Math Is Fun Forum

entony

Guest

differentiation

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

entony

Guest

Re: differentiation

i guess im right then? lol

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: differentiation

Well, I suppose you could simplify your answer a little further to 2sec[sup]2[/sup]xcos2x.