f=f'sinx+cosx
f'(x)=f'(π/2)cosx-sinx
当x=π/2时,
f'(π/2)=f'(π/2)cosπ/2-sinπ/2=-1
f'(x)=-cosx-sinx
∴f'(π/4)=-cosπ/4-sinπ/4
=-√2/2-√2/2=-√2