Differentiate `(logx)^x` with respect to `logx` .

Options

A. `(log x)^(x) [(1)/(logx) + log (logx)]`
B. `(log x)^(x) [logx+(1)/( log (logx))]`
C. `x(log x)^(x) [(1)/(logx)+ log (logx)]`
D. None of the above

Answer

C

Explanation

Let ` u = (log x)^(x)`
` rArr log u = x log (log x)`
`therefore (1)/(u) (du)/(dx) = x (1)/(log x).(1)/(x) + log (log x)`
` rArr (du)/(dx) = (log x)^(x) [(1)/(log x) + log(logx)]`
and ` v = log x rArr (dv)/(dx) = (1)/(x)`
`therefore (du)/(dv) = (log x)^(x) [ (1)/(log x) + log (log x)] xx x `
` = x (log x)^(x) [(1)/(logx) + log (log x)]`.