大一高数问题若y=f(t^2),且f''(x)存在,则dy\dt=,(d^2y)\dx^2d(2tf')/dt=2[f'
1个回答

dy/dt=(dy/dt^2)(dt^2/dt)=2tf'

(d^2y)dt^2=d(2tf')/dt=2[f'+t(df'/dt)]=2[f'+t(df'/dt^2)(dt^2/dt)]=2[f'+2(t^2)f"]

其中d(2tf')/dt=2[f'+t(df'/dt)]为d(2tf')/dt=2d(tf')/t,d(tf')/dt为先对t求导得1再乘f’,然后对f‘求导得df'/dt再乘t,相加为2[f'+t(df'/dt)]

相关问题