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Math Help - Derivative of Sin function Problem

  1. #1
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    Derivative of Sin function Problem

    If f(x)= sin²(3-x), then dy/dx(0)= ?

    A. -2cos3
    B. -2sin3cos3
    C. 6cos3
    D. 2sin3cos3
    E. 6sin3cos3

    This is a questions from my AP Calculus BC class, and I can't seem to get the correct answer. Any Thoughts?

    Thanks!
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by r2d2 View Post
    If f(x)= sin²(3-x), then dy/dx(0)= ?
    Any Thoughts?

    Thanks!
    Sure, start by differentiating f(x).

    It may be the case that you are confused because the problem contains two forms of notation.

    Understand that f'(x)=\frac{dy}{dx}. The fact that this question would mix and match notation is a strong indication that your teachers are crap. But, don't let it get ya down.

    So, you are asked to find f'(0)...
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  3. #3
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    so f'(x)= sin²(3-x)= sin(3-x) * (sin(3-x) = cos(3-x) * cos(3-x)

    Then plug in f'(0) to get cos(3) * cos(3)?

    Would that be the way to go about this problem? I think I am missing an important trigonometric derivative point.... Any Thoughts?
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  4. #4
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    Rewrite as (sin(3-x))²

    Then use chain rule:

    2(sin(3-x)) (cos(3-x)) (-1)
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