Qual é a derivada de $y = arcsec (x)$ $y = "arcsec" (x)$ ?

$\frac{d y}{d x} = \frac{1}{x^{2} \cdot \sqrt{1 - {(\frac{1}{x})}^{2}}}$ $dy/dx=1/[x^2*sqrt(1-(1/x)^2)]$

mostram que

$y = a r c sec x = \frac{1}{arccos x} = arccos (\frac{1}{x})$ $y=arcsecx=1/[arccosx]=arccos(1/x)$

$d/dx[arccosu]=1/sqrt(1-u^2)*u'$

$dy/dx=-1/[sqrt(1-(1/x)^2)]*[-1/x^2]$

$dy/dx=1/[x^2*sqrt(1-(1/x)^2)]$

Qual é a derivada de y = "arcsec" (x) y=arcsec(x) y = "arcsec" (x) ?