If `f(x)=|cosx-sinx|`,find `f^(prime)(pi/6)a n df^(prime)(pi/3)dot`

Explanation

When `0 < x< frac{pi}{4}
,cosx>sinx`, so that `cosx−sinx>0`,i.e.,

`f(x)=cosx−sinx`

`
⇒f
′
(x)=−sinx−cosx`
...