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Math Help - Quotient rule with Sin & Cos

  1. #1
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    Quotient rule with Sin & Cos

    f(x)/g(x) = f'(x) g(x) - f(x) g'(x) / g(x)^2
    How do I find the derivative of something like cos(x) / sin(1/x) ?

    = ( - sin x) sin(1/x) - cos (x) cos ( 1/x) / ( sin( 1/x )^ 2

    I know this is not completely correct. What is the derivative of sin ( 1/x )? Do I leave the value as 1/x ?
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  2. #2
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    Quote Originally Posted by bt12345 View Post
    f(x)/g(x) = f'(x) g(x) - f(x) g'(x) / g(x)^2
    How do I find the derivative of something like cos(x) / sin(1/x) ?

    = ( - sin x) sin(1/x) - cos (x) cos ( 1/x) / ( sin( 1/x )^ 2

    I know this is not completely correct. What is the derivative of sin ( 1/x )? Do I leave the value as 1/x ?
    g(x)=sin(\frac{1}{x}) is a a composite function, so you have to use the chain rule. Are you familiar with this?

    g'(x)=[\frac{d}{dx}(\frac{1}{x})]cos(\frac{1}{x})=-\frac{1}{x^2}cos(\frac{1}{x})

    In general, if you have a composite function:

    F(x)=f(g(x))

    The derivative of that function is:

    F'(x)=f'(g(x))g'(x)
    Last edited by adkinsjr; October 14th 2009 at 07:58 PM.
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