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Thread: Deriv

  1. #1
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    Deriv

    Hiya

    I need to deriv y=0.25*sin(4pi*T)

    Am new to this kinda derving but i think it should be somthing like

    0.25*cos(8pt*T)

    but i think am way off can anyone please give me a hand.
    I also need to find the curving amp. curving period
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by JBswe View Post
    Hiya

    I need to deriv y=0.25*sin(4pi*T)

    Am new to this kinda derving but i think it should be somthing like

    0.25*cos(8pt*T)

    but i think am way off can anyone please give me a hand.
    I also need to find the curving amp. curving period
    This is just the chain rule:

    $\displaystyle y=0.25*sin(4\pi*T)$

    constant stays out front, sin -> cos then find the derivative of what is inside of sin
    $\displaystyle \frac{dy}{dT}=0.25*cos(4\pi T)(\frac{dy}{dt}4\pi T)$


    $\displaystyle \frac{dy}{dT}=0.25cos(4\pi T)(4\pi)$

    And simplify
    $\displaystyle \frac{dy}{dT}=\pi cos(4\pi T)$


    Just look at it like this, c is the constant coefficient. f(x) is sin(x) and g(x) is $\displaystyle 4\pi T$ then you are taking the derivative of $\displaystyle c f[g(x)]$ which is $\displaystyle c * f\prime [g(x)] * g\prime (x)$

    If any of that was confusing, I can explain it in more detail.
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  3. #3
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    Quote Originally Posted by JBswe View Post
    Hiya

    I need to deriv y=0.25*sin(4pi*T)

    Am new to this kinda derving but i think it should be somthing like

    0.25*cos(8pt*T)

    but i think am way off can anyone please give me a hand.
    I also need to find the curving amp. curving period
    If $\displaystyle y = a \sin (kT)$ then $\displaystyle \frac{dy}{dT} = k a \cos (kT)$.

    In your question a = 0.25 and $\displaystyle k = 4 \pi$.
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