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

Mathematics
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y' = [(2x)/(ln(10)*sqrt(x)] + [2*log[10](sqrt(x))]
correct so far?
or no the derivative of log[10](sqrt(x)) would be (1/[sqrt(x)ln(10)])*(1/[2sqrt(x)])

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Other answers:

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

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