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Homework Help: Instantaneous rate of change and deriving a fraction

  1. Dec 7, 2018 #1
    1. The problem statement, all variables and given/known data
    Sorry for so many posts lately, hopefully this is allowed.

    What tangent points on ##g(x)=\frac {12} {x+1}## has an instantaneous rate of change of -3?

    2. Relevant equations


    3. The attempt at a solution

    I know that once I derive ##g(x)=\frac {12} {x+1}## I can set the derivative to -3 to get my x value then I can substitute the x value into the original equation and find my y. I can't figure out how to derive this.

    I know the rules of derivatives but I am very bad at applying them. This is what I have done so far.

    ##g(x)=\frac {12} {x+1}##

    ##g'(x)=12 (\frac {1} {x+1})## Removing the constant

    ## =12((x+1)^{-1})## Applying the power rule

    I think I would use the chain rule next because it states that if ##G(x)=(g(x))^{n}## then ##G'(x)=ng'(x)g(x)^{n-1}## which looks like what I have now.

    Writing out what is what... ##n = -1##, ##g'(x)=12(\frac {1} {x+1})##, and ##g(x)=\frac {12} {x+1}##

    applying it to the chain rule:

    ##=(-1)(12)(\frac {1} {x+1})(\frac {12} {x+1})^{-2}##

    Haven't moved on after this because I'm not confident that this is right.
     
    Last edited: Dec 7, 2018
  2. jcsd
  3. Dec 7, 2018 #2

    scottdave

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    You have 1/(x+1) then you have (x+1)^-1 which are equivalent. But then you somehow try to do [1/(x+1)]^-1 which is going to just be x+1. So you need to fix your g(x).
    Note you may try the quotient rule as well. If you do everything properly, both approaches will yield the same result.
     
  4. Dec 7, 2018 #3

    SammyS

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    None of the above three expressions is ##\ g'(x) ##. In fact each is identical to ##\ g(x)##.

    What may be useful is to notice that ##\ g(x)\ ## can be written as follows.
    ##\displaystyle g(x) = 12((x+1)^{-1}) ##​
    Your statement of the chain rule is fine, but you will need to state it differently for your problem, because the function ##\ g(x)\ ## is already defined and defined in such a way that it won't fit nicely into that form of the chain rule. Also, the coefficient, 12, makes putting your expression into a nice form for the chain rule a bit complicated.

    You could say that ##\displaystyle \ g(x) = 12\cdot F(x) \,, \text{ where } F(x) = (f(x))^{-1} \ \text{and } f(x)=(x+1) \ .##

    Now use the chain rule to differentiate ##\ F(x) \ .##
    By the way: The verb form for obtaining a derivative is differentiate, not derive.​
    .
     
  5. Dec 7, 2018 #4

    Mark44

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    @Specter, in the future, please post questions about derivatives and integrals in the Calculus & Beyond section. I have moved your thread.
    Absolutely.
     
  6. Dec 7, 2018 #5
    The definition of derivatives is
    $$f’(x) = \lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$
     
    Last edited: Dec 8, 2018 at 12:35 AM
  7. Dec 7, 2018 #6

    SammyS

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    No.
    The derivative does give the instantaneous rate of change.
     
  8. Dec 7, 2018 #7
    But doesn’t “ instant” means no time? or does it actually mean “almost no time” ?
     
  9. Dec 8, 2018 at 12:34 AM #8

    SammyS

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    That's what the limit gives you, the rate of change at an instant. After all, we have such quantities as instantaneous velocity and instantaneous acceleration. Right?
     
  10. Dec 8, 2018 at 12:36 AM #9
    Oops,it seems like my language understanding has a little problem.Previous post edited.
     
  11. Dec 8, 2018 at 8:46 AM #10

    scottdave

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    So were you able to solve it? What did you get for the derivative?
     
  12. Dec 13, 2018 at 3:47 PM #11
    It's been a bit since I posted but I haven't had a chance to solve it yet so here's my first attempt. Trying to follow what you said.

    ##\displaystyle \ g(x) = 12\cdot F(x) \,, \text{ where } F(x) = (f(x))^{-1} \ \text{and } f(x)=(x+1) \ .##

    So the chain rule is ## F'(x)=nf'(x)f(x)^{n-1}##

    ##g'(x)=12((x+1)^{-1})##

    ##g'(x)=12(-1)(1)(x+1)^{-2}##

    Is this correct so far?
     
  13. Dec 13, 2018 at 3:48 PM #12
    Haven't had a chance to until now, hopefully I can figure it out!
     
  14. Dec 13, 2018 at 4:30 PM #13

    Mark44

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    Yes, but you definitely should write it in simplest form.
     
  15. Dec 13, 2018 at 5:05 PM #14
    ##g'(x)=-12(x+1)^{-2}##

    I'm not sure what to do now, is there more to it? Do I expand?
     
  16. Dec 13, 2018 at 5:14 PM #15

    Mark44

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    No, not expand -- just write it as ##\frac{-12}{(x + 1)^2}##
     
  17. Dec 13, 2018 at 5:15 PM #16

    SammyS

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    Now go back to the original problem and continue.
     
  18. Dec 13, 2018 at 5:17 PM #17

    Mark44

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    BTW, for this problem, you could use either the product rule and chain rule (as you did) or the quotient rule. I never recommend the quotient rule when the numerator is just a constant.
     
  19. Dec 14, 2018 at 12:12 PM #18
    I need to set the derivative equal to -3.

    ##g'(x)=\frac {-12} {(x+1)^{2}} =-3##

    ##(x+1)^2(\frac {-12} {(x+1)^2}) =-3(x+1)^2##

    ##-12=-3(x+1)^2##

    ##\frac {-3(x+1)^2} {-3} = \frac {-12} {-3}##

    ##\sqrt (x+1)^2 = \sqrt 4##

    ##x+1=2##

    ##x=1##

    Now I plug the x value into the original equation to find a y value

    ##g(x)=\frac {12} {x+1}##

    ##g(1)=\frac {12} {1+1}##

    ##=\frac {12} {2}##

    ##=6##

    The tangent points (1,6) has an instantaneous rate of change of -3.
     
  20. Dec 14, 2018 at 12:35 PM #19

    Mark44

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    The step below is only partially correct.
    Instead of taking the square root of both sides as you did, solve the quadratic equation ##(x + 1)^2 = 4##. There are two solutions, one of which is x = 1.

    Also, in TeX, use braces for the argument of the square root, like this: \sqrt{ (x + 1)^2}. This renders as ##\sqrt{ (x + 1)^2}## so that the vinculum (the bar) extends over everything inside the radical.
     
  21. Dec 14, 2018 at 12:58 PM #20
    The two solutions that I got were x=1 and x=-3. How do I know which to use?

    I just started using TeX last week so thanks for the tips!
     
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