Question

求函数y=ln(arctantx+1/x-1)的导数

Answer 1

y =ln(arctan[(x+1)/(x-1)] )

y' = (1/arctan[(x+1)/(x-1)]) .d/dx (arctan[(x+1)/(x-1)])

=(1/arctan[(x+1)/(x-1)]) .1/{ 1+[(x+1)/(x-1)]^2 } .d/dx [(x+1)/(x-1)]

=(1/arctan[(x+1)/(x-1)]) .1/{ 1+[(x+1)/(x-1)]^2 } .[-2/(x-1)^2]

=(1/arctan[(x+1)/(x-1)]) .(x-1)^2/(2(x^2+1)) .[-2/(x-1)^2]

=-1/ { (x^2+1).artan[(x+1)/(x-1)] }