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calyneBest ResponseYou've already chosen the best response.0
y' = [(2x)/(ln(10)*sqrt(x)] + [2*log[10](sqrt(x))]
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
or no the derivative of log[10](sqrt(x)) would be (1/[sqrt(x)ln(10)])*(1/[2sqrt(x)])
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
well according to the textbook the answer is pretty different
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
alright well i've got 1/ln(10) + 2*log[10](sqrt(x))
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
that's not final though so help me from there
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
somehow the second term (after the +) needs to equal log[10](x) does that make sense
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
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

calyneBest ResponseYou've already chosen the best response.0
sure i guess it's something like that bro jesus you're a great help
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
but that's still not the textbook answer
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
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

calyneBest ResponseYou've already chosen the best response.0
alright here look the textbook answer is 1/ln(10) + log[10](x).
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
i got almost same answer ... except 1/ln(10) + 2log[10](x)
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
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

calyneBest ResponseYou've already chosen the best response.0
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

experimentXBest ResponseYou've already chosen the best response.0
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

calyneBest ResponseYou've already chosen the best response.0
i know that, and it still doesn't explain how 2*log[10](sqrt(x)) = log[10](x).
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
oh unless... that equals log[10]([sqrt(x)]^2) ???? is that it?
 2 years ago
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