Question

another question about the absolute value derivatives:

Answer 1

In the posts before, I have already come to an understanding that:
\[\frac{d}{dx}~\left| f(x) \right|~=~\frac{f(x) \cdot f'(x)}{\left| f(x) \right|}\]

So, now, my problem is
\[\frac{d}{dx} |\ln(x)|\]
\[=\frac{\ln(x)}{x|\ln(x)|}\]

Answer 2

is that correct?

Answer 3

looks good

Answer 4

okay, tnx for verifying...

Answer 5

\[\frac{d}{dx} |\ln(x)|=\frac{\ln(x)}{|\ln(x)|} \cdot (\ln(x))'=\frac{\ln(x)}{|\ln(x)|} \cdot \frac{1}{x}\]

Answer 6

I will actually go ahead and find the second derivative, that was a bit too easy even for me

Answer 7

I like to write it as \[\frac{d}{dx} |f(x)|=\frac{f(x)}{|f(x)|} \cdot f'(x)\]
so I don't get confused by all the funny fractions

Answer 8

I actually see it easier to write as I did, with f'(x), on the top but...

Answer 9

\[\frac{d}{dx} \frac{\ln(x)}{x|\ln(x)|}=\frac{d}{dx} ~~~\ln(x) ~~[~x|\ln(x)|~]^{-1}=\]
\[\frac{1}{x^2|\ln(x)|} \cdot \ln(x) \cdot (-x|\ln(x)|)^{-2} \cdot \left(\begin{matrix} \frac{x \ln(x)}{x|\ln(x)| }+|\ln(x)| \\ \end{matrix}\right)\]

Answer 10

then I will reduce this, to:

\[\frac{1}{x^2|\ln(x)|} \cdot \ln(x) \cdot (-x|\ln(x)|)^{-2} \cdot \left(\begin{matrix} \frac{ \ln(x)}{|\ln(x)| }+|\ln(x)| \\ \end{matrix}\right)\]

Answer 11

\[\frac{1}{x^2|\ln(x)|} \cdot \ln(x) \cdot (-x|\ln(x)|)^{-2} \cdot \left(\begin{matrix} \frac{ \ln(x)}{|\ln(x)| }+\frac{|\ln(x)|^2}{|\ln(x)| }\\ \end{matrix}\right)\]

\[\frac{1}{x^2|\ln(x)|} \cdot \ln(x) \cdot (-x|\ln(x)|)^{-2} \cdot \left(\begin{matrix} \frac{|\ln(x)|^2+ \ln(x)}{|\ln(x)|}\\ \end{matrix}\right)\]

Answer 12

the negative (in -x|ln(x)| ) is not squared, just wrote it like this

Answer 13

\[-\frac{1}{x^2|\ln(x)|} \cdot \ln(x) \cdot \left(\begin{matrix} \frac{|\ln(x)|^2+ \ln(x)}{x^2|\ln(x)|^3}\\ \end{matrix}\right)\]\[-\ln(x) \cdot \left(\begin{matrix} \frac{|\ln(x)|^2+ \ln(x)}{x^4|\ln(x)|^4}\\ \end{matrix}\right)\]

Answer 14

\[- \left(\begin{matrix} \frac{\ln(x) \times (|\ln(x)|^2+ \ln(x))}{x^4|\ln(x)|^4}\\ \end{matrix}\right)\]

Answer 15

I can re write it, but I will wait for a bit/./

Answer 16

I got to go