anonymous
  • anonymous
Differentiate f(x) = ln((e^(4 x)(x - 1))/(2 x + 1) i got 4+(1)/(x-1)-(1)/(2x+1) is that right?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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.Sam.
  • .Sam.
\[\text{Use the chain rule, } \frac{d}{dx}\left(\log \left(\frac{e^{4 x} (x-1)}{2 x+1}\right)\right)=\frac{d\log (u)}{du} \frac{du}{dx} \text{, where } u=\frac{e^{4 x} (x-1)}{2 x+1} \text{ and } \frac{d\log (u)}{du}=\frac{1}{u} \]
.Sam.
  • .Sam.
lol
anonymous
  • anonymous
i cant see what you typed lol

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.Sam.
  • .Sam.
answer is \[\frac{e^{-4 x} \left(\left(4 e^{4 x} (x-1)+e^{4 x}\right) (2 x+1)-2 e^{4 x} (x-1)\right)}{(x-1) (2 x+1)}\] try to figure out? :)
Callisto
  • Callisto
Do you mean ln((e^(4 x)(x - 1)) / (2 x + 1) or ln ((e^(4 x)(x - 1))/(2 x + 1) ?
.Sam.
  • .Sam.
Use chain and product rule
anonymous
  • anonymous
i mean the first one
anonymous
  • anonymous
ok i still get the same answer that i got
Callisto
  • Callisto
Is that your question (see the first line)
1 Attachment
anonymous
  • anonymous
yes that's the correct question
Callisto
  • Callisto
I made mistakes in the calculation...
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anonymous
  • anonymous
thats the right answer? if so, im way off..
Callisto
  • Callisto
Ask @.Sam.
.Sam.
  • .Sam.
\[\frac{-8 x^2+4 x+1}{-2 x^2+x+1}\]
anonymous
  • anonymous
thats the right answer?
.Sam.
  • .Sam.
yeah, wait ill type the working
.Sam.
  • .Sam.
\[\ln\frac{e^{4x}(x-1)}{2x+1}\] \[\ln(e^{4x}(x-1))-\ln(2x+1)\] differentiate \[\frac{1}{e^{4x}(x-1)}e^{4x}+(x-1)(4)(e^{4x})-\frac{1}{2x+1}(2)\] \[\frac{e^{4x}+(4x-4)(e^{4x})}{e^{4x}(x-1)}-\frac{2}{2x+1}\] \[\frac{e^{4x}(1+(4x-4)}{e^{4x}(x-1)}-\frac{2}{2x+1}\] \[\frac{4x-3}{x-1}-\frac{2}{2x+1}\] \[\frac{-8x^{2}+4x+1}{-2x^{2}+x+1}\]
Callisto
  • Callisto
|dw:1332776196866:dw| the question is 1 or 2?
anonymous
  • anonymous
2
.Sam.
  • .Sam.
my one's for (2)
Callisto
  • Callisto
Sam's answer, sorry
.Sam.
  • .Sam.
because your "ln" is on top
Callisto
  • Callisto
Yup, i'm mixing them up now but for \[\ln(e^{4x}(x-1))-\ln(2x+1)\] it can further be simplified to \[\ln(e^{4x}) +ln(x-1)-\ln(2x+1)\] \[=4x\ln(e) +ln(x-1)-\ln(2x+1)\]\[=4x+ln(x-1)-\ln(2x+1)\] That should be easier, sorry for all the mistakes i've made
anonymous
  • anonymous
|dw:1332776696290:dw|
anonymous
  • anonymous
|dw:1332776830440:dw|
anonymous
  • anonymous
do u guys see how i got that?
Callisto
  • Callisto
I think it is like this: |dw:1332777068552:dw| But ask .Sam.
anonymous
  • anonymous
ok youre right callisto! got it thanx
anonymous
  • anonymous
why was that so difficult? lol
.Sam.
  • .Sam.
@Callisto Same as mine
Callisto
  • Callisto
That's good then, finally, i got the answer right , thank you so much!

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