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Math Help - differentiate trig

  1. #1
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    differentiate trig

    Differentiate

    xsinx

    and

    sinxcosx

    any help thanks
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  2. #2
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    Quote Originally Posted by gracey View Post
    Differentiate

    xsinx

    and

    sinxcosx

    any help thanks
    Do you know product rule?

    It says if f(x) and g(x) are two functions of x, then (f(x) g(x))' = (f(x))'g(x) + f(x)(g(x))', where (f(x))' stands for differentiationof f(x).

    Now choose f(x) = x and g(x) = \sin x and apply the above rule. Do a similar substitution for the second problem too.
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  3. #3
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    The product rule states that if a function, f(x), is a product of two functions, u and v, then f'(x) = uv' + vu'

    For f(x) = x \sin^{2}x, let  u = x and v = \sin^{2} x.

    It follows that  u' = 1 and v' = 2 \sin x \cos x .

    Then  f'(x) = x (2 \sin x \cos x) + \sin^{2}x (1)

    = 2x \sin x \cos x + \sin^{2} x.

     = \sin x (2x \cos x + \sin x).

    Following the same procedure for the second function, you get

    f'(x) = \sin x (2 \cos x \cdot -\sin x) + \cos^{2} x (\cos x)

     = -2 \cos x \sin^2 x + \cos^{3} x

     = -2 \cos x (1 - \cos^{2} x) + \cos^{3} x

     = -2 \cos x + 2 \cos^{3} x + \cos^{3} x

     = \cos x (-2 + 2 \cos^{2} x + \cos^{2} x)

     = \cos x (3 \cos^{2} x - 2).

    or, if you instantly realise that \sin x \cos^{2} x = \sin x (1 - \sin^{2} x) = \sin x - \sin^{3} x, you do not need to use the product rule!

     f'(x) = \cos x - 3 \sin^{2} x \cos x

    = \cos x - 3 \cos x (1 - \cos^{2} x)

    = \cos x - 3 \cos x + 3 \cos^{3} x

    = -2 \cos x + 3 \cos^{3} x

    = \cos x (3 \cos^{2} x - 2) .

    I hope that helps.

    ILoveMaths07.
    Last edited by ILoveMaths07; January 17th 2009 at 06:41 AM.
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