Question

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)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

ok

Answer 6

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

Answer 7

alright

Answer 8

have you used substitution rule before?

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

du = =1/x^2 * dx

Answer 15

yes, addmaster?

Answer 16

ok sorry the substitution thing is confusing me

Answer 17

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

Answer 18

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

Answer 19

that is like a calc identity

Answer 20

ok i get it

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

yeah, you do less internal variable manipulation

Answer 25

alright got it sorry

Answer 26

cool, you got it?

Answer 27

yep that helped a lottt

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

so u = 1 + 1/x

Answer 32

du = (1/x) dx

Answer 33

er, (1/x)'

Answer 34

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

Answer 35

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

Answer 36

-1x^-2...?

Answer 37

yeah

Answer 38

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

Answer 39

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)

Answer 40

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

Answer 41

which you can simplify

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

thank you very much