两道高数问题,求解答,要过程
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d(y-x+1/3cosy)=0
dy-dx-1/3sinydy=0
dy/dx=3/(3-siny)
d²y/dx²=d[3/(3-siny)]/dx
=[3cosydy/(3-siny)²]/dx
=[3cosy/(3-siny)²]dy/dx
=[3cosy/(3-siny)²][3/(3-siny)]
=9cosy/(3-siny)³
dx/dt=f''(t) dy/dt=f'(t)+(t-1)f''(t)
dy/dx=[f'(t)+(t-1)f''(t)]/f''(t)
dy-dx-1/3sinydy=0
dy/dx=3/(3-siny)
d²y/dx²=d[3/(3-siny)]/dx
=[3cosydy/(3-siny)²]/dx
=[3cosy/(3-siny)²]dy/dx
=[3cosy/(3-siny)²][3/(3-siny)]
=9cosy/(3-siny)³
dx/dt=f''(t) dy/dt=f'(t)+(t-1)f''(t)
dy/dx=[f'(t)+(t-1)f''(t)]/f''(t)
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