anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
to start, do you know the derivative of \[\ln(x)\] ?
anonymous
  • anonymous
\[d/dx [ \ln (x) ] = 1/x\]
anonymous
  • anonymous
yes i just dont understand how it applies when you have things attached to x

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
such as the (lnx)^4 or the ln(x +1/x)
anonymous
  • anonymous
ok
anonymous
  • anonymous
let's try \[\ln((x+1)/x)\]
anonymous
  • anonymous
alright
anonymous
  • anonymous
have you used substitution rule before?
anonymous
  • 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
  • anonymous
No, i got that wrong, i was describing log.
anonymous
  • anonymous
I'm contradicting myself, I think i told you right
anonymous
  • anonymous
so, ln((x+1)/x)
anonymous
  • anonymous
using the substitution rule: let's have \[u = (x+1)/x , du = -1/x^2\]
anonymous
  • anonymous
du = =1/x^2 * dx
anonymous
  • anonymous
yes, addmaster?
anonymous
  • anonymous
ok sorry the substitution thing is confusing me
anonymous
  • 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
  • anonymous
outright, [ln((x+1)/x)]' looks difficult, but [ln(u)]'
anonymous
  • anonymous
that is like a calc identity
anonymous
  • anonymous
ok i get it
anonymous
  • 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
  • anonymous
ahhh ok so if you set it as a variable it complicates things less
anonymous
  • anonymous
so we have our first equation: ln((x+1)/x) dx
anonymous
  • anonymous
yeah, you do less internal variable manipulation
anonymous
  • anonymous
alright got it sorry
anonymous
  • anonymous
cool, you got it?
anonymous
  • anonymous
yep that helped a lottt
anonymous
  • anonymous
so, you want to solve it? i can stay to help if you need it...
anonymous
  • anonymous
but just to be sure once I have the x + 1 over x I have to then use quotient rule...?
anonymous
  • anonymous
so, u = (x+1)/x ... but that equals x/x + 1/x
anonymous
  • anonymous
so u = 1 + 1/x
anonymous
  • anonymous
du = (1/x) dx
anonymous
  • anonymous
er, (1/x)'
anonymous
  • anonymous
which is (\[x^-1\])'
anonymous
  • anonymous
so, whats the derivative with respect to x of 1 + x^-1 ?
anonymous
  • anonymous
-1x^-2...?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
so, ln(u) du ---> but du = (-1/x^2)*dx
anonymous
  • 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
  • anonymous
\[\frac{1}{(1+1/x)} * \frac{-1}{(x^2)}\]
anonymous
  • anonymous
which you can simplify
anonymous
  • anonymous
to : -1/(x^2 + x)
anonymous
  • anonymous
ok sorry my computers really crappy so i cant reply fast but i completely get it now thanks!
anonymous
  • anonymous
no problem, hopefully now you can use the book examples much better too!
anonymous
  • anonymous
also, here on OpenStudy, if someone gives you good help, you can award them with a medal here in the conversation
anonymous
  • anonymous
so, ask more questions and maybe you can help some other math learners if you see a question you know to answer
anonymous
  • anonymous
i'm off, but good luck with the calc ;)
anonymous
  • anonymous
thank you very much

Looking for something else?

Not the answer you are looking for? Search for more explanations.