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

  • Thread starter devilazy
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In summary, the original function is y=(sinx+cosx)secx and after taking the derivative, y' simplifies to cosx/-sinx - sinx/cosx. However, by simplifying (sinx+cosx)secx again, we can get a simpler answer of (1/cos2x)+1. The +1 can be kept or removed depending on the criteria for the answer.
  • #1
devilazy
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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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  • #2
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
 
  • #3
That simplifies quite a bit.

Hint/sanity check: Simplify (sinx+cosx)secx, then differentiate.
 
  • #4
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.
 
  • #5
Try simplifying (sinx+cosx)secx again. Hint: What is cos(x)*sec(x)?
 
  • #6
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?
 

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

The derivative of (sinx+cosx)secx is (cosx-sinx)secx + (sinx+cosx)secx tanx.

How do you solve for the derivative of (sinx+cosx)secx?

To solve for the derivative of (sinx+cosx)secx, you can use the product rule and the chain rule. First, differentiate sinx+cosx to get cosx-sinx. Then, differentiate secx to get secx tanx. Finally, multiply both results together and add them to get the final derivative.

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

The purpose of finding the derivative of (sinx+cosx)secx is to determine the instantaneous rate of change of the function at any given point. This can help in understanding the behavior of the function and its relationship with other functions.

Can the derivative of (sinx+cosx)secx be simplified?

Yes, the derivative of (sinx+cosx)secx can be simplified by using trigonometric identities. For example, you can rewrite cosx-sinx as -sin(x+π/4) and secx tanx as sec^2x. This will result in the simplified derivative of -sin(x+π/4)sec^2x.

Is it possible to find the derivative of (sinx+cosx)secx using other methods?

Yes, it is possible to find the derivative of (sinx+cosx)secx using other methods such as the quotient rule or the power rule. However, the product rule and the chain rule are often the most efficient methods for this particular function.

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