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Math Help - Differentiation of Sinusoidal Functions

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    Differentiation of Sinusoidal Functions

    Find the equation of the line that is tangent to y = x^2sin(2x) at x = -\pi.


    Shown work would be appreciated, thank you.
    Last edited by Eixo; April 13th 2010 at 05:44 PM.
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  2. #2
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    Quote Originally Posted by Eixo View Post
    Find the equation of the line that is tangent to y = x^2sin(2x) at x = pi.


    Shown work would be appreciated, thank you.

    why don't you show the work of finding \frac{dy}{dx}, first?
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    Quote Originally Posted by skeeter View Post
    why don't you show the work of finding \frac{dy}{dx}, first?
    That's unfortunately where I've run into my trouble. I know how to do the rest of the question, however I am unsure on how to find \frac{dy}{dx}
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    Quote Originally Posted by Eixo View Post
    That's unfortunately where I've run into my trouble. I know how to do the rest of the question, however I am unsure on how to find \frac{dy}{dx}
    product rule + chain rule for \sin(2x) ...

    \frac{d}{dx}(u \cdot v) = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}

    where u = x^2 and v = \sin(2x)


    this is a very basic process. give it a go ...
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    Thank you for the help, it was the mental roadblock that I was struggling to get past.

    \frac{dy}{dx}=x^2 \cdot 2cos(2x)+2x \cdot 2sin(2x)

    I then subbed in -\pi (typo from original post, will edit).

    =((-\pi)^2(2(1)))+(2(-\pi)(0))
    =2\pi^2

    y @  -\pi = (-\pi)^2 \cdot sin(2(-\pi))<br />
= 0

    Slope = Slope

    2\pi^2=\frac{y-0}{x+\pi} (Cross multiply)
    2\pi^2x+2\pi^3 = y
    Last edited by Eixo; April 13th 2010 at 05:50 PM.
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    Quote Originally Posted by Eixo View Post
    Thank you for the help, it was the mental roadblock that I was struggling to get past.

    \frac{dy}{dx}=(x^2)(2cos(2x))+(2x)(2sin(2x))

    close ... \textcolor{red}{\frac{dy}{dx} = x^2 \cdot 2\cos(2x) + \sin(2x) \cdot 2x}

    now sub in \pi for x to find the slope.

    use of the point-slope form is a bit easier ...

    \textcolor{red}{y - y_1 = m(x - x_1)}
    ...
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    Sorry, will correct. My answer remains unchanged though.
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