anonymous
  • anonymous
how do you find the log of a square root?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
log sqrt(x) = log x^(1/2) = (1/2)logx
anonymous
  • anonymous
ok, I have the log of the square root of x^2 +2 ?????
anonymous
  • anonymous
ok, then log sqrt(x^2 + 2) = log (x^2 + 2)^(1/2) = (1/2)log (x^2 + 2)

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anonymous
  • anonymous
do you know how to find the derivative of it?
anonymous
  • anonymous
yes, use the following, d/dx(logu) = 1/ulna
anonymous
  • anonymous
let u = x^2 + 2, then y = logu^(1/2) = (1/2)logu = (1/2)/ulna = (1/2)/(x^2 + 2)lna
anonymous
  • anonymous
u did not specify base a however
anonymous
  • anonymous
its the application of the chain rule to the derivative of a logarithm with base a , provided a > 0
anonymous
  • anonymous
i do not even know what you mean by that. the whole problem is to differentaite x^5 divided by (1-10x)(sqrt ofx^2 +2)
anonymous
  • anonymous
ok so u want me to differentiate x^5/(1-10x)(sqrt(x^2 + 2))?
anonymous
  • anonymous
srry, i thought u wanted me to to differentiate log sqrt(x^2 + 2) my bad
anonymous
  • anonymous
yes but I need to know specifically how to do the expression involving the square root. I cannot understand that; maybe it's the algebra involved to figure it out? /can you do the whole problem, focusing on how you get the derivative of the square root?
anonymous
  • anonymous
sure :)
anonymous
  • anonymous
ur my last client for this morning, i gotta go shower after this
anonymous
  • anonymous
well I sure am glad I found you! Thanks!
anonymous
  • anonymous
ok since, this is an ugly problem for application of the quotient rule, we r going to let y equal the expression and take the natural log of both sides
anonymous
  • anonymous
go on, pls
anonymous
  • anonymous
are you still there?
amistre64
  • amistre64
whats the question?
anonymous
  • anonymous
I need to find the derivative of x^5/(1-10X)(sqrtx^2+2) using logs.
anonymous
  • anonymous
My problem is the expression with the sqrt in it. the others I am comfortable with finding, but that sqrt is killing me!
amistre64
  • amistre64
\[\frac{x^5}{1-10x}*\sqrt{x^2 +2}\] is this the equation? or is that sqrt stuck under the bar?
anonymous
  • anonymous
the square root is stuck under the bar next to the (1-10x)
amistre64
  • amistre64
got it..... and why do we have to use logs to do this? is that in the directions? or something you thought might help?
anonymous
  • anonymous
it was in the instructions to do so.
amistre64
  • amistre64
odd instructions :)
anonymous
  • anonymous
are they? I am so new to this, 2 weeks into calc and I think the objective was teach us logs and how to differentiate them using all the forms of expressions possible.
amistre64
  • amistre64
\[\frac{\log(x^5)}{\log[(1-10x)(\sqrt{x^2+2})]}\]
anonymous
  • anonymous
yes!
amistre64
  • amistre64
anything with an exponent gets dragged to the fron....like this: 5 log(x) ..... that takes care of the top right? or at least is a step we can take..now for th e bottom
amistre64
  • amistre64
log(ab) = log(a) + log(b) sooooo..... log(1-10x) + log[sqrt(x^2 +2)] right so far?
anonymous
  • anonymous
yes
amistre64
  • amistre64
do you recall that radicals such as squaree roots and cube roots and the like are fraction exponentes? sqrt(4) = 4^(1/2) = 2 right?
anonymous
  • anonymous
yes
amistre64
  • amistre64
good; and after all that work i see a mistale that my brain made lol.... we first need to take the log of the whole entire fraction; not portions of it :)
amistre64
  • amistre64
then we can split it up lol
amistre64
  • amistre64
\[\log(\frac{a}{b*c}) = \log(a)-[\log(b)+\log(c)]\] like this :)
anonymous
  • anonymous
I trust you! go on! brain mistakes are allowed! ; )
amistre64
  • amistre64
\[5 \log(x) - [\log(1-10x) + \frac{1}{2}\log(x^2+2)]\]
amistre64
  • amistre64
\[5 \log(x) - \log(1-10x) - (1/2)\log(x^2+2)\]
amistre64
  • amistre64
thats as basic as you can get it; now take the derivatives :) and the "log" part doesnt have to be "log" it can also be "ln"; the natural log...ln migh tmake life easier with derivatives
amistre64
  • amistre64
\[Dx[5 \ln(x) - \ln(1-10x) - \frac{1}{2}\ln(x^2+2)]\]
amistre64
  • amistre64
5/x - 1/(1-10x) - 1/(x^2+2)
amistre64
  • amistre64
the derivative of ln(x) = Dx/x
amistre64
  • amistre64
that second term should be: -10/(1+10x) then :)
amistre64
  • amistre64
.....ack!!! +10/(1-10x) ill get it right eventually lol
anonymous
  • anonymous
youre great! Pls don't apologize!
amistre64
  • amistre64
did it make sense what I did :)
anonymous
  • anonymous
so far it makes sinse except where you said that it was a +10. I thought the whole equation was the expression -another expression-another expression. Why the +?
amistre64
  • amistre64
the middle term in logs is : - ln(1-10x) the derivative of ln(....) is: the derivative of (....) derivative of (1-10x) ------------------ = ------------------ (....) (1-10x) the second term the becomes.... dont forget the initial (-) in fron of it -10x - ------ = +10/(1-10x) :) (1-10x)
anonymous
  • anonymous
ok it's the "negative minus a negative gives a positive" thing, right?
amistre64
  • amistre64
exactly :)
anonymous
  • anonymous
ok, let's move to the sqrt thing...I cannot stand the suspense!
amistre64
  • amistre64
2 negative always gives a positive; -5(-3) = 15 7--3 = 10 -6/-3 = 2 -6 -2 = -8....except for that one lol
anonymous
  • anonymous
; )
amistre64
  • amistre64
the sqrt became the thrid term: - 1/2 ln(x^2+2) right? which derives to... 1 2x - -- ------- = -x/(x^2+2) right? 2 (x^2+2)
anonymous
  • anonymous
no how did you get that?
amistre64
  • amistre64
tell me the last part that makes sense to you and i can unravel the mystery from there :)
anonymous
  • anonymous
i understand where you got the third term -1/2 ln(x^2+2), but finding the derivative of it is where I am baffled.
amistre64
  • amistre64
i had to switch to firefox, IE acts wierd on this site
amistre64
  • amistre64
ok.... Do we have to derive the -1/2 part? or can we pull it aside and leave it alone?... visually that is.
amistre64
  • amistre64
D(5a^2) = 5 D(a^2) = 5 (2a) = 10a right?
anonymous
  • anonymous
leave it alone visually is fine. I understand that it has to be put back in when we are done.
amistre64
  • amistre64
good; then lets derive the ln(x^2+2) part :) D(x^2 +2) --------- right? so all we really have to do is derive the top (x^2 +2)
anonymous
  • anonymous
OMG! I get it!!!!!!!!!!!!!!!!! Can I try to explain it you so I know for sure?
amistre64
  • amistre64
you can try :)
anonymous
  • anonymous
Ok, give me a minute to write it all down then type it to you, but I am going to work only with the sqrt expression since that was the one giving me trouble. Can you hold for like 2 minutes?
amistre64
  • amistre64
illl be around :)
anonymous
  • anonymous
ok, here goes. i will do my best with the ^ sign and stuff, ok? Are yo with me? Here goes:::::: d/dx(ln(x^2+2)^1/2) = -1/2(ln(x^2+2)) = -1/2(1/x^2+2)d/dx(x^2+2) = -1/2(1/x^2+2)(2x) = -2x/2(x^2+2) = (after canceling out the 2) -x/(x^2+2) Am I right? Is that the derivative of the natural log of the sqrt of x^2+2?
amistre64
  • amistre64
that is very good :) except for the spurious (-) sign which is just a result of it being the third term subtracted from the others... it looks like you got a handle on it :)
anonymous
  • anonymous
should the 1/2 be then positive or negative?
amistre64
  • amistre64
it should be positive since if it were standing all alone by itself: Dx[ln(x^2+2)^(1/2))] = Dx[(1/2) ln(x^2+2)] = 1/2 Dx(ln(x^2+2)) 1/2 2x/x^2+2 the 2s cancel to give you... x/(x^2 + 2) :)
anonymous
  • anonymous
thank you, thank you, thank you!!! You helped me when no one else could make me understand! Can I give yuou a medal for this?
amistre64
  • amistre64
you can if you want ;)
anonymous
  • anonymous
how do I do that exactly?
amistre64
  • amistre64
you night have to press your refresh button on your browser...its usually just the f5 button on te keyboard. That will refresh the page and allow you to see a "give medal" option next to my name :)
anonymous
  • anonymous
ok, IO will do that now, although I agree that IE is weird with this; the page kept scrolling up and down on its own! i like Firefox so much better, but MathXL only works with IE.
amistre64
  • amistre64
is MathXL a program you are working with?
amistre64
  • amistre64
did the refresh work out for you?

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