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Homework Help: Derivatives of Exponential and Logarithmic

  1. Feb 2, 2006 #1

    I'm having a little bit of difficulty surrounding three problems I've been working on. All fall within the logarithmic/exponential derivative/limit category...

    http://www.synthdriven.com/images/deletable/calc01.jpg [Broken]
    The problem I'm having here is with what happens to -sinx to the left of zero... it's obviously positive, but less than one. I guess it's not exactly obvious as to what to do here, so I'm having a bit of difficulty. What direction do I go in from here?

    http://www.synthdriven.com/images/deletable/calc02.jpg [Broken]
    This one, I've nearly solved. I think I'm having difficulty surrounding the behavior of ...infinity... The answer I'm given for this is -1/3, but I'm not sure how to get from the second to last step depicted to -1/3. Any pointers?

    http://www.synthdriven.com/images/deletable/calc03.jpg [Broken]
    I'm not sure at all what to do with this. Chain rule? That can't be it, can it? How would you subtract one in that instance? just cosx-1?

    Any pointers?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 2, 2006 #2


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    1. -sinx is approaching zero from somewhere above it, but just below 1, so I'm not sure why you don't just plug in 0 for -sinx in the limit and evaluate the logarithm.

    3. First rewrite as:

    [tex] y = (e^{\ln x})^{\cos x} = e^{\ln x \cos x} [/tex]
  4. Feb 2, 2006 #3
    1.) Thank you. That was pretty stupid of me not to notice that. I guess I thought it was more complicated than it was.

    3.) I was told by my instructor to logarithmically differentiate that problem, and when I begin to do so, I find that trying to differentiate that is incredibly difficult. Is there an easier way?
  5. Feb 2, 2006 #4
    Wait, no. That wasn't so difficult. It just looked complicated.

    ...Though I don't quite understand what happened with that.

    And #2?... I know I'm right all up to the last step.
    Last edited: Feb 2, 2006
  6. Feb 2, 2006 #5


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    Yes, number 3 can be done as cepheid suggested.
    You can also do it a little bit differently. One can take log of both sides and differentiate both sides with respect to x. I'll give you an example:
    Example: Differentiate xx with respect to x. You can do it in two ways:
    The first way:
    [tex]x ^ x = {(e ^ {\ln (x)})} ^ x = e ^ {x \ln (x)}[/tex]. And differentiate [tex]e ^ {x \ln (x)}[/tex] with respect to x.
    The second way:
    [tex]y = x ^ x[/tex].
    [tex]\Leftrightarrow \ln (y) = \ln (x ^ x) = x \ln (x)[/tex].
    Differentiate both sides with respect to x gives:
    [tex]\frac{y'_x}{y} = \ln (x) + 1[/tex].
    [tex]\Rightarrow y'_x = y ( \ln (x) + 1) = x ^ x (\ln (x) + 1)[/tex].
    Now you can do this example using the first way to see if it returns the same result.
    For number 2,
    This limit:
    [tex]\lim_{x \rightarrow - \infty} \frac{e ^ {2x} - e ^ {-2x}}{2e ^ {2x} + 3e ^ {-2x}}[/tex], is of the indeterminate form: [tex]\frac{\infty}{\infty}[/tex]
    If you divide both numerator and denominator by e2x (which also tends to 0, as x tends to negative infinity), then you will again get the indeterminate form: [tex]\frac{\infty}{\infty}[/tex]. Do you see why?
    So why not consider divide both numerator, and denominator by e-2x?
    Last edited: Feb 2, 2006
  7. Feb 2, 2006 #6

    Well since the natural log is not defined at 0 he cant really just evaluate it like that.
  8. Feb 2, 2006 #7


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    Yeah, I was thinking of it backwards, ln1 = 0, which is fine. But ln0 = undefined. Whoops.

    But then does the limit in 1 even exist? I mean, as x -->0-, then -sinx goes down to zero, and is already < 1, so we have progressively larger and larger negative values for the logarithm.
    Last edited: Feb 2, 2006
  9. Feb 2, 2006 #8


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    Yes, that means the limit does not exist. Although you could write:
    [tex]/lim_{x\rightarrow\0^-} ln(-sin x)= -\infty[/tex]
    which just says the limit does not exist in a particular way.

    Verd, for 2, you don't want those negative exponentials! Multiply the numerator and denominator of that last fraction by e4x, then take the limit.
  10. Feb 2, 2006 #9

    Thank you, this is extremely helpful. ...For number 2 however, I see how that'll give us the desired result, but I'm not that sure as to why. I'm a bit confused with the behavior of exponential functions when they start dealing with infinity. ...It would make sense to me that [tex]e^{-4x} \rightarrow \infty[/tex], because a double negative would make that infinity positive... correct? And in that case, it would be [tex]\infty - 1[/tex]... I'm not quite sure how to get to -1/3. It would make the most sense for the [tex]e^{-4x}[/tex] to then equal 0...?

    Also, why divide by -2x as opposed to 2x??
    Last edited: Feb 2, 2006
  11. Feb 2, 2006 #10


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    This is correct.
    Define f(x) := ax.
    If 0 < a < 1, then f(x) is decreasing.
    If a > 1, then f(x) is increasing.
    Do you know this?
    Since e > 1. So f(x) := ex is increasing.
    If you graph that function, you will see that ex tends to 0, as x tends to negative infinity, and ex tends to positive infinity, as x tends to positive infinity.
    So you will have:
    [tex]\lim_{x \rightarrow + \infty} e ^ x \rightarrow \infty[/tex]
    [tex]\lim_{x \rightarrow - \infty} e ^ x = 0[/tex]
    We also have:
    [tex]\lim_{x \rightarrow - \infty} e ^ {-x} \rightarrow \infty[/tex]
    [tex]\lim_{x \rightarrow + \infty} e ^ {-x} = 0[/tex]

    I wouldn't say [tex]\infty - 1[/tex] really, since [tex]\infty[/tex] is not a number. In fact [tex]\infty - 1[/tex] is still [tex]\infty[/tex] :tongue:
    [tex]\lim_{x \rightarrow - \infty} e ^ {-4x} \rightarrow \infty[/tex], it tends to infinity, not 0!
    I remember telling you to divide both numerator, and denominator by e-2x, instead of e2x. Because at first, you get the indeterminate form [tex]\frac{\infty}{\infty}[/tex] if you divide both numerator, and denominator by e2x (which tends to 0 as x tends to negative infinity), so you will get:
    [tex]\frac{\frac{\infty}{0}}{\frac{\infty}{0}} \rightarrow \frac{\infty}{\infty}[/tex], again, you have the indeterminate form [tex]\frac{\infty}{\infty}[/tex], which is something that you obviously don't want, right?
    But if you divide everything by e-2x, you will get:
    [tex]\lim_{x \rightarrow - \infty} \frac{e ^ {2x} - e ^ {-2x}}{2e ^ {2x} + 3e ^ {-2x}} = \lim_{x \rightarrow - \infty} \frac{\frac{e ^ {2x}}{e ^ {-2x}} - \frac{e ^ {-2x}}{e ^ {-2x}}}{\frac{2e ^ {2x}}{e ^ {-2x}} + \frac{3e ^ {-2x}}{e ^ {-2x}}} = \lim_{x \rightarrow - \infty} \frac{e ^ {4x} - 1}{2e ^ {4x} + 3}[/tex]
    Hopefully, you can go from here, right? :smile:
    By the way, have you done number 1?
    Whoops, I didn't notice you've editted your post.
    To answer your last question:
    When you see the form [tex]\frac{\infty}{\infty}[/tex], you'd want to get rid of the [tex]\infty[/tex] thingy, right? To do that, you must divide the numerator, and denominator by something that also tends to infinity. In this case e-2x is the thing that causes the infinity in both numerator, and denominator, and you must get rid of that completely, so you divide everything by e-2x. Do you get this? :)
    Last edited: Feb 2, 2006
  12. Feb 2, 2006 #11
    I got that far, and that's actually where I'm confused, heheh. When you take the limit, you get infinity over infinity, don't you? e^-4x isn't going to come out to zero, so it wouldn't be -1/3, as the answer is advertised to be... (Confused!)

    As for Number 1, I understand that anything less than 0 DNE, correct?
  13. Feb 2, 2006 #12


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    :tongue: Look at my post again:
    [tex]\lim_{x \rightarrow - \infty} \frac{e ^ {2x} - e ^ {-2x}}{2e ^ {2x} + 3e ^ {-2x}} = \lim_{x \rightarrow - \infty} \frac{\frac{e ^ {2x}}{e ^ {-2x}} - \frac{e ^ {-2x}}{e ^ {-2x}}}{\frac{2e ^ {2x}}{e ^ {-2x}} + \frac{3e ^ {-2x}}{e ^ {-2x}}} = \lim_{x \rightarrow - \infty} \frac{e ^ {4x} - 1}{2e ^ {4x} + 3}[/tex]. Ohh, I feel so guilty for feeding you the answer not once but twice! :smile:
    It's not e-4x, it's e4x (which tends to 0, as x tends to negative infinity.)
    If you divide everything by e2x, you'll get e-4x, but if you divide everything by e-2x, you will get e4x. :wink:
    Look again at the above post for why you should divide by e-2x, and not e2x. :wink:
    What do you mean by that?
    A limit does not exists if it tends to infinity (positive, or negative).
    In your question:
    [tex]\lim_{x \rightarrow 0 ^ -} \ln (\sin (-x))[/tex]. sin(-x) will tend to 0 (but positive, or may I say 0+) when x tends to 0 from the left.
    Look at the graph of ln(x), one can say that:
    [tex]\lim_{x \rightarrow 0 ^ +} \ln (x) \rightarrow - \infty[/tex]
    Just graph it, and you'll see. From what I write above, what's:
    [tex]\lim_{x \rightarrow 0 ^ -} \ln (\sin (-x)) = ?[/tex] do you think?
    Last edited: Feb 2, 2006
  14. Feb 2, 2006 #13
    Haha! I see my mistake! Thank you!
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