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Tangent Line problem

  1. Sep 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Ok, how do i solve this question
    I've attached the picture.. The answer is A but i dont understand how to get that...

    3. The attempt at a solution

    The solution is also given , but i dont understand the adding and subtracting 5. the Derivative of

    y=ln (e^x + 5) y'= 1/ ( e ^ x + 5) , where is the extra e^x coming from in the first line of the solution. Also i already mention the adding and substracing 5.. I have no idea whats going on .. PLz Math guru's plz help me!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 23, 2008 #2

    CompuChip

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    You have to use the chain rule: y = ln(u) with u = e^x + 5; then y' = dy/du * du/dx

    For the rest of your question we'll have to wait to see the attachments.
     
  4. Sep 23, 2008 #3

    HallsofIvy

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    As CompuChip said, the "extra" ex is from the chain rule:
    ln(ex+ 5)= ln(u) with u= ex+ 5. dy/dx= (dy/du)(du/dx)
    dy/du= d(ln(u))/du= 1/u and du/dx= d(ex+ 5)/dx= ex
    dy/dx= (dy/du)(du/dx)= (1/u)(ex)= (1/(ex+5))ex)= ex/(ex+ 5).

    The first attachment asks what happens to the slope of the tangent lines to y= ln(ex[/sup+ 5) as x increases and answers that it increases to a limit of 1.

    The second explains that answer by noting that y'= ex/(ex+ 5), as above, and then does the following:
    1) add and subtract 5
    [tex]\frac{e^x}{e^x+ 5}+ 5- 5[/tex]
    which is incorrectly shown!

    Probably whoever did the "art" did not understand what was meant. It should be
    [tex]\frac{e^x+ 5- 5}{e^x+ 5}[/tex]
    with the "5, added and subtracted" in the numerator of the fraction. Of course, since 5- 5= 0, those are equal. Now separate into two fractions
    [tex]\frac{e^x+ 5}{e^x+ 5}- \frac{5}{e^x+ 5}= 1- \frac{5}{e^x- 5}[/tex]
    As x increases without bound, so does ex and so does the denominator of the second fraction: the fraction goes to 0 so the entire expression goes to 1.

    I would have done it a slightly different way, myself. From the original expression,
    [tex]\frac{e^x}{e^x+ 5}[/tex]
    instead of adding and subtracting 5, divide both numerator and denominator by ex:
    [tex]\frac{\frac{e^x}{e^x}}{\frac{e^x+ 5}{e^x}}= \frac{1}{1+ 5e^{-x}}[/tex]
    Now, as x increases without bound, 5e-x goes to 0 so that fraction goes to 1.
     
  5. Sep 23, 2008 #4

    CompuChip

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    And a final way would probably to use L'Hopitals rule, but I will not explain that unless the OP requests it because HallsofIvy shows the most common ways, apart from some general techniques with fractions which are quite important.
     
  6. Sep 23, 2008 #5
    Thankyou so much Guys.. I think i got the idea now.. BTW Compuchip , whats this L'Hopital rule and how is it applied to this question. I'm curious to find out.. Is it a faster method for doing this problem?

    Thanks in Advance.. You guys are awesome
     
  7. Sep 23, 2008 #6

    CompuChip

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    L'Hopitals rule can be used if you have to evaluate a limit
    [tex]\lim_{x \to a} \frac{f(x)}{g(x)}[/tex]
    where f and g separately tend to either 0 or (plus or minus) infinity. In other words, if just plugging in x = a would produce something like 0/0, or [itex]\infty / \infty[/itex] (or [itex]\infty / -\infty[/itex], [itex]-\infty / \infty[/itex] or [itex]-\infty / -\infty[/itex]).

    In this case, you may calculate
    [tex]\lim_{x \to a} \frac{f'(x)}{g'(x)}[/tex]
    and this is equal to original limit.

    For your current problem, e^x blows up as x goes to infinity, so the limit would become of the form [itex]\infty / \infty[/itex].
    Therefore, by L'Hopitals rule,
    [tex]\lim_{x \to \infty} \frac{e^x}{e^x + 5} = \lim_{x \to \infty} \frac{(e^x)'}{(e^x + 5)'} = \lim_{x \to \infty} \frac{e^x}{e^x} = \lim_{x \to \infty} 1 = 1.[/tex]

    However, I suggest one of the methods HallsofIvy has shown in this case.
     
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