anonymous
  • anonymous
Can someone help me find the derivative of the problem below? I'll give a medal! :)
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\frac{ 5 }{ (2x)^3 }\]
sleepyhead314
  • sleepyhead314
do you want to use quotient rule or product rule? personally I think product rule is easier xD
anonymous
  • anonymous
Whatever you think is easier, I'm okay with.

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sleepyhead314
  • sleepyhead314
mmmkk well do you know product rule? :) \[\frac{ d }{ dx } [f(x) * g(x)] ~~=~~ f(x)*g'(x) + g(x)*f'(x)\]right? :)
anonymous
  • anonymous
Yes, I am familiar with it
sleepyhead314
  • sleepyhead314
mmm I'm not awake we can rewrite the given equation as \[5*(2x)^{-3}\]right?
anonymous
  • anonymous
Yes
sleepyhead314
  • sleepyhead314
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 \[5*(2x)^{-3} \rightarrow 5(-3)*(2x)^{-4}*(\frac{ d }{ dx }2x)\]
sleepyhead314
  • sleepyhead314
tell me if you get confused xD
anonymous
  • anonymous
Sorry, my page keeps reloading. And that makes sense
anonymous
  • anonymous
You'd get -30(-2x)^4
anonymous
  • anonymous
-30(2x)^4 sorry the 2s not negative
sleepyhead314
  • sleepyhead314
mhmm then we simplify and get \[-15 * (2x)^{-4} * (2) \rightarrow -30 * (2x)^{-4} \rightarrow \frac{ -30 }{ (2x)^{4} } \]\(\huge \rightarrow \frac{-30}{16x^4} \rightarrow \frac{-15}{8x^4}\)
freckles
  • freckles
you can also just use power rule... if you write \[5(2x)^{-3} =5(2)^{-3}x^{-3} =\frac{5}{2^3}x^{-3}=\frac{5}{8}x^{-3} \\ \text{ then differentiate using power rule }\]
sleepyhead314
  • sleepyhead314
yay math thanks @freckles :)
anonymous
  • anonymous
Thank you, both of you. :)
sleepyhead314
  • sleepyhead314
glad we could helppp :)

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