Verified answer

y = arctan u(x)

y' = [ 1 / (1+u²(x) )] * u'(x)

entonces

y' = [ 1 / (1+(1/x)² )] * [ 1/x] '

y' = [ 1 / (1+(1/x)² )] * [ -1/ x²]

y' = [ 1 / ((x²+1)/x² )] * [ -1/ x²]

y' = [ x² / (x²+1)] * [ -1/ x²]

y' = [ - x² / x² (x²+1)]

y' = - 1 / (x²+1)

w = arc tan [G(z)]

dw / dz = {d[G(z)] / dz} * (1+[G(z)]`2}^-1

entonces:

f(x) = arc tan(1/x)

df(x) / dx = d(1/x) / dx * (1+1/x^2)^-1

f ' (x) = (-1/ x^2 ) x^2/(1+x^2)

f ' (x) = -1/(1+x^2)

Pos la derivada de arc tg (1/x) es:

y´= -1/(x²+1)

ni idea

saludos!