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x=f'(t),y=tf'(t)-f(t),f''(t)存在且不为0,求d2y/dx2.这里没有复合求导的问题吗?怎么就直接求导了。
请说详细一些,谢谢!我对这种东西不是很清楚。
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推荐答案 2012-10-26
当然是要复合求导的,
x=f'(t),y=tf'(t)-f(t)
所以
dx/dt=f "(t)
dy/dt=t *f "(t) +f '(t)- f '(t)=t *f "(t),
所以
dy/dx=(dy/dt)/(dx/dt)=t,
而
d²y/dx²
=d(dy/dx) /dx
= d(dy/dx) /dt * dt/dx
= dt/dt * dt/dx
=1/(dx/dt)
=1/f "(t)
所以y对x的二阶导数d²y/dx²=1/f "(t)
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其他回答
第1个回答 2012-10-23
参数方程求导:
y'(t)=f'(t)+tf''(t)-f'(t)=tf''(t)
x'(t)=f''(t)
dy/dx=y'(t)/x'(t)=t
y''(x)=[y'(t)/x'(t)]'(t)/x'(t)=1/f''(t)
第2个回答 2012-10-23
x=f'(t),y=tf'(t)-f(t),f''(t)存在且不为0,求d2y/dx2.这里没有复合求导的问题吗?怎么就直接求导了。 d2y/dx2=2dy/2dx=dy/dx, 因f''(t)存在且不为0,则 dy=tf''(t)+f'(t)-f'(t)=tf''(t), dx=f''(t); dx/dy=tf''(t)/f''(t)=t.
追问
是y对x的二阶导数。
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(t)-f(t),求d
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,d^
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/dt=f'(t)+tf''
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dx=(dy
/dt)/
(dx
/dt)=1/td^2y/dt^
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