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Math Help - differentiation of simple trig fuction

  1. #1
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    differentiation of simple trig fuction

    hi guys, can anyone differentiate this for me, 2sinxcosx, i'm sure it's very simple, i know how to use the chain and product rules etc, but the 2 is confusing me, do i include that in both parts? please someone show me how they would differentiate this, thanks everyone
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  2. #2
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    uh oh

    uh oh, this one aswell please, similar problem, 2xcos(x^2), thanks again anyone
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by captainlewis View Post
    hi guys, can anyone differentiate this for me, 2sinxcosx, i'm sure it's very simple, i know how to use the chain and product rules etc, but the 2 is confusing me, do i include that in both parts? please someone show me how they would differentiate this, thanks everyone
    This is product rule. The 2 is just a coefficient of the function sinxcosx. When you take the derivative of sinxcosx, the 2 stays out in front.

    (2sinxcosx)'
    = 2(sinxcosx)'
    = 2[(sinx)'cosx + sinx(cosx)']
    = 2[cos^2(x) - sin^2(x)]
    = 2[cos(2x)] = 2cos(2x)

    [Thanks for pointing out my mistake, captainlewis ]
    Last edited by ecMathGeek; May 3rd 2007 at 08:56 AM.
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by captainlewis View Post
    uh oh, this one aswell please, similar problem, 2xcos(x^2), thanks again anyone
    This one involves product and chain rule.

    (2xcosx^2)'
    = 2(xcosx^2)'
    = 2[(x)'cosx^2 + x(cosx^2)'(x^2)']
    = 2[cosx^2 - xsinx^2(2x)]
    = 2[cosx^2 - 2x^2*sinx^2]
    = 2cosx^2 - 4x^2*sinx^2
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  5. #5
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    thanks

    thanks man, megamathman that was really helpful and i can do this work now, i think though cos differentiated is -sin? that's what i learnt.
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  6. #6
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by captainlewis View Post
    thanks man, megamathman that was really helpful and i can do this work now, i think though cos differentiated is -sin? that's what i learnt.
    You're right, and I did that too, but I didn't "show" it. Notice how the sign in front of the term where I differentiated cosine changed from + to -.

    Oops, I just notice that on the first problem I made that mistake. Sorry. I'll go back and fix it.
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  7. #7
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by captainlewis View Post
    hi guys, can anyone differentiate this for me, 2sinxcosx, i'm sure it's very simple, i know how to use the chain and product rules etc, but the 2 is confusing me, do i include that in both parts? please someone show me how they would differentiate this, thanks everyone
    Here's a quicker way to do this problem:

    Notice that 2sinxcosx = sin(2x) <-- double angle formula for sine.

    (sin(2x))' = sin'(2x)*(2x)' = cos(2x)*2 = 2cos(2x)
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