If `f(x)=|x|^(|sinx|),` then find `f^(prime)(-pi/4)`

Explanation

In the neighbourhood of `-pi//4,` we have
`f(x)=(-x)^(-sin x)=e^(-sin x log(-x))`
`"or "f'(x)=_(e)^(- sin x log(-x))(-cos cdot log (-x) -(sin x)/(x))`
`=(-x)^(-sin x)(-cos x cdotlog(-x)-(sin x)/(x))`
`"or "f'(pi//4)=((pi)/(4))^(1//sqrt(2))((-1)/(sqrt(2))log""(pi)/(4)+(4)/(pi)xx((-1)/(sqrt(2)))`
`=((pi)/(4))^(1sqrt(2))((sqrt(2))/(2)log""(4)/(pi)-(2sqrt(2))/(pi))`