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anonymous

  • one year ago

derivative of y=x^(3/2)log(base3)squareroot(x+7)

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  1. zepdrix
    • one year ago
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    \(\large\rm y=x^{3/2}\cdot\log_3\sqrt{x+7}\) This guy? Hmm this looks like a fun one :) lol

  2. anonymous
    • one year ago
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    \[y= x ^{3/2}\log_{3}\sqrt{x+7}\]

  3. anonymous
    • one year ago
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    yeah lol

  4. zepdrix
    • one year ago
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    Do you remember your change of base formula for logs?\[\large\rm \log_{a}(b)=\frac{\ln(b)}{\ln(a)}\]We would choose natural log like this ^ Because it's easier to differentiate.

  5. zepdrix
    • one year ago
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    Unless you have a special formula memorized for log derivatives :) \[\large\rm y=x^{3/2}\cdot\frac{\ln\sqrt{x+7}}{\ln3}\]

  6. anonymous
    • one year ago
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    so it would be \[(3/2)x ^{1/2}*(\ln \sqrt{x+7}/\ln3)\]

  7. anonymous
    • one year ago
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    ?

  8. zepdrix
    • one year ago
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    No no no :O Product rule.

  9. anonymous
    • one year ago
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    what should i do then?

  10. zepdrix
    • one year ago
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    I'll do the "set up" and you can let me know if it makes sense:\[\large\rm y'=\color{royalblue}{\left(x^{3/2}\right)'}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}\]This is our product rule "set up". I haven't taken any derivatives yet. I'm just setting it up. We need to differentiate the blue parts.

  11. anonymous
    • one year ago
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    is the first blue part \[(3/2)x ^{(3/2)-1}\]

  12. zepdrix
    • one year ago
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    \[\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}\]Ok ya that sounds right!

  13. zepdrix
    • one year ago
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    This other guy is a bit tricky though, ya? :)

  14. anonymous
    • one year ago
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    \[1/((1/2)x ^{-1/2}+(1/2)7^{-1/2}) /(1/3)\]

  15. anonymous
    • one year ago
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    ?

  16. zepdrix
    • one year ago
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    No. And oh boy, did you ever make things complicated :D lol

  17. zepdrix
    • one year ago
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    Let's back up a sec, apply a helpful log rule before we take the derivative.

  18. anonymous
    • one year ago
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    sorry lol

  19. anonymous
    • one year ago
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    \[\ln \sqrt{x+7}-\ln 3\]

  20. anonymous
    • one year ago
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    ?

  21. zepdrix
    • one year ago
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    First notice that ln3 is just a constant, so we can pull it out of the differentiation process and ignore it.\[\large\rm \color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}=\frac{1}{\ln3}\color{royalblue}{\left(\ln\sqrt{x+7}\right)'}\]

  22. anonymous
    • one year ago
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    what about the x+7

  23. zepdrix
    • one year ago
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    Next rewrite your square root as a 1/2 power,\[\large\rm \frac{1}{\ln3}\color{royalblue}{\left(\ln(x+7)^{1/2}\right)'}\]And apply another log rule,\[\large\rm =\frac{1}{\ln3}\color{royalblue}{\left(\frac{1}{2}\ln(x+7)\right)'}\]And again pull the 1/2 out front because it's just a constant,\[\large\rm =\frac{1}{2\ln3}\color{royalblue}{\left(\ln(x+7)\right)'}\]The problem becomes a lot easier to deal with at this point.

  24. zepdrix
    • one year ago
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    This is one of the things that can make Calc a difficult subject, remember all of these silly Algebra rules :) They're really helpful though.

  25. anonymous
    • one year ago
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    does the x+7 become 8

  26. anonymous
    • one year ago
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    or does it become 2

  27. zepdrix
    • one year ago
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    I don't understand what you're asking

  28. anonymous
    • one year ago
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    does \[(\ln(x+7))\prime = \ln(2)\prime = 1/2\]

  29. zepdrix
    • one year ago
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    ?? 0_o

  30. zepdrix
    • one year ago
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    I have no idea where that 2 is coming from.. this is so confusing

  31. zepdrix
    • one year ago
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    Do you remember your derivative for ln(x) ?

  32. anonymous
    • one year ago
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    isnt the derivative of a constant = 1?

  33. anonymous
    • one year ago
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    so the derivative of x = 1 and the derivative of 7 = 1

  34. anonymous
    • one year ago
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    so 1+1 = 2

  35. anonymous
    • one year ago
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    ???

  36. zepdrix
    • one year ago
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    The derivative of a constant is = 0. x is not a constant. But let's not worry about that. That's not what we're dealing with here. I'm asking if you know the derivative of \(\large\rm \ln(x)\)

  37. anonymous
    • one year ago
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    is it 1?

  38. zepdrix
    • one year ago
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    ... No :(

  39. anonymous
    • one year ago
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    isnt the derivative of ln(x) = 1/x

  40. zepdrix
    • one year ago
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    Yes.

  41. zepdrix
    • one year ago
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    So then by that same rule,\[\large\rm \frac{d}{dx}\ln(x+7)=\frac{1}{(x+7)}\]

  42. anonymous
    • one year ago
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    wait so is the answer \[(3/2)x ^{1/2}-((\ln(\sqrt{x+7})/\ln3)*x ^{3/2})*(1/x+7)\]

  43. zepdrix
    • one year ago
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    No :(

  44. anonymous
    • one year ago
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    what is the right answer?

  45. zepdrix
    • one year ago
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    By the steps I was showing you before, the derivative of that second blue thing we were working on, would be,\[\large\rm \color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}=\color{orangered}{\frac{1}{2\ln3(x+7)}}\]

  46. zepdrix
    • one year ago
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    \[\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}\] So I guess our final result is something like this,\[\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{orangered}{\frac{1}{(2\ln3)(x+7)}}\]

  47. zepdrix
    • one year ago
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    Practice more problems cp :O Really really need to study!

  48. anonymous
    • one year ago
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    OH OKAY

  49. anonymous
    • one year ago
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    I GET IT NOW

  50. anonymous
    • one year ago
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    THANK YOU SO MUCH

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