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1st year calc, simple trig derivative

  1. May 25, 2006 #1
    Hello!
    This is probably real simple, just Im confused the question is Differentiate the following
    y=sin^2 x - cos^2 x
    I know you need to use just f'(x)+g'(x)
    and I know the derivate of sinx is cosx and derivative of cosx is -sinx.... but what do you do with the sin^2.... is the derivate of sin^2x just cos^2x??? and derivate of cos^2x just -sin^2x? THANKS
     
  2. jcsd
  3. May 25, 2006 #2

    berkeman

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    Staff: Mentor

    An easy way to think about it is cos^2(x) = cos(x) * cos(x), and use the chain rule (first times the derivative of the second, plus the second times the derivitave of the first)...
     
  4. May 26, 2006 #3
    You could make it even easier by substituting in:
    sin^2 x = 1 - cos^2 x
    Then you get:
    y = 1 - cos^2 x - cos^2 x = 1 - 2cos^2 x
    And then you only have to use the chain rule (or product rule) once. :)
     
  5. May 26, 2006 #4
    Unless I'm mistaken, [itex] cos^2 x - sin^2 x = cos 2x [/itex], so can you use that?
     
    Last edited by a moderator: May 26, 2006
  6. May 26, 2006 #5

    HallsofIvy

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    Yes, it can be done that way, but that's not the "chain rule", it is the "product rule".

    To use the chain rule, let u(x)= cos(x) and y(u)= u2. Then dy/dx= (dy/du)(du/dx). That will give the derivative of cos2(x).
    Also, of course, if you let u(x)= sin(x) and y(u)= u2 you can do exactly the same thing to get the derivative of sin2(x).

    I assume that you know that d(f(x)+ g(x))/dx= df/dx+ dg/dx.

    Looks like there are 3 different ways to do this!
     
  7. May 26, 2006 #6

    berkeman

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    Oopsies, thanks for the correction. :blushing:
     
  8. May 27, 2006 #7
    hello!
    Thanks alot for the help... I decided to use the product rule since I am most familiar with that.
    So once I did that I ended up with
    (sinx)(cosx)+(sinx)(cosx) and (cosx)(-sinx)+(cosx)(-sinx)
    so you add them together for f'(x)=h'(x)+g'(x)
    now i ALWAYS mess these big things up... so heres what i did:
    ((sinx)(cosx)+(sinx)(cosx)) + ((cosx)(-sinx)+(cosx)(-sinx))
    =2(sinxcosx)+2(cosx(-sinx))
    =2sinx+2cosx+2cosx-2sinx
    =0sinx+4cosx
    =4cosx
    Now that doesnt seem right can someone help me with what I am doing wrong? Or was the answer just 2(sinxcosx)+2(cosx(-sinx))... and I took it to far??? THANKS!
     
  9. May 27, 2006 #8

    VietDao29

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    Homework Helper

    There are some minor problems in your work.
    The plus sign is wrong, you can take a look at the problem again, it's
    y = sin2x - cos2x :)
    How can you go from the former expression to latter one? Are you trying to say that
    sin(x)cos(x) = sin(x) + cos(x)?
    ------------
    Apart from that, everything looks perfect. :)
     
  10. May 27, 2006 #9
    Ok thnx :)
    I see it was a stupid mistake...
    I went from
    =2(sinxcosx)+2(cosx(-sinx)) to
    =2sinx+2cosx+2cosx-2sinx Because i multiplied them by the 2 infront of the brackets.. guess thats wrong! haha but thanks. I always get confused when you do certain things and dont. Thanks for all your help!
     
  11. May 28, 2006 #10

    HallsofIvy

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    I don't mean to put you down but you really need to review your algebra. My experience is that people who cannot do algebra easily have a very hard time with calculus.

    Surely you cannot believe that
    2(sin x cos x)= 2sin x+ 2cos x!

    Because you were differentiating sin2x- cos2x you should have 2sin x cos x- (-2sin x cosx)= 2sin x cos x+ 2sin x cos x.

    Remember that finchie-88 pointed out that cos2x- sin2x= cos(2x)? Take the derivative of cos(2x) and compare it with the above.
     
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