You are not logged in.
Pages: 1
Hey everyone!
I just need someone to check if i have done this right. Thank you
If y=-2sec^2xtanx find dy/dx:
using the product rule; u=-2sec^2x, v=tanx
u=-2sec^2x
u=-2a^2, a=secx
du/da=-4a
da/dx=secxtanx
du/dx=du/da.da/dx
du/dx=-4u.secxtanx = -4sec^2xtanx
v=tanx
dv/dx=sec^2x
dy/dx=u.dv/dx + v.du/dx
dy/dx=-2sec^2x.sec^2x + tanx.-4sec^2xtanx
dy/dx=-2sec^4x -4sec^2xtan^2x
i guess im right then? lol
Well, I suppose you could simplify your answer a little further to 2sec[sup]2[/sup]xcos2x.
Offline
Pages: 1