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发表于 2019-11-6 10:00:28
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很简单啊。解法如下:
因为f''(x)存在
即f'(x)也存在
所以f''(x)=lim(h→0) [f'(x+h)-f'(x)]/h
又因为f'(x)=lim(h→0) [f(x+h)-f(x)]/h
所以f''(x)=lim(h→0) [f(x+2h)-2f(x+h)+f(x)]/h^2
lim(h→0) [f(x+2h)-2f(x+h)+f(x)]/h^2
=lim(h→0){[f(x+2h)-f(x+h)]/h-[f(x+h)+f(x)]/h}/h
=lim(h→0)[f'(x+h)-f'(x)]/h=f''(x)
dy/dx=2x变形为dy=2xdx两边同时积分则∫dy=∫2xdx
y=x^2+c,在由条件x=1是y=2,确定c=1
则y=x^2+1
f(x) = ∫<0, x> (x-t)e^(-t^2)dt = ∫<0, x> xe^(-t^2)dt - ∫<0, x> te^(-t^2)dt
= x∫<0, x> e^(-t^2)dt - ∫<0, x> te^(-t^2)dt (对 t 积分,x相对于常量,可提到积分号外)
f'(x) = ∫<0, x> e^(-t^2)dt + xe^(-x^2) - xe^(-x^2) = ∫<0, x> e^(-t^2)dt
df(x) = f'(x)dx = [∫<0, x> e^(-t^2)dt] dx
2. dy/dx = y'<t>/x'<t> = 3t^2/(2t) = (3/2)t, t = 2 时, 切线斜率 k = (3/2)t = 3,
切点 (5,8), 切线方程 y-8 = 3(x-5), 即 3x-y-7 = 0
∫<0, 2π>|cosx|dx
= ∫<0, π/2>cosxdx + ∫<π/2, 3π/2> -cosxdx + ∫<π/2, 2π>cosxdx
= [sinx]<0, π/2> - [sinx]<π/2, 3π/2> + [sinx]<3π/2, 2π>
= 1 + 2 + 1 = 4
∫tcoswtdt=∫(t/w)dsinwt=1/w(tsinwt-∫sinwt dt)=1/w〔tsinwt+(coswt)/w〕
=1/w〔-1/w-1/w〕=-2/(w^2)
∫<-1, 1>(x^3+x+1)dx = ∫<-1, 1>(x^3+x)dx + ∫<-1, 1>dx
根据题意,应有lim(x->0)f(x)=f(0)=a
lim(x->0)f(x)=lim(x->0)[∫(x,0)sin2t/tdt]/x
=lim(x->0)(-sin2x/x)
=lim(x->0)(-2x/x)
=-2
又因为√(n^2+4n+5)-(n-1)
=n^2+4n+5)-(n-1)^2]/[√(n^2+4n+5)+(n-1
=(6n+6)/[√(n^2+4n+5)+(n-1)]
=(6+6/n)/[√(1+4/n+5/n^2)+(1-1/n)]
(6+0)/[√(1+0+0)+(1-0)]=5
f(a+x)-f(a-2x)]/h=[f(a+x)-f(a)+f(a)-f(a-2x)]/x
=[f(a+x)-f(a)]/x+2*[f(a)-f(a-2x)]/2x
再用L 'Hopital's Rule
3f'(a)=30
太简单了。
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