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Math Help - Need Derivitive Help

  1. #1
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    Need Derivitive Help

    Well, i have two problems, and i'm not really sure if the first is derivitive or not. Please help!

    1. At time t a particle is moving along the x-axis is at position x. The relationship between t and x is given by tx = x(squared) + 8. At x = 2, the velocity of the particle is:

    A. 1 B. 2 C. 6 D. -2 E. -1


    I don't know what to do here. I tried just plugging 2 into the equation, solving to t, and doing velocity equals distance over time, but that didn't work.


    2. If x(squared) + 2xy - 3y = 3, dy/dx of x = 2 equals:

    A. 1 B. 2 C. -2 D. 10/3 E. -1

    I tried the product rule of 2x and y, but it didn't work out right.

    Thanks in advance guys!

    Actually there is another problem that's even harder

    If d/dx of f(x) = g(x) and d/dx of g(x) = f(3x), then d(squared)/dx(squared) of f(x(squared)) equals:

    A. 4x(squared)f(3x(squared)) + 2g(x(squared)) B. f(3x(squared)) C. f(x^4) D. 2xf(3x(squared)) + 2g(x(squared)) E. 2xf(3x(squared))

    I'm pretty sure its not B or C, but i really have no idea how to do this problem. My teacher did mention using the chain rule as a hint, but i stil really don't get it.
    Last edited by Blue Griffin; November 29th 2007 at 02:33 PM. Reason: another problem
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  2. #2
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    Hello, Blue Griffin!

    1. At time t a particle is moving along the x-axis is at position x.
    The relationship between t and x is given by: . tx \:= \:x^2 + 8

    At x = 2, the velocity of the particle is:

    . . (A)\;1\qquad(B)\;2\qquad(C)\;6\qquad(D)\;-2\qquad(E)\;-1
    We have: . t \;=\;x + 8x^{-1}

    . . Then: . \frac{dt}{dx} \:=\:1-8x^{-2} \:=\:\frac{x^2-8}{x^2}

    . . Hence: . \frac{dx}{dt} \:=\:\frac{x^2}{x^2-8}

    At x = 2\!:\;\frac{dx}{dt} \;=\;\frac{2^2}{2^2-8} \;=\;\frac{4}{\text{-}4} \;=\;-1 . Answer (E)



    2)\text{ If }\:x^2 + 2xy - 3y \:=\: 3,\;\text{ find: }\frac{dy}{dx}\;\text{ at }x = 2.
    . . (A)\;1\qquad(B)\;2\qquad(C)\;-2\qquad(D)\;\frac{10}{3}\qquad(E)\;-1
    Differentiate implicitly: . 2x + 2x\!\cdot\!\frac{dy}{dx} + 2y - 3\!\cdot\!\frac{dy}{dx}\:=\:0

    . . 2x\!\cdot\!\frac{dy}{dx} - 3\!\cdot\!\frac{dy}{dx} \:=\:-2x-2y

    . . (2x-3)\!\cdot\!\frac{dy}{dx}\:=\:-2(x+y)

    . . \frac{dy}{dx}\:=\:\frac{-2(x+y)}{2x-3} \:=\:\frac{2(x+y)}{3-2x}


    When x = 2, the original equation is: . 4 + 4y - 3y \:=\:3\quad\Rightarrow\quad y = \text{-}1


    Therefore: . \frac{dy}{dx} \:=\:\frac{2+(\text{-}1)}{3-2(2)} \:=\:-1 . Answer (E)




    3)\;\text{If }\:f'(x)\:=\:g(x)\;\text{ and }\;g'(x)\:=\:f(3x),\;\text{find: }f''(x^2)

    (A)\;4x^2\!\cdot\!f(3x^2) + 2\!\cdot\!g(x^2)\quad B)\;f\left(3x^2\right)\quad (C)\;f\left(x^4\right)\quad (D)\;2x\!\cdot\!f(3x^2)\: + 2\!\cdot\!g(x^2)\quad (E)\;2x\!\cdot\!f(3x^2)
    We have: . y\;=\;f(x^2)

    Then: . y' \;=\;f'(x^2)\cdot2x . . . . but f'(x^2) \,=\,g(x^2)

    So we have: . y' \:=\:2x\cdot g(x^2)


    Product Rule: . y'' \:=\:2x\!\cdot\!g'(x^2)\!\cdot\!2x + 2\!\cdot\!g(x^2) . . . . but g'(x^2) \:=\:f(3x^2)

    Therefore: . y'' \;=\;4x^2\!\cdot\!f(x^2) + 2\!\cdot\!g(x^2) . Answer (A)

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  3. #3
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    Thanks a lot. It all makes sense now, lol.
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