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calyne

Differentiate the function y = 2x*log[10](sqrt(x))

  • 2 years ago
  • 2 years ago

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  1. calyne
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    y' = [(2x)/(ln(10)*sqrt(x)] + [2*log[10](sqrt(x))]

    • 2 years ago
  2. calyne
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    correct so far?

    • 2 years ago
  3. calyne
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    or no the derivative of log[10](sqrt(x)) would be (1/[sqrt(x)ln(10)])*(1/[2sqrt(x)])

    • 2 years ago
  4. calyne
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    no it's not

    • 2 years ago
  5. calyne
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    well according to the textbook the answer is pretty different

    • 2 years ago
  6. calyne
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    alright well i've got 1/ln(10) + 2*log[10](sqrt(x))

    • 2 years ago
  7. calyne
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    that's not final though so help me from there

    • 2 years ago
  8. calyne
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    somehow the second term (after the +) needs to equal log[10](x) does that make sense

    • 2 years ago
  9. experimentX
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    isn't it something like this http://www.wolframalpha.com/input/?i=d%2Fdx%28+2x*log+base+10+%28sqrt%28x%29%29%29

    • 2 years ago
  10. calyne
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    sure i guess it's something like that bro jesus you're a great help

    • 2 years ago
  11. calyne
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    but that's still not the textbook answer

    • 2 years ago
  12. experimentX
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    let me test, dy/dx = 2{dx/dx*log[10](sqrt(x))+x*d(log[10](sqrt(x)))/dx} = 2{log[10](sqrt(x))+x*d(ln(sqrt(x)))/dx*1/ln10} = 2{log[10](sqrt(x))+x*1/sqrt(x)*1/2*1/sqrt(x)*1/ln10} = 2{log[10](sqrt(x))+1/2ln10} my answer.

    • 2 years ago
  13. calyne
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    alright here look the textbook answer is 1/ln(10) + log[10](x).

    • 2 years ago
  14. experimentX
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    i got almost same answer ... except 1/ln(10) + 2log[10](x)

    • 2 years ago
  15. calyne
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    i got as far as 1/ln(x) + 2(log[10](x)), the latter of which = 2[log(sqrt(x))/log(10)], so [2*log(sqrt(x))] / log(10) = log(sqrt(x))^2/log(10) ?? which = log(x)/log(10)? which = log[10](x) ???? is that correct? does that make sense? i'm reeeal rusty on the log business

    • 2 years ago
  16. calyne
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    nah bro according to the product rule that other end of the + sign is gonna be the derivative of 2x * the log[10](sqrt(x)). unless that's not the case. but i'm pretty sure it is. don't see any way around it really. and you got the derivative of 2x multiplying the other factor which is supposed to be left as is, so.... you tell me..... if and how i'm wrong.....

    • 2 years ago
  17. experimentX
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    use this property of log to convert it into natural log. \[ \log_ax = \frac{\ln x}{\ln a} \] I think there's a better example at wikipedia.

    • 2 years ago
  18. calyne
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    i know that, and it still doesn't explain how 2*log[10](sqrt(x)) = log[10](x).

    • 2 years ago
  19. calyne
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    oh unless... that equals log[10]([sqrt(x)]^2) ???? is that it?

    • 2 years ago
  20. calyne
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    oh so it is

    • 2 years ago
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