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Math Help - derivatives

  1. #1
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    derivatives

    I picked up a new textbook and this exercise was in it

    x u(x) v(x) u'(x) v'(x)
    2 -3 4 6 5
    3 2 7 -2 1


    then it wants me to find \frac{dy}{dx} for \frac{u(x)}{v(x)} for x = 2 and v(u(x)) when x = 3

    the first one was easy and I got \frac{39}{16}

    the second one I figured would be easy as well. the book has -10 but I got

    dy/dx = v'(u(x)) * u'(x) \Rightarrow (1)(2)(-2)=-4
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  2. #2
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    Re: derivatives

    The book says that v'(u(x)) = v'(2) which is 5 in the table. still unclear about this.
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  3. #3
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    Re: derivatives

    I'm not connecting your table and your workings. Aka. I don't understand what you're supposed to do. Could you explain more clearly?
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  4. #4
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    Re: derivatives

    you take the given values in the table to calculate the derivative asked.

    such as,

    the first one asked is find dy/dx of u(x)/v(x) when x = 2 which is

    \frac{u'(x)v(x) - v'(x)u(x)}{[v(x)]^2} \Rightarrow \frac{6(4) - 5(-3)}{4^2} \Rightarrow dy/dx = \frac{39}{16}


    for the second one is dy/dx for v(u(x)) when x = 3.

    I get  v'(u(x)) * u'(x) = 1(2)*(-2) = -4 = dy/dx

    the book says it is -10 though.
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  5. #5
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    Re: derivatives

    I think I get it v'(u(x)) isn't 1(2) but instead its v'(u(x)) \Rightarrow v'(2) because u(x) = 2, when\, x = 3
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  6. #6
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    Re: derivatives

    I think there's something wrong in your working for x=3 but I can't find what is wrong. Srry!
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