Hey guys I'm having problems with derivatives if someone could explain the logx/lnx derivatives to me that would be great heres a few examples y = -2x^2(lnx)^4 or y = ln((x +1)/x)

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- anonymous

- katieb

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- anonymous

to start, do you know the derivative of \[\ln(x)\] ?

- anonymous

\[d/dx [ \ln (x) ] = 1/x\]

- anonymous

yes i just dont understand how it applies when you have things attached to x

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## More answers

- anonymous

such as the (lnx)^4 or the ln(x +1/x)

- anonymous

ok

- anonymous

let's try \[\ln((x+1)/x)\]

- anonymous

alright

- anonymous

have you used substitution rule before?

- anonymous

That example is very simple, though the process might be tedious. Simply put a 1 over the expression then multiply by derivative of the expression

- anonymous

No, i got that wrong, i was describing log.

- anonymous

I'm contradicting myself, I think i told you right

- anonymous

so, ln((x+1)/x)

- anonymous

using the substitution rule: let's have \[u = (x+1)/x , du = -1/x^2\]

- anonymous

du = =1/x^2 * dx

- anonymous

yes, addmaster?

- anonymous

ok sorry the substitution thing is confusing me

- anonymous

ok, well, the concept is thus: essentially you can simply things by taking a difficult expression to immeditaely differentitate, like here ln((x+1)/x), and simplify the first steps a bit

- anonymous

outright, [ln((x+1)/x)]' looks difficult, but [ln(u)]'

- anonymous

that is like a calc identity

- anonymous

ok i get it

- anonymous

the trick is keeping track of this substitution, you'll need to find the derivtative of u. as above, u=(x+1)/x, while du = (-1/x^2)*dx

- anonymous

ahhh ok so if you set it as a variable it complicates things less

- anonymous

so we have our first equation: ln((x+1)/x) dx

- anonymous

yeah, you do less internal variable manipulation

- anonymous

alright got it sorry

- anonymous

cool, you got it?

- anonymous

yep that helped a lottt

- anonymous

so, you want to solve it? i can stay to help if you need it...

- anonymous

but just to be sure once I have the x + 1 over x I have to then use quotient rule...?

- anonymous

so, u = (x+1)/x ... but that equals x/x + 1/x

- anonymous

so u = 1 + 1/x

- anonymous

du = (1/x) dx

- anonymous

er, (1/x)'

- anonymous

which is (\[x^-1\])'

- anonymous

so, whats the derivative with respect to x of 1 + x^-1 ?

- anonymous

-1x^-2...?

- anonymous

yeah

- anonymous

so, ln(u) du ---> but du = (-1/x^2)*dx

- anonymous

ln(u)' --> 1/u, where u is (1 + 1/x) ....so you have \[1/[(1 + 1/x)]\]....then, you can't forget that du was -1/x^2 * dx, so when you now substitute back in for u so you have the original variable x, you have to multiply by (-1/x^2)

- anonymous

\[\frac{1}{(1+1/x)} * \frac{-1}{(x^2)}\]

- anonymous

which you can simplify

- anonymous

to : -1/(x^2 + x)

- anonymous

ok sorry my computers really crappy so i cant reply fast but i completely get it now thanks!

- anonymous

no problem, hopefully now you can use the book examples much better too!

- anonymous

also, here on OpenStudy, if someone gives you good help, you can award them with a medal here in the conversation

- anonymous

so, ask more questions and maybe you can help some other math learners if you see a question you know to answer

- anonymous

i'm off, but good luck with the calc ;)

- anonymous

thank you very much

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