If `f(x)=|x|^(|sinx|)`, then `f'((pi)/(4))` equals

Options

A. `((pi)/(4))^(1//sqrt(2))((sqrt(2))/(2)"ln"(4)/(pi)-(2sqrt(2))/(pi))`
B. `((pi)/(4))^(1//sqrt(2))((sqrt(2))/(2)"ln"(4)/(pi)+(2sqrt(2))/(pi))`
C. `((pi)/(4))^(1//sqrt(2))((sqrt(2))/(2)"ln"(pi)/(4)-(2sqrt(2))/(pi))`
D. `((pi)/(4))^(1//sqrt(2))((sqrt(2))/(2)"ln"(pi)/(4)+(2sqrt(2))/(pi))`

Answer

A

Explanation

In the neighbourhood of `-pi//4`, we have
`f(x)=(-x)^(-sinx)=e^(-sinxlog(-x))`
`implies" "f'(x)-e^(-sinxlog(-x))(-cosx.log(-x)-(sinx)/(x))`
`implies" "f'(x)=(-x)^(-sinx)(-cosx.log(-x)-(sinx)/(x))`
`implies" "f'(-(pi)/(4))=((pi)/(4))^(1//sqrt(2))((-1)/(sqrt(2))"log"(pi)/(4)+(4)/(pi)xx(-1)/(sqrt(2)))`
`implies" "f'(-(pi)/(4))=((pi)/(4))^(1//sqrt(2))((sqrt(2))/(2)"log"(4)/(pi)-(2sqrt(2))/(pi))`