简单计算一下即可，答案如图所示
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第1个回答 2017-12-20
x=∫(0->t) f(u^2) du
y=[f(t^2)]^2
f''(u) exists
f(u)≠0 , y''(x) =?
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solution:
x=∫(0->t) f(u^2) du
dx/dt = f(t^2)
y=[f(t^2)]^2
dy/dt = 4t.f(t^2).f'(t^2)
y' =dy/dx = (dy/dt)/ (dx/dt )= 4t.f'(t^2)
dy'/dt = 4[f'(t^2) +2t^2.f''(t^2) ]
d^2y/dx^2
=(dy'/dt )/ (dx/dt)
=4[f'(t^2) +2t^2.f''(t^2) ] /f(t^2)本回答被网友采纳
y=[f(t^2)]^2
f''(u) exists
f(u)≠0 , y''(x) =?
-------------
solution:
x=∫(0->t) f(u^2) du
dx/dt = f(t^2)
y=[f(t^2)]^2
dy/dt = 4t.f(t^2).f'(t^2)
y' =dy/dx = (dy/dt)/ (dx/dt )= 4t.f'(t^2)
dy'/dt = 4[f'(t^2) +2t^2.f''(t^2) ]
d^2y/dx^2
=(dy'/dt )/ (dx/dt)
=4[f'(t^2) +2t^2.f''(t^2) ] /f(t^2)本回答被网友采纳
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