queenfatso
Badges: 3
Rep:
?
#1
Report Thread starter 6 years ago
#1
Given that y=secx+tanx, find dy/dx and show that it can be written in the form 1/1-sinx

My working out:

dy/dx = secxtanx+sec2x

then I converted them
= 1/cosx (√ sec2x -1 ) + 1/cos2x

Am I on the right track? If not please help mee!
0
reply
Blazy
Badges: 8
Rep:
?
#2
Report 6 years ago
#2
(Original post by queenfatso)
Given that y=secx+tanx, find dy/dx and show that it can be written in the form 1/1-sinx

My working out:

dy/dx = secxtanx+sec2x

then I converted them
= 1/cosx (√ sec2x -1 ) + 1/cos2x

Am I on the right track? If not please help mee!
Write the whole thing as one fraction:  \displaystyle \frac{dy}{dx} = \sec(x)\tan(x) + \sec^2(x) = \frac{\sin(x)+1}{\cos^2(x)} . Rewrite the denominator in terms of a familiar identity and then think about difference of two squares.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (173)
14.62%
I'm not sure (54)
4.56%
No, I'm going to stick it out for now (345)
29.16%
I have already dropped out (35)
2.96%
I'm not a current university student (576)
48.69%

Watched Threads

View All