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mathcalculus

  • one year ago

can someone explain this simple derivative to me please: sort(6x)

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  1. mathcalculus
    • one year ago
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    isn't it 1/2(6x)^(-1/2)

  2. mathcalculus
    • one year ago
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    how does it go from there to the answer: sqrt(6)/2 sqrt(x)

  3. Psymon
    • one year ago
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    It's chain rule so you also need to multiply by the derivative of the inside. So it would be what you put, but then multiply by 6.

  4. Psymon
    • one year ago
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    \[\sqrt{6x} = \frac{ 1 }{ 2 }(6x)^{\frac{ -1 }{ 2 }}(6)\]

  5. mathcalculus
    • one year ago
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    ok

  6. mathcalculus
    • one year ago
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    im no sure about the chain rule its confusing

  7. mathcalculus
    • one year ago
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    i know its: [f(g(x))]'= f'(g(x))*g'(x)

  8. Psymon
    • one year ago
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    Well, I have a unique way of showing it, so maybe it will help maybe not. Were you one of the ones I sent the derivative files to?

  9. mathcalculus
    • one year ago
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    no

  10. mathcalculus
    • one year ago
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    oh wait, i think so a while ago.

  11. mathcalculus
    • one year ago
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    let me open

  12. Psymon
    • one year ago
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    Ah xD Yeah, one of them had the chain rule in it. I teach it in a unique way, so maybe it helps maybe it doesn't.

  13. mathcalculus
    • one year ago
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    does it always work?

  14. Psymon
    • one year ago
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    Yeah. As long as you recognize what layers you have then yeah. Chain rule is multiplying the derivatives of each layer you have, making sure to not disturb what was originally inside of each layer.

  15. mathcalculus
    • one year ago
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    thanks im going to read it right now

  16. Psymon
    • one year ago
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    Yeah, just let me know.

  17. mathcalculus
    • one year ago
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    ugh i gave it a shot. wrong. :(

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  18. Psymon
    • one year ago
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    How'd it become 1/4?

  19. mathcalculus
    • one year ago
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    i did 1/2 *-1/2

  20. Psymon
    • one year ago
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    Well, you bring the power down and then you made the power become -1/2. You're done with that layer, now you just go to the inner layer. \[\frac{ 1 }{ 2 }(----)^{\frac{ -1 }{ 2 }}\] Thats it, no more to that part.

  21. mathcalculus
    • one year ago
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    huh?

  22. Psymon
    • one year ago
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    Well, when you do the outer layer. All you do is bring the power down as a multiplication then lower the power by 1. After that you are done with that layer. There's no other 1/2 to multiply by.

  23. mathcalculus
    • one year ago
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    down in the denominator?

  24. mathcalculus
    • one year ago
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    can you show me the steps to the answer? just want to see how you did it so i can ask

  25. Psymon
    • one year ago
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    Alright, so we have two layers. The first layer is simply ( )^1/2 and the inner layer 6x. So the chain rule says that we take the derivative of each layer and multiply the results. So following the normal derivative rule of \[\frac{ d }{ dx }k ^{n} = nk ^{n-1}\], I'll do that with the first layer. \[\frac{ 1 }{ 2 }(---)^{-\frac{ 1 }{ 2 }}\]I just left the inner part blank for now, but that is the derivative of the first layer. Now I do the second layer, which is just 6x. So the derivativeof 6x is simply 6. So now that I have the derivative of both layers, I now multiply both of these derivatives \[\frac{ 1 }{ 2 }(---)^{-\frac{ 1 }{ 2 }}*(6)\] This of course becomes: \[\frac{ 3 }{ (---)^{\frac{ 1 }{ 2 }}}\] Now all that is left to do is plug back in what was originally inside of that layer, giving us: \[\frac{ 3 }{ \sqrt{6x} }\]

  26. mathcalculus
    • one year ago
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    oh the answer is sqrt (6)/2 sqrt(x)

  27. Psymon
    • one year ago
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    It's the same thing actually xD

  28. Psymon
    • one year ago
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    I'll show ya why: \[\frac{ 3 }{ \sqrt{6x} }=\frac{ 3 }{ (\sqrt{6})(\sqrt{x)} }\]Now multiply top and bottom by sqrt(6) \[\frac{ 3(\sqrt{6)} }{ (\sqrt{6})(\sqrt{x})(\sqrt{6}) }\]This then becomes finally: \[\frac{ \sqrt{6} }{ 2\sqrt{x} }\]

  29. mathcalculus
    • one year ago
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    alrigght thanks ! :)))

  30. Psymon
    • one year ago
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    Mhm, np ^_^ Hope that made sense xD

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