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Math Help - Derivative of Secant and Sine (w/ Chain Rule)

  1. #1
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    Derivative of Secant and Sine (w/ Chain Rule)

    Evaluate the derivative of the function at the indicated points:
    y=37-sec^3(2x), point (0,36)

    Does the first equation become y=[37-sec(2x)]^3 and then use chain rule? After I do so, I plug in the point to find a value?

    Find an equation of the tangent line to the graph of f at the indicated point.
    f(x)=sin2x, point (\pi,0)

    What do I do here? :[

    Thanks for the help in advance.
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  2. #2
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    Quote Originally Posted by DarkestEvil View Post
    Evaluate the derivative of the function at the indicated points:
    y=37-sec^3(2x), point (0,36)

    Does the first equation become y=[37-sec(2x)]^3 and then use chain rule? After I do so, I plug in the point to find a value?

    You have to take the 37 out of the brackets. You can't write it like that. You can just ignore the 37 all together since the derivative of a constant is zero. The derivative of your function is the same as the derivative of:

    y=-sec^3(2x)

    Find an equation of the tangent line to the graph of f at the indicated point.
    f(x)=sin2x, point (\pi,0)

    What do I do here? :[

    Thanks for the help in advance.

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  3. #3
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    Find an equation of the tangent line to the graph of f at the indicated point.
    , point


    In general, the equation of a tangent is given by y-f(a)=f'(a)(x-a) where "a" is the x-coordinate of the point through which the tangent passes. In your case this would be \pi. It should be easy to apply this formula.
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  4. #4
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    So for the first one, my derivative became: y'=3[-sec(2x)]^2[2 sec(2x) tan(2x)]
    Last edited by DarkestEvil; October 20th 2009 at 04:38 PM. Reason: forgot the math tag
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  5. #5
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    Quote Originally Posted by DarkestEvil View Post
    So for the first one, my derivative became: y'=3[-sec(2x)]^2[2 sec(2x) tan(2x)]
    Yes, so now just find the value of the derivative at the point you were given, and that is the slope f'(a).
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