how do you find the log of a square root?

- anonymous

how do you find the log of a square root?

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- anonymous

log sqrt(x) = log x^(1/2) = (1/2)logx

- anonymous

ok, I have the log of the square root of x^2 +2 ?????

- anonymous

ok, then log sqrt(x^2 + 2) = log (x^2 + 2)^(1/2) = (1/2)log (x^2 + 2)

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## More answers

- anonymous

do you know how to find the derivative of it?

- anonymous

yes, use the following, d/dx(logu) = 1/ulna

- anonymous

let u = x^2 + 2, then y = logu^(1/2) = (1/2)logu = (1/2)/ulna = (1/2)/(x^2 + 2)lna

- anonymous

u did not specify base a however

- anonymous

its the application of the chain rule to the derivative of a logarithm with base a , provided a > 0

- 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

ok so u want me to differentiate x^5/(1-10x)(sqrt(x^2 + 2))?

- anonymous

srry, i thought u wanted me to to differentiate log sqrt(x^2 + 2) my bad

- 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

sure :)

- anonymous

ur my last client for this morning, i gotta go shower after this

- anonymous

well I sure am glad I found you! Thanks!

- 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

go on, pls

- anonymous

are you still there?

- amistre64

whats the question?

- anonymous

I need to find the derivative of x^5/(1-10X)(sqrtx^2+2) using logs.

- 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

\[\frac{x^5}{1-10x}*\sqrt{x^2 +2}\] is this the equation? or is that sqrt stuck under the bar?

- anonymous

the square root is stuck under the bar next to the (1-10x)

- 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

it was in the instructions to do so.

- amistre64

odd instructions :)

- 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

\[\frac{\log(x^5)}{\log[(1-10x)(\sqrt{x^2+2})]}\]

- anonymous

yes!

- 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

log(ab) = log(a) + log(b)
sooooo.....
log(1-10x) + log[sqrt(x^2 +2)] right so far?

- anonymous

yes

- 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

yes

- 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

then we can split it up lol

- amistre64

\[\log(\frac{a}{b*c}) = \log(a)-[\log(b)+\log(c)]\] like this :)

- anonymous

I trust you! go on! brain mistakes are allowed! ; )

- amistre64

\[5 \log(x) - [\log(1-10x) + \frac{1}{2}\log(x^2+2)]\]

- amistre64

\[5 \log(x) - \log(1-10x) - (1/2)\log(x^2+2)\]

- 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

\[Dx[5 \ln(x) - \ln(1-10x) - \frac{1}{2}\ln(x^2+2)]\]

- amistre64

5/x
- 1/(1-10x)
- 1/(x^2+2)

- amistre64

the derivative of ln(x) = Dx/x

- amistre64

that second term should be: -10/(1+10x) then :)

- amistre64

.....ack!!!
+10/(1-10x) ill get it right eventually lol

- anonymous

youre great! Pls don't apologize!

- amistre64

did it make sense what I did :)

- 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

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

ok it's the "negative minus a negative gives a positive" thing, right?

- amistre64

exactly :)

- anonymous

ok, let's move to the sqrt thing...I cannot stand the suspense!

- 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

; )

- 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

no how did you get that?

- amistre64

tell me the last part that makes sense to you and i can unravel the mystery from there :)

- 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

i had to switch to firefox, IE acts wierd on this site

- 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

D(5a^2) = 5 D(a^2) = 5 (2a) = 10a right?

- anonymous

leave it alone visually is fine. I understand that it has to be put back in when we are done.

- 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

OMG! I get it!!!!!!!!!!!!!!!!! Can I try to explain it you so I know for sure?

- amistre64

you can try :)

- 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

illl be around :)

- 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

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

should the 1/2 be then positive or negative?

- 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

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

you can if you want ;)

- anonymous

how do I do that exactly?

- 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

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

is MathXL a program you are working with?

- amistre64

did the refresh work out for you?

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