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Thread: 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:
    $\displaystyle y=37-sec^3(2x)$, point $\displaystyle (0,36)$

    Does the first equation become $\displaystyle y=[37-sec(2x)]^3$ 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.
    $\displaystyle f(x)=sin2x$, point $\displaystyle (\pi,0)$

    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:
    $\displaystyle y=37-sec^3(2x)$, point $\displaystyle (0,36)$

    Does the first equation become $\displaystyle y=[37-sec(2x)]^3$ 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:

    $\displaystyle y=-sec^3(2x)$

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

    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 $\displaystyle y-f(a)=f'(a)(x-a)$ where "a" is the x-coordinate of the point through which the tangent passes. In your case this would be $\displaystyle \pi$. 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: $\displaystyle y'=3[-sec(2x)]^2[2 sec(2x) tan(2x)]$
    Last edited by DarkestEvil; Oct 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: $\displaystyle y'=3[-sec(2x)]^2[2 sec(2x) tan(2x)]$
    Yes, so now just find the value of the derivative at the point you were given, and that is the slope $\displaystyle f'(a)$.
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