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calyne

  • 2 years ago

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

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

  2. calyne
    • 2 years ago
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    correct so far?

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

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

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

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

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

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

  9. experimentX
    • 2 years ago
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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

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

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

  12. experimentX
    • 2 years ago
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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.

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

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

  15. calyne
    • 2 years ago
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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

  16. calyne
    • 2 years ago
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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.....

  17. experimentX
    • 2 years ago
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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.

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

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

  20. calyne
    • 2 years ago
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    oh so it is

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