Question

Differentiate: ln(x-1/x^3+1)

Answer 1

I really need help with this one! natural logs always throw me off.

Answer 2

so I know I can split this equation because of the natural logs rule like this:
\[(lnx-1)-(lnx ^{3}+1)\]

Answer 3

but then do i take the derivative of each term? which rule do I use?

Answer 4

\[\Large \ln(x-1)-\ln(x^3+1)\]Yah looks good so far :)
Hmm, yes we want to take the derivative of each term separately.

We'll use the rule for the derivative of natural log :O remember that one?

Answer 5

which would be inverse right?

Answer 6

so I would have something like 1/(x-1)

Answer 7

Gives us the reciprocal thing? yah that looks good for the first one.
Don't forget to apply the `chain rule` on the second term.

Answer 8

after I do the reciprocal thing for the second term right?

Answer 9

also how do I know I'm suppose to do the chain rule for the second only?

Answer 10

You can apply the chain rule to the first term also, it might be good practice.\[\Large \left[\ln(x-1)\right]' \qquad=\qquad \frac{1}{x-1}\color{royalblue}{(x-1)'} \qquad=\qquad \frac{1}{x-1}\color{royalblue}{(1)}\]

Answer 11

You want to try and think ahead, "If I apply the chain rule, will it give me anything more than just 1?"

If the answer is yes, then you need to apply the chain rule.

Answer 12

See how applying it to the first term gave us 1 for the derivative of the inner function? :o

Answer 13

so we didnt' need to use the chain rule

Answer 14

Didn't need to, correct ^^

Answer 15

ok so the second term needs to use the chain rule right?

Answer 16

Yes, because the derivative of the inner function will end up giving us more than just 1.

Answer 17

right. can u check if I have the right answer?

Answer 18

sure c: whatcha got?

Answer 19

for the second term I got 3x/x^3+1

Answer 20

Woops, small mistake with your chain rule there! :)
\[\large \left[\ln(x^3+1)\right]' \qquad=\qquad \frac{1}{x^3+1}\color{royalblue}{(x^3+1)'} \qquad=\qquad \frac{1}{x^3+1}\color{royalblue}{(3x^2)}\]

Answer 21

oh yeah I see it :) haha

Answer 22

so then is this my final answer?

Answer 23

Yes.
You could combine the fractions into one big messy thing.
But given that the problem worked out the way it did, with the logs splitting up, this is probably where you're expected to stop.

Answer 24

YAY! Thanks!

Answer 25

Oh I have one quick question

Answer 26

its not about natural logs..

Answer 27

Uh oh +_+ what kind of logs is it about then?
I'm not a beaver.. i don't know much about wood besides the natural log D:

Answer 28

oh my... lol its implicit differentiation problem

Answer 29

Oh those are fun :)

Answer 30

so I have y-x=sinxy

Answer 31

yes haha sometimes?

Answer 32

Like this? both x and y inside the sine?
\[\Large y-x=\sin(xy)\]

Answer 33

I'm wondering when I differentiate each term for sinxy do I just (cosxy)*y'?

Answer 34

yup!

Answer 35

It's gonna be a smidge more complex than that I'm afraid :u

Answer 36

\[\Large \left[\sin(xy)\right]' \qquad=\qquad \left[\cos(xy)\right]\color{royalblue}{(xy)'}\]Here is our chain rule setup.
Looks like we'll have to apply the product rule on this blue term.

Answer 37

ok!

Answer 38

so for the blue term its y+x?

Answer 39

what do I do with y'? I just assumed that became 1.

Answer 40

\[\Large =\cos(xy)\left[\color{royalblue}{(x)'}y+x\color{royalblue}{(y)'}\right]\]
So you've got the right idea with your product rule setup.
But noooo, y' doesn't become 1! :O
That is the thing we're trying to solve for.
x' becomes 1, because we're taking our derivative `with respect to x`.
y' is the derivative term we're looking for.

\[\Large =\cos(xy)\left[\color{royalblue}{(1)}y+x\color{royalblue}{y'}\right]\]

Answer 41

ok .. so its just y+xy

Answer 42

xy' i mean

Answer 43

now what?

Answer 44

\[\Large y-x=\sin(xy)\]Taking the derivative gives us,\[\Large y'-1=\cos(xy)\left(y+xy'\right)\]We want to distribute the cosine to each term in the brackets.

Answer 45

ok~???

Answer 46

\[\Large y'-1=y\cos(xy)+x \cos(xy)y'\]

Understand that step ok? I know it looks a little messy.

Answer 47

We need to get our (y')'s onto one side and this is the only way we can do it :o

Answer 48

yeah I got it! but how are we gonna get our y's to one side..

Answer 49

Let's subtract the (big ugly cosine with the y' attached to it) from each side.\[\Large y'-x \cos(xy)y'-1=y \cos(xy)\]Then we'll add 1 to each side,\[\Large y'-x \cos(xy)y'=y \cos(xy)+1\]Then we'll factor y' out of each term on the left,\[\Large y'\left[1-x \cos(xy)\right]=y \cos(xy)+1\]

Understand that last step there? That one is pretty important.
We can factor derivative terms just like any other variable.

Answer 50

yup i'm following!

Answer 51

to solve for y' we simply divide by that big mess on the left,\[\Large y'=\frac{y \cos(xy)+1}{1-x \cos(xy)}\]

Answer 52

so ur just trying first to get y' on each side not just any y's.

Answer 53

Correct just `y'` 's, we don't care about the y's :)

Answer 54

Stupid apostrophe :P man that's confusing lolol

Answer 55

yup got it!

Answer 56

wow thanks for the help! ok one last question lol this one you don't have to walk through the whole thing with me cuz you've been such a big help but i was just wondering...

Answer 57

heh

Answer 58

when you have three terms like this (xsomething)(xsomething)(ysomething) do you just do the product rule? Implicit differentiation again.

Answer 59

Your text book might suggest that you apply logarithmic differentiation.
Because:\[\large \ln(x^3e^{2x}\cos x) \qquad=\qquad \ln x^3+ \ln(e^{2x})+\ln(\cos x)\]
That would allow you to split up terms easily and take their derivatives individually.

But yah another option is to apply the product rule.
It follows the same pattern with 3 terms.\[\Large (uvw)' \qquad=\qquad u'vw+uv'w+uvw'\]

Answer 60

Oh sorry I misread your comment :3 thought the last one was (xsomething).
No big deal though, same idea applies.
Probably would want to do product rule if it involves (ysomething).

Answer 61

ok thanks so much for ur help!!