Question

高数求微分

y＝ln²（1-x）

Answer 1

这是个复合函数，
令y=u²，u=lnt，t=1-x
dy/dx=dy/du · du/dt · dt/dx
=2u · (1/t) · (-1)
=-2ln(1-x)/(1-x)
=2ln(1-x)/(x-1)
故dy=2ln(1-x)/(x-1) dx

Answer 2

y'=[ln²(1-x)]'
=2ln(1-x)*[ln(1-x)]'
=2ln(1-x)*[1/(1-x)]*(1-x)'
=2[ln(1-x)]/(x-1)
所以dy=2[ln(1-x)]/(x-1)*dx