Qual é a derivada de # y = "arcsec" (x) #?
Responda:
#dy/dx=1/[x^2*sqrt(1-(1/x)^2)]#
Explicação:
mostram que
#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)]#
#dy/dx=1/[x^2*sqrt(1-(1/x)^2)]#
mostram que
#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)]#