Question

Differentiate the function.
f(x)= ln[(2x+1)^3/(3x-1)^4]

Answer 1

Oh wow, you have everything here. Quotient & Chain rule! (or product rule if you want to use that).

Answer 2

Try the chain rule:\[\frac{da}{db}=\frac{da}{dc} \times\frac{dc}{db}\]

Answer 3

How far did you get with this problem?

Answer 4

So I was trying to use the chain rule first but now I am thinking that you are supposed to start with the log and then move to the Quotient rule. Is this correct?

Answer 5

so could you next write it as ln[(3(2x+1)^2)/(4(3x-1)^3)

Answer 6

Is this the function? \[\huge f(x)= \ln\left(\frac{(2x+1)^3}{(3x-1)^4}\right)\]

Answer 7

Yes

Answer 8

Ok, so the key here is to use log rules to simplify the expression to the sum or difference of logs.

Answer 9

f(x)= ln[(2x+1)^3/(3x-1)^4]
you can write it as..
f(x) = ln(2x+1)^3 - ln(3x-1)^4
then apply derivative...

Answer 10

http://www.purplemath.com/modules/logrules.htm

Answer 11

@AJW99 got it??

Answer 12

Still trying to wrap my brain around it

Answer 13

First log rule you'll use here...
\[\huge \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\]

Answer 14

In this case

a = (2x+1)^3
and
b = (3x-1)^4

Answer 15

okay so then it would look like this correct? ln(2x+1)^3 - ln(3x-1)^4

Answer 16

so next would you use the rule d/dx [f(x)]^b=b[f(x)]^b-1 f'(x)?

Answer 17

I am stuck.

Answer 18

Can you still help me I have a really hard time with implicit differentiation

Answer 19

sure..

Answer 20

@AJW99 can you do it now... or i'll solve it till the end??

Answer 21

Thank you so much for all of your help! I really appreciate the help!

Answer 22

no problem... :)

Answer 23

There is a simpler way about it...

Once you have

\[\huge \ln((2x+1)^2)-\ln((3x-1)^4)\] you can use another log rule...

\[\huge \ln((a)^b)=b\cdot\ln(a)\]

In other words,

\[\huge f(x) = 2\cdot \ln(2x+1)-4\cdot \ln(3x-1)\]

Differentiating this is much easier, in my opinion. :)

Answer 24

@mathteacher1729 I think I have done the same way...

Answer 25

@Kashan Yes, our final answers agree. But you differentiated using the exponent rule and then in your final result did simplification to yield the same answer.

This solution (simplifying as much as possible using logs) avoids having to do any canceling at the end.

Both are correct, though. :)

Answer 26

Yes I agree.. thanks :)