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Math Help - trig derivative help...

  1. #1
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    trig derivative help...

    h(x) = sin2xcos2x
    h'(x) = -4cos2xsin2x <--- doubt this is correct
    --------------------
    f(x) = cotx/sinx (unsure on how to do this....)

    -thx
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  2. #2
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    Quote Originally Posted by maybnxtseasn View Post
    h(x) = sin2xcos2x
    h'(x) = -4cos2xsin2x <--- doubt this is correct
    --------------------
    f(x) = cotx/sinx (unsure on how to do this....)

    -thx

    For the first one try using <br />
2\sin{2x}\cos{2x}=sin{4x}

    If you know the derivatives of cotx and sinx, then use the quotient rule on the second one
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  3. #3
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    Quote Originally Posted by maybnxtseasn View Post
    h(x) = sin2xcos2x
    h'(x) = -4cos2xsin2x <--- doubt this is correct
    --------------------
    f(x) = cotx/sinx (unsure on how to do this....)

    -thx
    you need to use the product rule ...

    h(x) = \sin(2x) \cos(2x)

    h'(x) = \sin(2x)[-2\sin(2x)] + \cos(2x)[2\cos(2x)]

    h'(x) = 2[\cos^2(2x) - \sin^2(2x)] = 2\cos(4x)

    or use an identity ...

    h(x) = \frac{1}{2} \cdot 2\sin(2x)\cos(2x)

    h(x) = \frac{1}{2} \sin(4x)

    h'(x) = 2\cos(4x)



    note that \frac{\cot{x}}{\sin{x}} = \cot{x}\csc{x}

    product rule again ...

    \cot{x}[-\csc{x}\cot{x}] + \csc{x}[-\csc^2{x}]

    -\csc{x}[\cot^2{x} + \csc^2{x}]

    -\csc{x}[2\csc^2{x} - 1]

    or ...

    \frac{\cot{x}}{\sin{x}} = \frac{\cos{x}}{\sin^2{x}}

    use the quotient and chain rules ...

    \frac{\sin^2{x}(-\sin{x}) - \cos{x}[2\sin{x}\cos{x}]}{\sin^4{x}}

    \frac{-\sin{x}[\sin^2{x} + 2\cos^2{x}]}{\sin^4{x}}

    -\frac{1+\cos^2{x}}{\sin^3{x}}
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