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101Ryan101

  • 4 years ago

what's the derivative of cosxsinx

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  1. lalaly
    • 4 years ago
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    use the product rule

  2. anonymous
    • 4 years ago
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    product rule \[(fg)'=f'g+g'f\] use \[f(x)=\sin(x), f'(x)=\cos(x), g(x)=\cos(x), g'(x)=-\sin(x)\]

  3. lalaly
    • 4 years ago
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    f(x)=cos x g(x)=sin x product rule: d/dx ( f(x)g(x) ) = f'(x)g(x) + f(x) g'(x)

  4. 101Ryan101
    • 4 years ago
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    dy/dx of cosxsinx = (-sinx)(sinx) + cosxcosx = cos^2x - sin^2x = ??

  5. lalaly
    • 4 years ago
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    yupp thats right....and cos^2x-sin^2x = cos(2x)

  6. 101Ryan101
    • 4 years ago
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    = (1+cos2x)/2 - (1-cos2x)/2 = 2(cos2x)2 = yeaaaaeee!

  7. lalaly
    • 4 years ago
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    hehe

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