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Math Help - Differentiation

  1. #1
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    Differentiation

    12.5 sin(0.06 t)


    What rule would I use to differentiate it?
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  2. #2
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    Re: Differentiation

    Quote Originally Posted by Thepiman1 View Post
    12.5 sin(0.06 t)


    What rule would I use to differentiate it?
    I presume you know what the differential of the sine function is. Is that correct?

    Now you use a substitution and the chain rule.

    $y = 12.5sin(0.06t).$

    $u = 0.06t \implies \dfrac{du}{dx} = what?$

    $Also\ u = 0.06t \implies y = 12.5sin(u) \implies \dfrac{dy}{du} = what?$

    How do you calculate $\dfrac{dy}{dx}$ knowing $\dfrac{du}{dx}$ and $\dfrac{dy}{du}$?
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  3. #3
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    Re: Differentiation

    du/dx= 0.06
    dy/du= 12.5 cos (0.06)
    dy/dx= 0.06 x 12.5 cos (0.06)

    Look right?
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  4. #4
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    Re: Differentiation

    Quote Originally Posted by Thepiman1 View Post
    du/dx= 0.06
    dy/du= 12.5 cos (0.06) No. 12.5cos(u) = 12.5cos(0.06t)
    dy/dx= 0.06 x 12.5 cos (0.06)

    Look right? NO, see below
    First, I see that I made an error in my first post. I should have been using dt, not dx. That may have confused you. If so, I apologize.

    $y = 12.5sin(0.06t).$

    $Let\ u = 0.06t \implies \dfrac{du}{dt} = 0.06.$ You were right here.

    $Also\ u = 0.06t \implies y = 12.5sin(u) \implies \dfrac{dy}{du} = 12.5cos(u) = 12.5cos(0.06t).$

    $\therefore \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dt} = 0.06 * 12.5cos(0.06t) = 0.75cos(0.06t).$

    It is not technically incorrect to leave an arithmetic expression in an answer, but if the arithmetic expression is easily computable, I think it is good form to do the computation. So I would not give an answer of 0.06 * 12.5cos(0.06t).


    EDIT: A lot of the mechanics in differential calculus is just using substitutions to reduce a complex expression into one or more simple expressions that have a derivative you have memorized and then using the chain rule. The formal step of doing the substitutions explicitly is often skipped to save time (although it increases the risk of errors), but whenever you see something really complex or unknown to you, think about explicitly trying one or more substitutions.
    Last edited by JeffM; May 8th 2014 at 09:25 AM.
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