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anonymous

  • one year ago

Can someone help me find the derivative of the problem below? I'll give a medal! :)

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  1. anonymous
    • one year ago
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    \[\frac{ 5 }{ (2x)^3 }\]

  2. sleepyhead314
    • one year ago
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    do you want to use quotient rule or product rule? personally I think product rule is easier xD

  3. anonymous
    • one year ago
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    Whatever you think is easier, I'm okay with.

  4. sleepyhead314
    • one year ago
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    mmmkk well do you know product rule? :) \[\frac{ d }{ dx } [f(x) * g(x)] ~~=~~ f(x)*g'(x) + g(x)*f'(x)\]right? :)

  5. anonymous
    • one year ago
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    Yes, I am familiar with it

  6. sleepyhead314
    • one year ago
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    mmm I'm not awake we can rewrite the given equation as \[5*(2x)^{-3}\]right?

  7. anonymous
    • one year ago
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    Yes

  8. sleepyhead314
    • one year ago
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    turns out we don't need product rule xD (cuz the numeator is a constant...) ingore what me said before we use power rule and chain rule \[5*(2x)^{-3} \rightarrow 5(-3)*(2x)^{-4}*(\frac{ d }{ dx }2x)\]

  9. sleepyhead314
    • one year ago
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    tell me if you get confused xD

  10. anonymous
    • one year ago
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    Sorry, my page keeps reloading. And that makes sense

  11. anonymous
    • one year ago
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    You'd get -30(-2x)^4

  12. anonymous
    • one year ago
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    -30(2x)^4 sorry the 2s not negative

  13. sleepyhead314
    • one year ago
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    mhmm then we simplify and get \[-15 * (2x)^{-4} * (2) \rightarrow -30 * (2x)^{-4} \rightarrow \frac{ -30 }{ (2x)^{4} } \]\(\huge \rightarrow \frac{-30}{16x^4} \rightarrow \frac{-15}{8x^4}\)

  14. freckles
    • one year ago
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    you can also just use power rule... if you write \[5(2x)^{-3} =5(2)^{-3}x^{-3} =\frac{5}{2^3}x^{-3}=\frac{5}{8}x^{-3} \\ \text{ then differentiate using power rule }\]

  15. sleepyhead314
    • one year ago
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    yay math thanks @freckles :)

  16. anonymous
    • one year ago
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    Thank you, both of you. :)

  17. sleepyhead314
    • one year ago
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    glad we could helppp :)

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