- #1

elitespart

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- Thread starter elitespart
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- #1

elitespart

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- #2

rootX

- 465

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for negative x?

using just log |x-3| defines the differentiation function for pos x only

using just log |x-3| defines the differentiation function for pos x only

- #3

exk

- 119

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- #4

elitespart

- 95

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Alright so I'm guessing there was a typo. Thanks for your help.

- #5

rootX

- 465

- 4

d/dx (log |x-3|) != 1/(x-3)

for x element of (-inf, inf) for sure! edit: wrong

ooops.. my mistake. I was ignoring the absolute part

for x element of (-inf, inf) for sure! edit: wrong

ooops.. my mistake. I was ignoring the absolute part

Last edited:

- #6

rock.freak667

Homework Helper

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d/dx (log |x-3|) != 1/(x-3)

for x element of (-inf, inf) for sure!

For [itex]x \neq 3[/itex]

- #7

exk

- 119

- 0

the original function is undefined for x=3 also btw.

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