设 f（0）=1 ,f'(0)= -1 求------------------ X→0时,( cosX - f(x) )

设f(x)具有一阶连续导数,f(0)=0,f'(0)=2,求了lim(x→0)f(1-cosx)/tan(x^2) 设函数f(x)=sinx-cosx+x+1,0 设f(x)导数在【-1,1】上连续,且f(0)=1,计算∫【f（cosx）cosx-f‘（cosx）sin^2x】dx（ 设函数 F(x)在x=0处可导 又F(0)=0,求lim(x→0) F(1-cosx)/tan(x²) 设f(x)=(1+cosx)^(x+1)*sin(x^2-3x),求f(0)的导数等于多少 设f=[sin(2/x)]=1+cosx,求f(x),f[cos(2/x)]. 设F（x）为f(x)的原函数,当x≥0时,有f(x)F(x)=(sin2x)^2,且F(0)=1,F(x)≥0,求f(x 设函数f(x）{xe^(x^2),x>=0 {1/cosx ,-π 设f（sin x/2）=cosx+1,求f（x）及f（cos x/2）. ①设f(x)=x+2∫(0,1)f(t)dt,求f(x). 设函数f(x)=1+sin2x,求f'(0) f(x) = x - ∫(0~π) f(x) * cosx dx f'(x) = 1