高数证明d/dx(x∫(0~x)f(t)dt)=∫(0~x)f(t)dt+xf(x)

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微积分基本定理：d/dx ∫(a(x)→b(x)) ƒ(t) dt = b'(x)ƒ[b(x)] - a'(x)ƒ[a(x)]
导数乘尺锋手基掘法则：(uv)' = vu' + uv'

d/陵嫌dx [x∫(0→x) ƒ(t) dt]
= x' * ∫(0→x) ƒ(t) dt + x * [∫(0→x) ƒ(t) dt]'
= ∫(0→x) ƒ(t) dt + x * [x' * ƒ(x) - 0' * ƒ(0)]
= ∫(0→x) ƒ(t) dt + xƒ(x)

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这腊磨卜个一步就出游链答案了轮穗,
就是一个求导运算
把F(X)=x ,G(X)=∫(0~x)f(t)dt
F(X)' =1 ,G(X)'=f(X)
d/dx(F(X).G(X))=F(X)'.G(X)+F(X).G(X)'

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