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

Options

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

Answer

C

Explanation

Let u = `(log x)^(x)`
`implies log u = x log (log x )`
`therefore (1)/(u) (du)/(dx) = x"" (1)/(log x) * (1)/(x) + log (log x )`
`implies (du)/(dx) = (log x)^(x) [ (1)/(log x) + log (log x)]`
and `v = log x `
`therefore (dv)/(dx) = (1)/(x)`
Now ,
`(du)/(dv) = (logx )^(x) [ (1)/(log x ) xx log (log x)] xx x`
= `x (log x )^(x) [ (1)/(log x) + log (log x) ]`