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Math Help - Derivative Help

  1. #1
    Member ~berserk's Avatar
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    Derivative Help

    find the derivative:
    s(x)=sin^2⁡((3x-pi))

    could someone show me how to find the derivative of this using both methods of the chain rule: F'(x) = f '(g(x)) g '(x) and (dy/du)*(du/dx).
    I keep getting 2(sin(3x-pi)(-3cos(3x)) which i know to be incorrect
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by ~berserk View Post
    find the derivative:
    s(x)=sin^2⁡((3x-pi))

    could someone show me how to find the derivative of this using both methods of the chain rule: F'(x) = f '(g(x)) g '(x) and (dy/du)*(du/dx).
    I keep getting 2(sin(3x-pi)(-3cos(3x)) which i know to be incorrect
    s=u^2 \quad u=\sin(y) \quad y=3x-\pi

    Now by the chain rule we get

     \frac{ds}{dx}=\frac{ds}{du}\frac{du}{dy}\frac{dy}{  dx}=2u\cos(y)(3)=6\sin(y)\cos(y)=6\sin(3x-\pi)\cos(3x-\pi)
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  3. #3
    Member ~berserk's Avatar
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    Is there any way you could show your process in a little more detail?
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by ~berserk View Post
    Is there any way you could show your process in a little more detail?
     \frac{ds}{dx}=\underbrace{\frac{ds}{du}}_{2u} \underbrace{\frac{du}{dy}}_{\cos(y)} \underbrace{ \frac{dy}{dx} }_{3}=2u\cos(y)(3)

    Now we need to back substitute the variable in the first post

    6u\cos(y)=6\sin(y)\cos(y)=6\sin(3x-\pi)\cos(3x-\pi)
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