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Math Help - Easy question: i'm new too calc. (urgent)

  1. #1
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    Easy question: i'm new too calc. (urgent)

    y= u^5 +u^3

    u= 3/v - 4v

    v= 3 - x^2

    Use the chain rule in Leibniz notation to find (dy/dx) when x = 2

    Text gives the solution: -48608

    do i use:

    (dy/dx) = (dy/du)(du/dv)(dv/dx) because when I do, I don't get the right answer.
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  2. #2
    Newbie Mr. Engineer's Avatar
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    Quote Originally Posted by white_cap View Post
    y= u^5 +u^3

    u= 3/v - 4v

    v= 3 - x^2

    Use the chain rule in Leibniz notation to find (dy/dx) when x = 2

    Text gives the solution: -48608

    do i use:

    (dy/dx) = (dy/du)(du/dv)(dv/dx) because when I do, I don't get the right answer.

    you do use the chain rule, but first you have to substitute to get your function in terms of x
    u=\frac{3}{3-x^2}-4(3-x^2)
    then distrubute
    u=\frac{3}{3-x^2}-(12-4x^2)

    y= u^5 +u^3
    y=(\frac{3}{3-x^2}-(12-4x^2))^5+(\frac{3}{3-x^2}-(12-4x^2))^3
    now u can differentiate this function using the chain rule in terms of \frac{dy}{dx}
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  3. #3
    o_O
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    Well take the derivative:
    \frac{dy}{dx} = {\color{blue}\frac{dy}{du}} \cdot {\color{red}\frac{du}{dv}} \cdot {\color{magenta}\frac{dv}{dx}}
    \frac{dy}{dx} = {\color{blue} \left(5u^{4} + 3u^{4}\right) } \cdot {\color{red} \left(-3v^{-2} - 4\right)} \cdot {\color{magenta}-2x} (Just the application of the power rule over and over again)

    Now use x = 2 to find v(2) and u(2):
    v(2) = 3 - (2)^{2}
    u(2) = \frac{3}{v(2)} - 4v(2)

    Then plug all these values into your expression for \frac{dy}{dx} at x = 2
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