If `y=x^(n)logx+x(logx)^(n)," then "(dy)/(dx)` is equal to

Options

A. `x^(n-1)(1+nlogx)+(logx)^(n+1)[n+logx]`
B. `x^(n-2)(1+nlogx)+(logx)^(n-1)[n+logx]`
C. `n^(n-1)(1+nlogx)+(logx)^(n-1)[n-logx]`
D. None of the above

Answer

A

Explanation

Given `y=x^(n) log x+(log x)^(n)`
`(dy)/(dx)=nx^(n-1) log x+x^(n)*(1)/(x)+xn (log x)^(n-1)((1)/(x))+1*( log x)^(n)`
`=x^(n-1)(1+n log x)+(log x)^(n-1)[ n+log x]`