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    Start by thinking of the quotient rule, it's (f'(x)g(x)-f(x)g'(x))/(g(x))^2

  2. frx
    • 3 years ago
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    Then you just replace the derivitives using the chainrule

  3. frx
    • 3 years ago
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    \[\frac{ f'(x)*(x-3)^{5} - (3x-2)^{3}g'(x) }{(x-3)^{10} }\]

  4. marytheshade
    • 3 years ago
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    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?

  5. frx
    • 3 years ago
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    You have to do the chainrule when replacing the derivitives in the function i wrote above

  6. frx
    • 3 years ago
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    I think it's easier to start of writing it as above

  7. marytheshade
    • 3 years ago
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    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?

  8. frx
    • 3 years ago
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    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.

  9. Chlorophyll
    • 3 years ago
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    You've done with derivative u' and v' Plug them into the formula ( u'v - uv' ) / v²

  10. marytheshade
    • 3 years ago
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    Thank you!!

  11. frx
    • 3 years ago
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    np

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