Question

CALCULUS - Derivatve of natural log

i need to find the derivzative of ln x(3x-2).
i know its the chain rule but i cant figure it out; this is what i have so far:

Answer 1

Is everything inside of that log?\[\large\rm \ln\left[x(3x-2)\right]\]Like that?

Answer 2

i got the outside function which is \[\frac{ 1 }{ x(3x-2) }\]

Answer 3

yes

Answer 4

but i dont know if i have to treat x as g(x) and (3x-2) as h(x) in the chain rule. idk i got stuck there, this is part of a bigger natural log

Answer 5

Ah ok good.
First step looks good.
We'll chain rule into product rule.

It might feel a little weird at first.\[\large\rm \frac{d}{dx}\ln\left[x(3x-2)\right]\quad=\quad \frac{1}{\left[x(3x-2)\right]}\left[x(3x-2)\right]'\]

Answer 6

oh i see, alright

Answer 7

\[\large\rm \frac{d}{dx}\ln\left[x(3x-2)\right]\quad=\quad \frac{1}{\left[x(3x-2)\right]}\left[x'(3x-2)+x(3x-2)'\right]\]

Answer 8

I would simplify the (x(3x-2)) So you dont have to use the product rule

Answer 9

Saves you an extra step

Answer 10

oh thats an awsome idea tehHota. hadnt think of that

Answer 11

Yes, simplifying the inside of the log before differentiation is a good option.
Or expanding the expression into two separate log using log rule is another option as well.

Answer 12

how would you separate it? x and (3x-2) right?

Answer 13

Sometimes you cant simplify the inside so its always good practice to do it the long way

Answer 14

yeah i know

Answer 15

Yes

Answer 16

x and (3x-2)

Answer 17

ok, lets do it without simplifying for practice. got a final in a week

Answer 18

with the product rule

Answer 19

Oh boy :D fun stuff

Answer 20

I prefer anti-derivatives

Answer 21

yeah lol fun stuff and this is only part of a BIGGER problem lol. trying to logarithmic differentiation on \[\frac{ x^2 \sqrt{3x-2}}{ (x+1)^2 }\]

Answer 22

thats a nasdty problem lol

Answer 23

ok so for the product rule i got this

Answer 24

\[\frac{ 1 }{ x(3x-2) } * (6x-2)\]

Answer 25

please check it, im not sure if its right or not

Answer 26

\[\large\rm \frac{d}{dx}\ln\left[x(3x-2)\right]\quad=\quad \frac{1}{\left[x(3x-2)\right]}\left[x'(3x-2)+x(3x-2)'\right]\]\[\large\rm \frac{d}{dx}\ln\left[x(3x-2)\right]\quad=\quad \frac{1}{\left[x(3x-2)\right]}\left[1(3x-2)+x(3)\right]\]\[\large\rm \frac{d}{dx}\ln\left[x(3x-2)\right]\quad=\quad \frac{1}{\left[x(3x-2)\right]}\left[6x-2\right]\]Mmm ya looks correct.

Answer 27

Confirmed

Answer 28

thank you!! this is my last math before i transfer schools. not a big fan of calculus

Answer 29

It can get confusing

Answer 30

I hated the chain rule

Answer 31

WHUT Calculus is the best!! :D

Ya chain rule is definitely the hardest of the simple differentiation rules.
Takes some practice.

Answer 32

its just hard to see the different functions inside the chain rule. its hard when you have to combine it with product rule and others

Answer 33

one last question. i could simplify the last answer like this right?
\[\frac{ 6x-2 }{ x(3x-2) }\]

Answer 34

Technically even with simple \[5x^2\] equations you still use the chain rule, but it only multiplies by 1 so we leave that part out

Answer 35

Yes.
You could even factor a 2 out of the terms in the numerator,
but that's unnecessary.

The way you wrote it is preferable.

Answer 36

The only good part of calculus, leaving things expanded.

Answer 37

alright, thanks

Answer 38

yeah my professor dont mind if i dont simply to the end

Answer 39

Another way is to make use of log properties first before differentiating
i.e \[\ln [x(3x-2)]=\ln x+\ln (3x-2)\]
now we launch into derivative
\[\frac{d}{dx}[\ln(x(3x-2))]=\frac{d}{dx}[\ln x]+\frac{d}{dx}[\ln (3x-2) ]\]

Answer 40

just an alternative if you like