anonymous
  • anonymous
Differentiate the function y = 2x*log[10](sqrt(x))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
y' = [(2x)/(ln(10)*sqrt(x)] + [2*log[10](sqrt(x))]
anonymous
  • anonymous
correct so far?
anonymous
  • anonymous
or no the derivative of log[10](sqrt(x)) would be (1/[sqrt(x)ln(10)])*(1/[2sqrt(x)])

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anonymous
  • anonymous
no it's not
anonymous
  • anonymous
well according to the textbook the answer is pretty different
anonymous
  • anonymous
alright well i've got 1/ln(10) + 2*log[10](sqrt(x))
anonymous
  • anonymous
that's not final though so help me from there
anonymous
  • anonymous
somehow the second term (after the +) needs to equal log[10](x) does that make sense
experimentX
  • experimentX
isn't it something like this http://www.wolframalpha.com/input/?i=d%2Fdx%28+2x*log+base+10+%28sqrt%28x%29%29%29
anonymous
  • anonymous
sure i guess it's something like that bro jesus you're a great help
anonymous
  • anonymous
but that's still not the textbook answer
experimentX
  • experimentX
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.
anonymous
  • anonymous
alright here look the textbook answer is 1/ln(10) + log[10](x).
experimentX
  • experimentX
i got almost same answer ... except 1/ln(10) + 2log[10](x)
anonymous
  • anonymous
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
anonymous
  • anonymous
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.....
experimentX
  • experimentX
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.
anonymous
  • anonymous
i know that, and it still doesn't explain how 2*log[10](sqrt(x)) = log[10](x).
anonymous
  • anonymous
oh unless... that equals log[10]([sqrt(x)]^2) ???? is that it?
anonymous
  • anonymous
oh so it is

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