If `f(x)=(log)_(x^2)(logx),` then `f^(prime)(x)` at `x=e` is 0 (b) 1 (c) `1/e` (d) `1/2e`

Explanation

`"Given "f(x)=log_(x)(log x)=(log_(e)(log_(e)x))/((log_(e)x))`
`"or "f'(x)=((1)/(log_(e)x)(1)/(x)log_(e)x-(1)/(x)log_(e)(log_(e)x))/((log_(e)x)^(3))`
`=((1)/(x)[1-log_(e)(log_(e)e)])/((log_(e)x)^(2))`
`"At "x=e," we get"`
`f'(e)=((1)/(e)[1-log_(e)(log_(e)e)])/((log_(e)e)^(2))=((1)/(e)[1-log_(e)1])/((1)^(2))`
`=(1)/(e)(1-0)=(1)/(e)`