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Math Help - f(x)=cos^2(sinx)

  1. #1
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    f(x)=cos^2(sinx)

    I think I know how to differentiate it, this is what I did:

    f(x)=cos^2(sinx)
    f'(x)=[(2)(cos)(-sin)(sinx)](cosx)
    and the answer is:
    =-2cos(sinx)[sin(sinx)](cosx)

    Why are there so many brackets being used in the answer? Couldn't I have went from my step 2 to:

    =-2cos(sinx)sin(sinx)(cosx)?
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  2. #2
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    Quote Originally Posted by kmjt View Post
    I think I know how to differentiate it, this is what I did:

    f(x)=cos^2(sinx)
    f'(x)=[(2)(cos)(-sin)(sinx)](cosx)
    and the answer is:
    =-2cos(sinx)[sin(sinx)](cosx)

    Why are there so many brackets being used in the answer? Couldn't I have went from my step 2 to:

    =-2cos(sinx)sin(sinx)(cosx)?
    You need to use the chain rule twice.

    y = [\cos{(\sin{x})}]^2.


    To find \frac{dy}{dx} let u = \cos{(\sin{x})} so that y = u^2.

    \frac{dy}{du} = 2u = 2\cos{(\sin{x})}.


    To find \frac{du}{dx} let v = \sin{x} so that u = \cos{v}.

    \frac{dv}{dx} = \cos{x}

    \frac{du}{dv} = -\sin{v} = -\sin{(\sin{x})}.

    Therefore \frac{du}{dx} = -\cos{x}\sin{(\sin{x})}.


    Thus \frac{dy}{dx} = -2\cos{(\sin{x})}\cos{x}\sin{(\sin{x})}.
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  3. #3
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    So my answer was actually right too?
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  4. #4
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    Wouldn't that be the same as my answer of:

    =-2cos(sinx)sin(sinx)(cosx)?
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