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Differentiate y=x sinx cosx

  1. Feb 17, 2004 #1
    I'm not sure how to do this one. Only way I can think of is using the Product Rule but I don't know how to apply it when there are more than two functions.

    Something like:

    y=(x)(sinx)(cosx)

    --separate it into three different functions--

    f(x)=x, g(x)=sinx, z(x)=cosx

    --use Product Rule--

    ???

    Are those the right steps to differentiate the function and if they are how do I apply the Product Rule to three functions instead of just two?

    Any help would be appreciated.
     
  2. jcsd
  3. Feb 17, 2004 #2

    Hurkyl

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    whats wrong with

    f(x) = x
    g(x) = (sin x)(cos x)

    ?
     
  4. Feb 17, 2004 #3
    Ah. Didn't think of it that way. Sorry, brand new to Trig and Calc.
     
  5. Feb 17, 2004 #4

    Hurkyl

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    There is a generalization of the product rule; I just wanted to realize this other approach to the problem because this idea it may be useful in the future. (and, IMHO, is easier to remember!)

    Anyways, the generalization is:

    [tex]
    \begin{array}{l}
    (f_1 f_2 f_3 \ldots f_n)' = \\
    f_1' f_2 f_3 \ldots f_n +
    f_1 f_2' f_3 \ldots f_n +
    f_1 f_2 f_3' \ldots f_n +
    \ldots +
    f_1 f_2 f_3 \ldots f_n'
    \end{array}
    [/tex]
     
    Last edited: Feb 17, 2004
  6. Feb 18, 2004 #5
    Differentiation Product Rule...

    Differentiate:
    y = x(sin(x))(cos(x))

    f(x) = x
    g(x) = sin(x)
    h(x) = cos(x)

    Product Rule:
    (d/dx)[f(x)g(x)h(x)] = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)

    Factor g(x):
    (d/dx)[f(x)g(x)h(x)] = f(x)h(x)g'(x) + g(x)(h(x)f'(x) + f(x)h'(x))
    (d/dx)[f(x)g(x)h(x)] = x*cos(x)*(d/dx)[sin(x)] + sin(x)(cos(x)(d/dx)[x] + x(d/dx)[cos(x)])

    Derivative:
    (d/dx)[f(x)] = (d/dx)[x] = 1
    (d/dx)[g(x)] = (d/dx)[sin(x)] = cos(x)
    (d/dx)[h(x)] = (d/dx)[cos(x)] = -sin(x)

    = x*cos(x)*cos(x) + sin(x)(cos(x)(1) + x*-sin(x))
    = x*cos^2(x) + sin(x)(cos(x) - x*sin(x))
    = x*cos^2(x) + sin(x)*cos(x) - x*sin^2(x)
    = sin(x)*cos(x) + x*cos^2(x) - x*sin^2(x)

    Factor x:
    = cos(x)sin(x) + x(cos^2(x) - sin^2(x))
    x(cos^2(x) - sin^2(x)) = x*cos(2x)

    Solution:
    = x*cos(2x) + cos(x)*sin(x)

    ---

    (d/dx)[f(x)g(x)h(x)] = (d/dx)[x(sin(x))(cos(x))]

    Use Product Rule:
    [tex]\frac{d(u,v)}{dx} = \frac{du}{dx}v + u \frac{dv}{dx}[/tex]
    u = x, v = cos(x)sin(x)
    (d/dx)[x] = 1
    = x(d/dx)[cos(x)sin(x)] + cos(x)sin(x)(d/dx)[x]
    = x(d/dx)[cos(x)sin(x)] + cos(x)sin(x)

    Use Product Rule:
    [tex]\frac{d(u,v)}{dx} = \frac{du}{dx}v + u \frac{dv}{dx}[/tex]
    u = cos(x), v = sin(x)
    = cos(x)sin(x) + x(cos(x)(d/dx)[sin(x)] + sin(x)(d/dx)[cos(x)])
    (d/dx)[sin(x)] = cos(x)
    (d/dx)[cos(x)] = -sin(x)
    = cos(x)sin(x) + x(cos^2(x) - sin^2(x))
    x(cos^2(x) - sin^2(x)) = x*cos(2x)

    Solution:
    = x*cos(2x) + cos(x)sin(x)
     
  7. Mar 27, 2004 #6

    h2

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    xcosxsinx = xsin(2x)/2 so
    (xcosxsinx)' = sin(2x)/2 + xcos(2x)
     
  8. Mar 27, 2004 #7
    As h2 said, use the identity.

    [tex]f(x) = x\cos x\sin x = \frac{x\sin 2x}{2}[/tex]
    [tex]f'(x) = \frac{1}{2}(\sin 2x + 2x\cos 2x) = \frac{\sin 2x}{2} + x\cos 2x = \cos x\sin x + x\cos 2x[/tex]
     
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