设f(x)导数在x=2时连续

lim(x-->0)[xf(x)+x+ln(1+x)-x]/x^2=3/2==>lim(x-->0)[f(x)+1]/x+lim(x-->0)[ln(1+x)-x]/x^2=3/2==>lim(x--

f(0)=f(x)+f'(x)(0-x)+0.5f''(a)(0-x)^2f(1)=f(x)+f'(x)(1-x)+0.5f''(b)(1-x)^2两式相减,移项,取绝对值得|f'(x)|=|f(1)

用分部积分法.∫^(0,1)x(1-x)f"(x)dx(u=x(1-x)v'=f''(x)u'=1-2xv=f'(x)=[x(1-x)f'(x)](0,1)-∫^(0,1)(1-2x)f'(x)dx再

∫(0,π/2)[f(cosx)cosx-f'(cosx)sin^2x]dx=∫(0,π/2)d[sinxf(cosx)]=sinxf(cosx)|(0,π/2)=1*f(0)-0*f(1)=f(0)

嗯··用洛比塔法则求解,对式子上下同时求导可得在x趋向2时limf'(x)/1=2也就是说f'(2)=2咯哈哈,不晓得方法对不,我数学一般仅供参考

分子上第1个负号应为正号,否则极限不存在

1=lim(x→0)F(x)所以lim(x→0)f(x)=01=lim(x→0)F(x)=lim(x→0)f(x)/x+lim(x→0)3ln(1+x)/x=lim(x→0)(f(x)-f(0))/(

tanx-sinx=tanx(1-cosx)=1/2x^3,f(x)=f(0)+f'(0)x+(1/2)f''(x)^2+1/6f'''(x)x^3+o(x^3),f'''(x)=3

如下图不知能看明白否

令F(x)=(积分(从0到x)f(t)dt)^2-积分(从0到x)f(t)^2dt,00,g(x)严格递增.故g(x)>g(0)=0,于是F'(x)=f(x)*g(x)>0.故F(x)递增,故F(1)

用分部积分法.∫^(0,1)x(1-x)f"(x)dx(u=x(1-x)v'=f''(x)u'=1-2xv=f'(x)=[x(1-x)f'(x)](0,1)-∫^(0,1)(1-2x)f'(x)dx再

用分部积分法.∫^(0,1)x(1-x)f"(x)dx(u=x(1-x)v'=f''(x)u'=1-2xv=f'(x)=[x(1-x)f'(x)](0,1)-∫^(0,1)(1-2x)f'(x)dx再

Taylor展式：对任意的x,f(0)＝f(x)+f'(x)(0－x)+f''(c1)(0－x)^2/2,f(1)＝f(x)+f'(x)(1－x)+f''(c2)(1－x)^2/2.两式相减,得f'(

应该是证g(x)在R上有一阶连续导数吧?当x≠0时,g(x)=f(x)/x∴g'(x)=[xf'(x)-f(x)]/x²g'(x)在x≠0时连续x=0时,g'(0)=lim(x→0)[g(x

∵对任意的x,f(0)＝f(x)+f'(x)(0－x)f(1)＝f(x)+f'(x)(1－x)两式相加得∴2f(x)＝（2x-1)f'(x)即f(x)=(x-1/2)f'(x)且0≤x≤1∴l∫f(x

关键在于将y=2x在求导中按复合函数来处理,首先在f(x,2x)=x两边对x求导数,根据复合函数求导法则,有f'x+f'y*(2x)'=1,即f'x+2f'y=1,由于f'x=x^2,所以f'y=(1

x→0时,1/2√x→∞.要把sin√x与1/√x合在一起讨论,这是个等价无穷小再问：为什么趋于无穷啊？不好意思我高数刚学很多不明白，能解释详细点吗谢谢再答：分子是1，分母趋向于0，分式不就是趋向于∞

lim(x→0+)(d/dx)f(cos√x)　　=lim(x→0+)f'(cos√x)*(-sin√x)*[1/(2√x)]　　=(-1/2)*lim(x→0+)f'(cos√x)*lim(x→0+

d[f(cos√(x-1))]/dx=f'(x)*(-sin√(x-1))*1/2*1/√(x-1)=-1/2*f'(x)*sin√(x-1)/√(x-1)limx->1+时的d[f(cos√(x-1