What is the derivative of (sinx+cosx)secx?

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1. As the title states, I need to fine d/dx(sinx+cosx)secx.

2. I am given pretty much most of the derivatives and trig functions.

3. Here's my attempt to solve the question:
y=(sinx+cos)secx
y'=(sinx+cosx)(secxtanx)+(secx)(cosx-sinx)
=sinxsecxtanx+cosxsecxtanx+cosxsecx-sinxsecx
=secx(sinxtanx+cosxtanx+cosx-sinx)
and that's where I've gotten up to.


So now I'm curious if my last step is the final answer, or can I simplify it more. I know some can be switched around but I'm not sure if that would help make the answer. Please advice and thanks for the help in advance!
 
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Try turning the secants and tangents into sines and cosines before any re-arranging (after taking the derivative though) and see if you get something that you consider 'nicer'. I think most people would accept

y'=(sinx+cosx)(secxtanx)+(secx)(cosx-sinx)

As an answer, but obviously I don't know the criteria on which you're graded
 
That simplifies quite a bit.

Hint/sanity check: Simplify (sinx+cosx)secx, then differentiate.
 
Ok, after reading DH's reply, I tried the question again and here's what I did.
y=(sinx+cosx)1/cosx
y'=(sinx+cosx)(-1/sinx)+(1/cosx)(-sinx+cosx)
=cosx/-sinx + sinx/-sinx + cosx/cosx - sinx/cosx
=cosx/-sinx - sinx/cosx

does that seem simple enough for you guys? I wish I were given the answers to check with.
 
Try simplifying (sinx+cosx)secx again. Hint: What is cos(x)*sec(x)?
 
D H said:
Try simplifying (sinx+cosx)secx again. Hint: What is cos(x)*sec(x)?

Oh, I see where you're going...
(sinx)(secx)+1 <----since cosx*secx = cosx*1/cosx
(sinx)(1/cosx)+1
[[(cosx)(cosx)-(sinx)(-sinx)]/(cosx)^2]+1
[(cos2x+sin2x)/cos2x]+1
(1/cos2x)+1?
is the +1 supposed to stay?
 
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