ft ln tan 1 tSolution ddtlogtan1t Use the chain rule ddtlo

f(t) = ln (tan^ -1 t)

d/dt(log(tan^(-1)(t))) | Use the chain rule, d/dt(log(tan^(-1)(t))) = ( dlog(u))/( du) ( du)/( dt), where u = tan^(-1)(t) and ( dlog(u))/( du) = 1/u: = | (d/dt(tan^(-1)(t)))/(tan^(-1)(t)) | The derivative of tan^(-1)(t) is 1/(t^2+1): = | (1/(t^2+1))/(tan^(-1)(t))