设f(x)连续,Y=∫0~X tf（x^2-t^2）dt 则dy/dx=?
2个回答
y = ∫[0,x] t f(x² - t²) dt
令u = x² - t²,du = -2t dt
当t = 0,u = x²；当t = x,u = 0
y = ∫[x²,0] t f(u) * du/(-2t)
= 1/2 ∫[0,x²] f(u) du
dy/dx = 1/2 [2x * f(x²) - 0]
= x f(x²)
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楼上的方法也不错.
令z² = x² - t²,2z dz = -2t dt => dt = -z/t dz
当t = 0,z = x；当t = x,z = 0
∫[0,x] t f(x² - t²) dt
= ∫[x,0] t f(z²) * (-z/t) dz
= ∫[x,0] -z f(z²) dz
= ∫[0,x] z f(z²) dz
dy/dx = x f(x²) - 0
= x f(x²)
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两个方法也行,但楼上的替换怎么又会涉及因变量y呢?
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