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If f(x)=|log(e)|x||," then "f'(x) equals...

If `f(x)=|log_(e)|x||," then "f'(x)` equals

A

`(1)/(|x|),xne0`

B

`(1)/(x)" for "|x|gt1" and "-(1)/(x)" for "|x|lt1`

C

`-(1)/(x)" for "|x|gt1" and "(1)/(x)" for "|x|lt1`

D

`(1)/(x)" for "xgt0" and "-(1)/(x)" for "xlt0`

Text Solution

Verified by Experts

The correct Answer is:
B
For `xgt1`, we have
`f(x)=|log|x||=logximpliesf'(x)=(1)/(x)`
For `xlt-1`, we have
`f(x)=|log|x||=log(-x)impliesf'(x)=(1)/(x)`
For `0ltxlt1`, we have
`f(x)=|log|x||=-logximpliesf'(x)=-(1)/(x)`
For`-1ltxlt0`, we have
`f(x)=-log(-x)" "impliesf'(x)=-(1)/(x)`
Hence, `f'(x)={{:(" "(1)/(x)",",|x|gt1),(-(1)/(x)",",|x|lt1):}`
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