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anonymous

  • one year ago

Help with the Chain Rule for a medal?

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  1. FireKat97
    • one year ago
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    do you have a question in particular?

  2. anonymous
    • one year ago
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    \[-3\sqrt[4]{2-9x}\]

  3. anonymous
    • one year ago
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    ok so rerewrite: \[-3(2-9x)^{\frac{1}{4}}\]

  4. anonymous
    • one year ago
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    so we have something of the form of f(u(x)) thus the derivative will be: \[ \frac{d}{dx}f(u(x))=f'(u(x))*u'(x)\]

  5. anonymous
    • one year ago
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    so we have: \[\frac{d}{dx}(-3(2-9x)^{\frac{1}{4}})=\frac{d}{dx}(-3(u(x))^{\frac{1}{4}})*\frac{d}{dx}(2-9x) \] where I defined: \[ u(x)=2-9x \]

  6. anonymous
    • one year ago
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    Ok so that gives us: \[\frac{d}{dx}(-3(2-9x)^{\frac{1}{4}})=\frac{d}{dx}(-3(u(x))^{\frac{1}{4}})*\frac{d}{dx}(2-9x) = -3*\frac{1}{4}(u(x))^{-\frac{3}{4}}*(-9)\] The back sub and simplify: \[ ... = \frac{27}{4}(2-9x)^{-\frac{3}{4}}\]

  7. anonymous
    • one year ago
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    Does my notation make sense? I was trying to make it more explicit, but looking back it may look more confusing than it actually is... did this procedure make sense?

  8. anonymous
    • one year ago
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    I'm reading over it again to see if I can better understand it.

  9. anonymous
    • one year ago
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    My goal was for you to identify the composition of functions here... one function was raising something to the 1 quarter power... the other function was (2-9x)

  10. anonymous
    • one year ago
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    The goal of the chain rule is to look at the outermost function of the composition and apply the simple power rule for derivatives... then I can look at (separately) the interior function and differentiate that (also using the deriv power rule)

  11. anonymous
    • one year ago
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    I understand how you rewrote the function and separated out the composite function...

  12. anonymous
    • one year ago
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    Maybe I shouldve wrote one of the steps instead as: \[\frac{d}{dx}( -3*(u(x))^{\frac{1}{4}})*\frac{d}{dx}(u(x)) \]

  13. anonymous
    • one year ago
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    The first three parts of it make sense, but I'm lost at the 4th

  14. anonymous
    • one year ago
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    i.e.: \[f(u(x))=(u(x))^{\frac{1}{4}}\]

  15. anonymous
    • one year ago
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    well -3 times that anyways

  16. anonymous
    • one year ago
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    Oh! Okay, I went back and read it, and now it's making sense. So on the fourth part, you used the power rule on the outside function. That's where I got lost.

  17. anonymous
    • one year ago
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    So basically, you separate out the composite functions, then find the derivatives of each, next multiply them together, and finally simplify. Correct?

  18. anonymous
    • one year ago
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    correct.... the only trick is correctly identifying the composite function

  19. anonymous
    • one year ago
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    Ok challenge question:

  20. anonymous
    • one year ago
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    Okay

  21. anonymous
    • one year ago
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    what is the answer: \[\frac{d}{dx}(a(b(c(d(e(f(x)))))))\]

  22. anonymous
    • one year ago
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    you can use ' to just denote prime

  23. anonymous
    • one year ago
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    1?

  24. anonymous
    • one year ago
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    and a(x) can stand in for everything inside of it....ah hah gotcha... try this:

  25. anonymous
    • one year ago
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    \[\frac{d}{dx}(a(b(x)))\]

  26. anonymous
    • one year ago
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    Just a second

  27. anonymous
    • one year ago
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    i.e. a(x)=a(b(x))

  28. anonymous
    • one year ago
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    just for shorthand's sake

  29. anonymous
    • one year ago
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    Yeah, I'm going to be honest. I haven't a clue.

  30. anonymous
    • one year ago
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    sure you do :D:D

  31. anonymous
    • one year ago
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    Its just as before: \[\frac{d}{dx}(a(b(x)))=a'(b(x))*b'(x) \]

  32. anonymous
    • one year ago
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    im just using a and b instead of f and u (pardon the last two letters total accident)

  33. anonymous
    • one year ago
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    so what about: \[\frac{d}{dx}(a(b(c(x)))) \]

  34. anonymous
    • one year ago
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    Well if the function is x, you would end up taking the derivative of both a'(b(x) and b'(x). Both of them are the function x which would mean both would have a derivative of 1. Multiplying them together 1 x 1 = 1. Or am I completely off track?

  35. anonymous
    • one year ago
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    Wut?

  36. anonymous
    • one year ago
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    Where did you get that the function was x?... a(x) means the function is a... and a(b(x)) means the function a composed with the function b

  37. anonymous
    • one year ago
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    Just as before where I said f(u(x)) were the functions f and u both functions of the independent variable x

  38. anonymous
    • one year ago
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    I am being a bit liberal with my function definitions, but this is good if you are confused believe it or not because if it were easy you wouldn't have learned anything... no pain no gain :D

  39. anonymous
    • one year ago
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    I am trying to get you to a very powerful point about the chain rule using a really abstract example.... follow the logic through

  40. anonymous
    • one year ago
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    Trust me it isnt as complicated as you think it is

  41. anonymous
    • one year ago
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    So \[\frac{ d }{ dx } a(b(x))\] would be \[a'(b(x)) * b'(x)\]

  42. anonymous
    • one year ago
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    BINGO!!! :D :D and what about: \[\frac{d}{dx}(a(b(c(x)))) \]

  43. anonymous
    • one year ago
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    \[a'(b(c(x))) * b'(c(x))* c'(x)\]

  44. anonymous
    • one year ago
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    BINGO :D :D :D..... Take a moment to take this in.... this is actually a really REALLY useful property about derivatives that you can learn to exploit in a lot of different ways to make your life a LOT easier.... but remember if you don't correctly identify the compositions then it all falls apart

  45. anonymous
    • one year ago
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    I will definitely remember this. I'll take any help I can get with calculus.

  46. anonymous
    • one year ago
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    Whenever you get a really complicated function that you don't think you can do in one go... just try and break it up into a couple of functions and exploit the chain rule.... as long as you define your composite functions correctly you should never have issues

  47. anonymous
    • one year ago
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    Thank you so much for all your help! :D

  48. anonymous
    • one year ago
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    No sweat cya later :D

  49. anonymous
    • one year ago
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    @PlasmaFuzer Sorry to summon you back, but I have one last question. Do you always just fill the "u" back in after you're done?

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