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amonoconnor

  • 2 years ago

When using the Chain Rule, and taking the second piece of derivative (dy/dx of the term for "theta"), would this be correct... (?) In my particular situation: The "theta" = (1/y) So... Because it's not the first step in the chain rule, and I'm "re-deriving" as my math teacher says, the constant (1) is affectable. So... Is the following derivation correct? 1/y = (0)/[y^(1-1)(dy/dx)] = 0/y^(0)(dy/dx) = 0/(1)(dy/dx) = 0/(dy/dx) = (0)

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  1. amonoconnor
    • 2 years ago
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    @agent0smith Yes... I know. I'm a thorn in your side ;)

  2. zepdrix
    • 2 years ago
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    "theta" = (1/y) ?? I don't understand :(

  3. amonoconnor
    • 2 years ago
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    Yep, in he function I'm taking the derivative of, the quantity "1/y" is in the "inside" spot, in Chain Rule terminology, or the spot of theta in regard to the trig function also in the function. I have taken the derivative of the whole "outside", finding the derived/ new trig function (theta term untouched), but now I must take the derivative of the theta term itself, and multiply that by the first part; to complete the derivation process, and find the final answer. I'm just wondering if this is in fact the right derivative. :)

  4. zepdrix
    • 2 years ago
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    Ok so for the derivative of the inner function (1/y), You're thinking you get zero for some reason? Did you apply the quotient rule or power rule or something?

  5. amonoconnor
    • 2 years ago
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    My problem: "ySIN(1/y) = 1 - xy"; Find dy/dx. ( Thought this would be helpful to see)

  6. amonoconnor
    • 2 years ago
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    Umm, no.... I should have, shouldn't I have though? Used the quotient rule?

  7. zepdrix
    • 2 years ago
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    You can rewrite it like this:\[\Large \frac{1}{y}\quad=\quad y^{-1}\]And apply the power rule as you normally would. I'm not quite sure where your 0 is coming up in the denominator. If you use the quotient rule you should get something like:\[\Large \frac{0y-1(1)}{y^2}\]

  8. zepdrix
    • 2 years ago
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    the zero in the numerator, i meant to say :)

  9. zepdrix
    • 2 years ago
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    Oh and a \(\Large y'\) should show up when you chain again.

  10. zepdrix
    • 2 years ago
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    But it looks like you understood that part :o i see your dy/dx in there somewhere (although it shouldn't be in the denominator XD heh )

  11. amonoconnor
    • 2 years ago
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    The "Power Rule"? I'm not familiar with that... This is what we learned in Calc, at least from my teacher: (?) y' = (y^(1-1))*(dy/dx) = (y^(0))*(dy/dx) = (1)*(dy/dx) = dy/dx

  12. zepdrix
    • 2 years ago
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    Is that a y to the first power on the left there?

  13. amonoconnor
    • 2 years ago
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    Y prime

  14. zepdrix
    • 2 years ago
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    You've learned about the chain rule before the power rule...?\[\Large \frac{d}{dx}x^n\quad=\quad n x^{n-1}\]That doesn't look familiar? +_+

  15. amonoconnor
    • 2 years ago
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    Oh... Yes, I know of that. Whoops! I guess we don't know the terminology... I will bring that up to him!

  16. amonoconnor
    • 2 years ago
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    Ahhhh, I follow your post from earlier... :)

  17. zepdrix
    • 2 years ago
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    Using rules of exponents allows us to write the inner function like this:\[\Large \frac{1}{y}\quad =\quad y^{-1}\]From here we don't need to use the quotient rule, we use the power rule instead:\[\Large \frac{d}{dx}y^{-1}\quad=\quad -1y^{-1-1}\frac{dy}{dx}\quad=\quad -y^{-2}\frac{dy}{dx}\]

  18. amonoconnor
    • 2 years ago
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    Alright.... You're so amazing. I do believe I'm good now. I have what I need, and most importantly, I fully understand what happened, and what I need to do. Thank you so much!

  19. amonoconnor
    • 2 years ago
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    (Would you mind medaling me?)

  20. zepdrix
    • 2 years ago
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    oh got it from there? cool c:

  21. amonoconnor
    • 2 years ago
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    Yep, I do! Thanks you so much again; you really helped me look at that from a better angle.

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