f(x)=f'(兀/4)cosx+sinx
f'(x)=f'(兀/4)*(-sinx)+cosx
=-f'(兀/4)sinx+cosx
f'(兀/4)=-f'(兀/4)*√2/2+√2/2
f'(兀/4)+f'(兀/4)*√2/2=√2/2
f'(兀/4)(1+√2/2)=√2/2
f'(兀/4)=√2/2 /(1+√2/2)
=√2/(2+√2)
=√2(2-√2)/(2+√2)(2-√2)
=(2√2-2)/2
=√2-1
f(x)=(√2-1)cosx+sinx请问为什么f'(π/4)是常数