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Schrodinger

This isn't the question itself, but I need to find the derivative of something to do the question (Problem below)

  • one year ago
  • one year ago

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  1. Schrodinger
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    \[y = x ^{2/3}(2+x).\] What i'm doing is applying the Product Rule for Derivatives like such: \[(x ^{2/3})*\frac{ dy }{ dx }(2+x) + (2+x)\frac{ dy }{ dx }(x ^{2/3})\] From there, I get\[x ^{2/3}+\frac{ 2(2+x) }{ 3\sqrt[3]{x} }\]What i'm doing is then multiplying everything by the denominator and ending up with\[5x+4\] after simplifying, but the answer should be\[\frac{ 5x+4 }{ 3\sqrt[3]{} }\]

    • one year ago
  2. Schrodinger
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    Whoops. Denominator should be \[3\sqrt[3]{x}\]

    • one year ago
  3. Schrodinger
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    Why am I getting \[5x+4\] as opposed to\[\frac{ 5x+4 }{ 3\sqrt[3]{x} }\] ?

    • one year ago
  4. Ahmad_Shadab
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    just take the LCM of denominator and u wl get the answer

    • one year ago
  5. Schrodinger
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    That didn't really help, but let me specifically go through multiplying through the denominator so somebody can see if i'm doing something wrong: \[x ^{2/3}(3\sqrt[3]{x}) = x ^{2/3}*x ^{1/3}*3=x ^{1}*3=x*3=3x\]Right? Then, since the numerator has been multiplied out, I just have \[3x+2(x+2)\]which just simplifies to\[5x+4\]?

    • one year ago
  6. Ahmad_Shadab
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    u got the correct numerator but u forgot the denominator

    • one year ago
  7. Schrodinger
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    What does that mean, dude? I know that. You're kind of unclear.

    • one year ago
  8. Schrodinger
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    I'm just curious why, unless I broke math, lol, my method isn't as valid.

    • one year ago
  9. stamp
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    \[y=x^{2/3}(2+x)\] u = x^2/3, du = 2/3 x^-1/3 v = 2 + x, dv = 1 u dv + v du = y' x^2/3 (1) + (2 + x) 2/3 x^-1/3 \[y'=x^{2/3}+\frac{2(2+x)}{3x^{1/3}}\]\[y'=\frac{3x+4+2x}{3x^{1/3}}\]\[y'=\frac{5x+4}{3x^{1/3}}\]verification - http://www.wolframalpha.com/input/?i=derivative+of+x^%282%2F3%29%282%2Bx%29 post a line number if you want to go over any steps more in depth

    • one year ago
  10. Schrodinger
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    Line number?

    • one year ago
  11. Schrodinger
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    (I dunno what that means.) I checked the wolfram page to see if you were referring to the step number in a step-by-step solution, but I don't see one. PS, thank you

    • one year ago
  12. Schrodinger
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    OH. Derp. Just got how your method works. But why is mine still wrong? Shouldn't it just be another way to arrive at the same answer?

    • one year ago
  13. stamp
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    yours is ok but you combine the terms, so you need to get a common denominator of 3x ^ 1/3

    • one year ago
  14. ZeHanz
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    I would do this:\[y'=\frac{2(2+x)}{3\sqrt[3]{x}}+\sqrt[3]{x^2}=\frac{2(2+x)}{3\sqrt[3]{x}}+\frac{\sqrt[3]{x^2}}{1}\cdot \frac{3\sqrt[3]{x}}{3\sqrt[3]{x}}=\frac{2(2+x)}{3\sqrt[3]{x}}+\frac{3x}{3\sqrt[3]{x}}=\]\[y'=\frac{ 5x +4}{ 3\sqrt[3]{x} }\]

    • one year ago
  15. stamp
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    \[x^{2/3}=\frac{3x}{3x^{1/3}}\]

    • one year ago
  16. stamp
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    3x + 2(2+x) = 3x + 4 + 2x = 5x + 4 thats you numerator

    • one year ago
  17. Schrodinger
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    Give me just a minute to mull over this, mind being blown.

    • one year ago
  18. stamp
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    lolk

    • one year ago
  19. Schrodinger
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    Ugugugugugugugugh. I still don't understand what's wrong with multiplying out the denominator. If I multiply out to the denominator like this: \[3\sqrt[3]{x}[\frac{ x ^{2/3} }{ 1 }+\frac{ 2(2+x) }{ 3\sqrt[3]{x} }]\]Shouldn't the denominator of the second term disappear, and the denominator of the first term stay the same since you're multiplying by \[\frac{ 3\sqrt[3]{x} }{ 1 }\]?

    • one year ago
  20. stamp
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    \[x^{2/3}*3x^{1/3}/3x^{1/3}=3x/3x^{1/3}\]

    • one year ago
  21. stamp
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    \[3x/3x^{1/3}+2(2+x)/3x^{1/3}=(5x+4)/3x^{1/3}\]

    • one year ago
  22. Meepi
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    The problem with multiplying it out like you did \[ y' = x ^{2/3}+\frac{ 2(2+x) }{ 3\sqrt[3]{x} }\] Multiplying by \(\Large \frac{ 3\sqrt[3]{x} }{ 1 }\) gives \[\frac{ 3\sqrt[3]{x} }{ 1 }y' = \frac{ 3\sqrt[3]{x} }{ 1 }\left(x ^{2/3}+\frac{ 2(2+x) }{ 3\sqrt[3]{x} }\right)\] So instead of the y' you were looking for, you calculated \(\Large \frac{ 3\sqrt[3]{x} }{ 1 }y' \)!

    • one year ago
  23. Schrodinger
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    I'm confused about why you multiplied by \[\frac{ 3x ^{1/3} }{ 3x ^{1/3} } \]As opposed to\[\frac{ 3x ^{1/3} }{ 1 }\]? If I multiplied by the second term, wouldn't the denominator of the second term cancel out, and since the term i'm multiplying by is over one, it should only affect the numerator of the second term? I feel like I fundamentally learned something wrong in Algebra, and i'm just now figuring it out.

    • one year ago
  24. Schrodinger
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    @Meepi: Just saw your reponse, gimme a sec to read, my bad

    • one year ago
  25. Schrodinger
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    Okay. World class screwup. I'm pretty sure over literally the history of my doing math i've been constantly assuming functions to be equal to zero even though they're clearly not and it's usually not consequential unless something like this happens. Jesus. Thanks.

    • one year ago
  26. Schrodinger
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    *headdesk*

    • one year ago
  27. stamp
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    it is ok

    • one year ago
  28. Schrodinger
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    Yup, better just not do that EVER again, lmao.

    • one year ago
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