f'(x)=f'(丌/4)cosx+sinx则f(x)?

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f''(x)=f''(丌/4)cosx+sinx则f(x)?f''(x)=f''(丌/4)cosx+sinx则f(x)?f''(x)=f''(丌/4)cosx+sinx则f(x)?f''(π/4)是常数当x=π/4

f'(x)=f'(丌/4)cosx+sinx则f(x)?
f'(x)=f'(丌/4)cosx+sinx则f(x)?

f'(x)=f'(丌/4)cosx+sinx则f(x)?
f'(π/4)是常数
当x=π/4时
f'(π/4)=f'(π/4)cos(π/4)+sin(π/4)
f'(π/4)[1-cos(π/4)]=sin(π/4)
f'(π/4)=1/(√2-1)=√2+1
所以原式两边积分得
f(x)=(√2+1)sinx-cosx+C

f'(x)=-f'(兀/4)sinx+cosx
f'(兀/4)=-f'(兀/4)sin兀/4+cos兀/4
f'(兀/4)=√2-1
f(兀/4)=f'(兀/4)cos兀/4+sin兀/4
=(√2-1)*√2/2+√2/2=1