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First and second derivatives. Please shed some light!

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Hey forum, hope everyone is having a good day!
    If someone could check this question out for me that'd be great!

    Determine the first and second derivative:

    g(x) = (2x - 3)/(x + 4)


    3. The attempt at a solution

    g(x) = (2x – 3)(x + 4)^(-1)
    g’(x) = (2x – 3)’(x + 4)^(-1) + (2x – 3)[(x + 4)^(-1)]’
    g’(x) = (2)(x + 4)^(-1) + (2x – 3)(-1)(x+4)^(-2)(1)
    g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)

    g’’(x) = [2(x + 4)^(-1)]’ – (2x – 3)’(x + 4)^(-2) + (2x – 3)[(x + 4)^(-2)]’
    g’’(x) = (-1)2(x + 4)^(-2)(1) – (2)(x + 4)^(-2) + (2x – 3)(-2)(x + 4)^(-3)(1)
    g’’(x) = -2(x + 4)^(-2) – 2(x + 4)^(-2) +(–4x + 6)(x + 4)^(-3)

    Is this right?
    Also should I be dividing my entire final answer by -2

    if so it should be:

    g''(x) = (x+4)^(-2) + (x + 4)^(-2) + (2x + 3)(x+4)^(-3)


    2. Relevant equations

    also.. when using an online derivative calculator.. it told me that the first derivative for g(x) is g'(x) = 11/(x^2+8x+16) which initially appears much different than my answer g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)

    However! when I put each first derivative into my graphing calculator the same graph appeared.. implying they are equivalent.. can someone PLEASEEEE explain to me how this is possible or why this makes sense..


    ***BONUS:confused:

    for the same question I had two more functions to find the first and second derivative for:

    a) y = (3x + 2)^2
    y’ = 2(3x + 2)^1(3)
    y’ = 6(3x + 2)
    y’ = 18x + 12

    y’’ = 18


    b) f(x) = 5x^2 – 2x
    f’(x) = 10x – 2

    f’’(x) = 10



    I'm pretty sure both a) and b) are right.. so I'll get to my question..
    both graphs are a concave up parabola.. neither of which go below 0 on the y-axis

    I understand that the 1st derivative, related to the graph of the original function..
    where the original function was decreasing.. the 1st derivative graph is below 0 on the y-axis.
    Where the original function was increasing.. the 1st derivative graph is above 0 on the y-axis

    Now my understanding of the 2nd derivative's graph is somewhat shaky.. but from what I know.. I believe it to be..
    where the original function is concave up.. the 2nd derivative graph is above 0 on the y-axis
    where the original function is concave down.. the 2nd derivative graph is below 0 on the y-axis

    So it would make sense that the 2nd derivative graph for both a) and b) are horizontal lines at y=18 and y=10, respectively. Because both graphs are never concave down.. and a slanted line would eventually go below 0 on the y-axis.

    MY QUESTION.. is why y=18 and y=10 is there some connection from the original graph to these numbers I'm not seeing? or is it as simple as 'thats just what the 2nd derivative works out to be'

    Thanks so much to anyone who can shed some light on this!
     
  2. jcsd
  3. Mar 26, 2012 #2


    They use [itex]\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}[/itex]

    [itex](x+4)^2=x^2+8x+16[/itex]
     
  4. Mar 26, 2012 #3
    where does the 11 come from?
     
  5. Mar 26, 2012 #4
    [itex]u'v-uv'=?[/itex]
     
  6. Mar 26, 2012 #5
    What do u and v represent?
     
  7. Mar 26, 2012 #6
    Well, in your case, [itex]u=2x-3[/itex] and [itex]v=x+4[/itex].

    Look at this formula again [itex]\left(\frac{u}{v}\right)^′=\frac{u^′v−uv^′}{v^2}[/itex] and at [itex]\left(\frac{2x-3}{x+4}\right)'[/itex].
     
  8. Mar 26, 2012 #7
    oh wow okay.. I just worked my answer of g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)
    through and got the same answer of g'(x) = 11/(x^2+8x+16)

    I see what you mean!
    so is the second answer (in red) better than my other answer? obviously they are equivalent but you can see I got mine from using the product rule

    Ill definitely keep you're rule in mind! AHA the quotient rule I just looked it up! wicked..
    but yeah which should I use for my final answer? :S

    Also any thoughts on my second derivative answer?
     
  9. Mar 26, 2012 #8

    SammyS

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    For your first derivative, use a common denominator.

    [itex]\displaystyle g'(x)=2(x + 4)^{-1} – (2x – 3)(x + 4)^{-2}=\frac{2}{x + 4}-\frac{2x – 3}{(x + 4)^{2}}=\dots=\frac{11}{(x + 4)^{2}}[/itex]
     
  10. Mar 26, 2012 #9
    yes thank you SammyS :) so far I'm at caught up with you to this point
    so I'm thinking that g'(x) = 11/(x^2+8x+16

    is a better answer than

    g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)

    yes?

    Also any thoughts on my second derivative answer?


    g’’(x) = -2(x + 4)^(-2) – 2(x + 4)^(-2) +(–4x + 6)(x + 4)^(-3)


    that was taken from my
    g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)

    I'm thinking it might be way off..
    what do you think?

    even if thats way wrong..
    I was a little confused on how to take the derivative from 2(x+4)^-1
    just for my knowledge.. is the derivative of 2(x+4)^-1

    -2(x+4)^(-2)

    im just confused as to what happens with the 2 outside the bracket.. I'm thinking it stays?
     
  11. Mar 26, 2012 #10

    Mark44

    Staff: Mentor

    You started with a coefficient of 2, and multiplied it by -1, resulting in -2.
     
  12. Mar 26, 2012 #11
    yes, so I did it right then?
     
  13. Mar 26, 2012 #12

    Mark44

    Staff: Mentor

    Yes.

    d/dx( 2(x + 4)-1) = -2(x + 4)-2.
     
  14. Mar 26, 2012 #13

    SammyS

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    Now that you have that [itex]\displaystyle g'(x)=\frac{11}{(x + 4)^{2}}\,,[/itex] why not find g''(x) using that ?
     
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