Question

find the derviative of 4-ln(x)/2+ln(x) and its domain.

Answer 1

this is
\[f(x)=\frac{4-\ln(x)}{2+\ln(x)}\] right?

Answer 2

yes

Answer 3

so far i have \[2+\ln(x)\times0-1 \div(x)-4-\ln(x)\times1\div(x) all divide by (2+(\ln(x))^{2}\]

Answer 4

use "quotient rule"

Answer 5

yes i think you have it. denominator is
\[(2+\ln(x))^2\]

Answer 6

i did I just need help on simplifying

Answer 7

numerator is
\[(2+\ln(x))\times \frac{1}{x}-(4-\ln(x))\times \frac{1}{x}\]

Answer 8

and the domain

Answer 9

multiply out in the numerator and you get
\[\frac{2+\ln(x)-4+\ln(x)}{x}\]

Answer 10

oh damn i messed up sorry

Answer 11

it should be numerator of
\[(2+\ln(x))\times -\frac{1}{x}-(4+\ln(x))\times \frac{1}{x}\]

Answer 12

now multiply out and get
\[\frac{-2-\ln(x)-4+\ln(x)}{x}\]
\[=\frac{-6}{x}\]

Answer 13

so your "final answer" is
\[-\frac{6}{x(2+\ln(x))^2}\]

Answer 14

okay i got it now could you explain the domain

Answer 15

answer is right, but i made yet another typo above
should be
\[2+\ln(x))\times -\frac{1}{x}-(4-\ln(x))\times \frac{1}{x}\]

Answer 16

as for the domain you are not allowed to divide by 0

Answer 17

and you are not allowed to take the log of a negative number

Answer 18

so you have to make sure that
\[\ln(x)+2\neq 0\]
\[\ln(x)\neq -2\]
\[x\neq e^{-2}\]

Answer 19

therefore domain is
\[(0,e^{-2})\cup (e^{-2},\infty)\] which is a long winded way of saying that x must be positive and not equal to e^-2

Answer 20

okay thank you so much!

Answer 21

yw

Answer 22

could you also help me with f(x)=[ ln(x) ]^2 / x^2