Répondre :
y=ln[sec(x)+tan(x)]
=ln(1/cos(x)+cos(x)/sin(x))
=ln((sin(x)+cos²(x)))(sin(x)cos(x))
alors y'=sec(x)=1/cos(x)
=ln(1/cos(x)+cos(x)/sin(x))
=ln((sin(x)+cos²(x)))(sin(x)cos(x))
alors y'=sec(x)=1/cos(x)