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Derivative of a fractional function without quotient rule

  1. Nov 9, 2015 #1
    1. The problem statement, all variables and given/known data
    The displacement of a particle can be modeled by the function [itex]x(t)=\frac{2x-5}{4x^2+2x}[/itex], where [itex]t[/itex] is in seconds, [itex]x[/itex] is in meters, and [itex]t ∈ [1,10][/itex]

    a) Determine the derivative of the function without using the quotient rule.

    b) Hence, find exactly when the particle is stationary.

    c) Determine when the particle is moving at a constant velocity. You can use your calculator to assist you.


    d) A jolt is defined as being a change in acceleration over time. With the help of your calculator, determine the time for which the jolt is equal to the displacement.

    2. Relevant equations
    I'm not sure, besides the differentiation rules. However, I can't use the quotient rule.

    3. The attempt at a solution
    I'm honestly not sure how to even begin here. I've never been taught how to solve this sort of question without the quotient rule. Can anyone please give me an idea on where I should start? Thank you for your time.
     
  2. jcsd
  3. Nov 9, 2015 #2

    Krylov

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    Use the product rule and the chain rule instead?
     
  4. Nov 9, 2015 #3
    I have never learnt the chain rule method. Also, I'm not sure how the product rule would work here: the [itex](4x^2+2x)[/itex] term would be to the power of negative 1, and I'm not sure how to apply to product rule then. Sorry for being stupid, but I'm really new (not to mention bad) when it comes to calculus...
     
  5. Nov 9, 2015 #4

    Krylov

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    Lack of knowledge does not imply stupidity. Do you have a good calculus book? I recommend that you make a few hundred exercises from such a book (not all at once!) differentiating all kinds of functions using the chain rule, product rule and quotient rule. This is essential material and it should not be an obstruction.

    Also, now that I take a better look, I think your function should read ##x(t) = \frac{2 t - 5}{4 t^2 + 2 t}## instead, so ##t## instead of ##x## in the right-hand side.
     
  6. Nov 9, 2015 #5

    Ray Vickson

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    [QUpOTE="Saracen Rue, post: 5283378, member: 521193"]1. The problem statement, all variables and given/known data
    The displacement of a particle can be modeled by the function [itex]x(t)=\frac{2x-5}{4x^2+2x}[/itex], where [itex]t[/itex] is in seconds, [itex]x[/itex] is in meters, and [itex]t ∈ [1,10][/itex]

    a) Determine the derivative of the function without using the quotient rule.

    b) Hence, find exactly when the particle is stationary.

    c) Determine when the particle is moving at a constant velocity. You can use your calculator to assist you.


    d) A jolt is defined as being a change in acceleration over time. With the help of your calculator, determine the time for which the jolt is equal to the displacement.

    2. Relevant equations
    I'm not sure, besides the differentiation rules. However, I can't use the quotient rule.

    3. The attempt at a solution
    I'm honestly not sure how to even begin here. I've never been taught how to solve this sort of question without the quotient rule. Can anyone please give me an idea on where I should start? Thank you for your time.[/QUOTE]

    The product rule gives
    [tex] \frac{d}{dx} (2x-5)(4x^2 +2x)^{-1} = \left[ \frac{d}{dx} (2x-5) \right] (4x^2 + 2x)^{-1} + (2x - 5) \left[ \frac{d}{dx} (4x^2 + 2x)^{-1} \right] [/tex]
    Getting ##d(2x-5)/dx## is easy; where you need to use the chain rule is in the second differentiation:
    [tex] \frac{d(4x^2 + 2x)^{-1}}{dx} = \frac{d(4x^2 + 2x)^{-1}}{d (4x^2 + 2x)} \cdot \frac{d(4x^2 + 2x)}{dx} [/tex]
    This means that if we have a function of the form ##f(x) = g(h(x))## we can let ##u = h(x)## and get
    [tex] \frac{df}{dx} = \left. \frac{dg}{du} \right|_{u = h(x)} \cdot \frac{dh}{dx} [/tex]
    For f = g(h) (with h = h(x)), a way of remembering this is to think of
    [tex] \frac{df}{dx} = \frac{dg}{dh} \cdot \frac{dh}{dx} [/tex]
    (cancelling the dh's).

    Anyway, if ##h = 4x^2 + 2x## and ##g(h) = h^{-1}##, can you calculate ##dg/dh## and ##dh/dx##? That's all there is to it.
     
  7. Nov 9, 2015 #6

    Mark44

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    If your function is ##x(t) = \frac{2t - 5}{4t^2 + 2t}## (with change as noted by Krylov), and the objective is to find x'(t) without using the quotient rule, there are three other possibilities.
    1) Use the definition of the derivative as the limit of a quotient.
    2) Use the product rule and chain rule, writing the function as ##x(t) = (2t - 5) (4t^2 + 2t)^{-1}##
    3) Use the product rule and a rule you might have learned about the derivative of the reciprocal of a function, with ##x(t) = (2t - 5) \left(\frac{1}{4t^2 + 2t}\right)## .

    Choice 1 above would be pretty tricky, so it seems to me that one of the other two choices is what you're expected to use.
     
  8. Nov 9, 2015 #7

    epenguin

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    Or you could express the fraction as

    ## \frac{(2x + 1) - 6}{2x (2x + 1)} ##

    and I hope easy to see ways to proceed.
     
  9. Nov 9, 2015 #8

    SammyS

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    You could use partial fraction decomposition. Then you only need to deal with ##\displaystyle \ \frac{A}{t}+ \frac{B}{2t+1}\ ## .

    Alternatively, you could multiply both sides by the denominator, then do implicit differentiation.
     
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