Question

Differentiate f(x) = ln((e^(4 x)(x - 1))/(2 x + 1)

i got 4+(1)/(x-1)-(1)/(2x+1)
is that right?

Answer 1

\[\text{Use the chain rule, } \frac{d}{dx}\left(\log \left(\frac{e^{4 x} (x-1)}{2 x+1}\right)\right)=\frac{d\log (u)}{du} \frac{du}{dx} \text{, where } u=\frac{e^{4 x} (x-1)}{2 x+1} \text{ and } \frac{d\log (u)}{du}=\frac{1}{u} \]

Answer 2

lol

Answer 3

i cant see what you typed lol

Answer 4

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? :)

Answer 5

Do you mean
ln((e^(4 x)(x - 1)) / (2 x + 1)
or ln ((e^(4 x)(x - 1))/(2 x + 1) ?

Answer 6

Use chain and product rule

Answer 7

i mean the first one

Answer 8

ok i still get the same answer that i got

Answer 9

Is that your question (see the first line)

Answer 10

yes that's the correct question

Answer 11

I made mistakes in the calculation...

Answer 12

thats the right answer? if so, im way off..

Answer 13

Ask @.Sam.

Answer 14

\[\frac{-8 x^2+4 x+1}{-2 x^2+x+1}\]

Answer 15

thats the right answer?

Answer 16

yeah, wait ill type the working

Answer 17

\[\ln\frac{e^{4x}(x-1)}{2x+1}\]
\[\ln(e^{4x}(x-1))-\ln(2x+1)\]
differentiate
\[\frac{1}{e^{4x}(x-1)}e^{4x}+(x-1)(4)(e^{4x})-\frac{1}{2x+1}(2)\]
\[\frac{e^{4x}+(4x-4)(e^{4x})}{e^{4x}(x-1)}-\frac{2}{2x+1}\]
\[\frac{e^{4x}(1+(4x-4)}{e^{4x}(x-1)}-\frac{2}{2x+1}\]
\[\frac{4x-3}{x-1}-\frac{2}{2x+1}\]
\[\frac{-8x^{2}+4x+1}{-2x^{2}+x+1}\]

Answer 18

the question is 1 or 2?