Question

求y=[ln(1-x)^2]^2的微分

Answer 1

y=[ln(1-x)^2]^2

y'= 2[ln(1-x)^2] * [ln(1-x)^2]'

= 2[ln(1-x)^2] * [2ln(1-x)]'

= 2[ln(1-x)^2] * 2*1/(1-x)

= 4*[ln(1-x)^2]/(1-x)