# Derivatives of Exponential and Logarithmic

1. Feb 2, 2006

### verd

Hi,

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...

1.)

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?

2.)

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?

3.)

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?

2. Feb 2, 2006

### cepheid

Staff Emeritus
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:

$$y = (e^{\ln x})^{\cos x} = e^{\ln x \cos x}$$

3. Feb 2, 2006

### verd

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?

4. Feb 2, 2006

### verd

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
5. Feb 2, 2006

### VietDao29

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:
$$x ^ x = {(e ^ {\ln (x)})} ^ x = e ^ {x \ln (x)}$$. And differentiate $$e ^ {x \ln (x)}$$ with respect to x.
The second way:
$$y = x ^ x$$.
$$\Leftrightarrow \ln (y) = \ln (x ^ x) = x \ln (x)$$.
Differentiate both sides with respect to x gives:
$$\frac{y'_x}{y} = \ln (x) + 1$$.
$$\Rightarrow y'_x = y ( \ln (x) + 1) = x ^ x (\ln (x) + 1)$$.
Now you can do this example using the first way to see if it returns the same result.
-------------
For number 2,
This limit:
$$\lim_{x \rightarrow - \infty} \frac{e ^ {2x} - e ^ {-2x}}{2e ^ {2x} + 3e ^ {-2x}}$$, is of the indeterminate form: $$\frac{\infty}{\infty}$$
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: $$\frac{\infty}{\infty}$$. Do you see why?
So why not consider divide both numerator, and denominator by e-2x?

Last edited: Feb 2, 2006
6. Feb 2, 2006

### d_leet

Well since the natural log is not defined at 0 he cant really just evaluate it like that.

7. Feb 2, 2006

### cepheid

Staff Emeritus
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
8. Feb 2, 2006

### HallsofIvy

Staff Emeritus
Yes, that means the limit does not exist. Although you could write:
$$/lim_{x\rightarrow\0^-} ln(-sin x)= -\infty$$
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.

9. Feb 2, 2006

### verd

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 $$e^{-4x} \rightarrow \infty$$, because a double negative would make that infinity positive... correct? And in that case, it would be $$\infty - 1$$... I'm not quite sure how to get to -1/3. It would make the most sense for the $$e^{-4x}$$ to then equal 0...?

Also, why divide by -2x as opposed to 2x??

Last edited: Feb 2, 2006
10. Feb 2, 2006

### VietDao29

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:
$$\lim_{x \rightarrow + \infty} e ^ x \rightarrow \infty$$
$$\lim_{x \rightarrow - \infty} e ^ x = 0$$
We also have:
$$\lim_{x \rightarrow - \infty} e ^ {-x} \rightarrow \infty$$
$$\lim_{x \rightarrow + \infty} e ^ {-x} = 0$$

I wouldn't say $$\infty - 1$$ really, since $$\infty$$ is not a number. In fact $$\infty - 1$$ is still $$\infty$$ :tongue:
Noooo.
$$\lim_{x \rightarrow - \infty} e ^ {-4x} \rightarrow \infty$$, 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 $$\frac{\infty}{\infty}$$ if you divide both numerator, and denominator by e2x (which tends to 0 as x tends to negative infinity), so you will get:
$$\frac{\frac{\infty}{0}}{\frac{\infty}{0}} \rightarrow \frac{\infty}{\infty}$$, again, you have the indeterminate form $$\frac{\infty}{\infty}$$, which is something that you obviously don't want, right?
But if you divide everything by e-2x, you will get:
$$\lim_{x \rightarrow - \infty} \frac{e ^ {2x} - e ^ {-2x}}{2e ^ {2x} + 3e ^ {-2x}} = \lim_{x \rightarrow - \infty} \frac{\frac{e ^ {2x}}{e ^ {-2x}} - \frac{e ^ {-2x}}{e ^ {-2x}}}{\frac{2e ^ {2x}}{e ^ {-2x}} + \frac{3e ^ {-2x}}{e ^ {-2x}}} = \lim_{x \rightarrow - \infty} \frac{e ^ {4x} - 1}{2e ^ {4x} + 3}$$
Hopefully, you can go from here, right?
-------------
By the way, have you done number 1?
-------------
Whoops, I didn't notice you've editted your post.
When you see the form $$\frac{\infty}{\infty}$$, you'd want to get rid of the $$\infty$$ 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
11. Feb 2, 2006

### verd

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?

12. Feb 2, 2006

### VietDao29

:tongue: Look at my post again:
$$\lim_{x \rightarrow - \infty} \frac{e ^ {2x} - e ^ {-2x}}{2e ^ {2x} + 3e ^ {-2x}} = \lim_{x \rightarrow - \infty} \frac{\frac{e ^ {2x}}{e ^ {-2x}} - \frac{e ^ {-2x}}{e ^ {-2x}}}{\frac{2e ^ {2x}}{e ^ {-2x}} + \frac{3e ^ {-2x}}{e ^ {-2x}}} = \lim_{x \rightarrow - \infty} \frac{e ^ {4x} - 1}{2e ^ {4x} + 3}$$. Ohh, I feel so guilty for feeding you the answer not once but twice!
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.
Look again at the above post for why you should divide by e-2x, and not e2x.
What do you mean by that?
A limit does not exists if it tends to infinity (positive, or negative).
$$\lim_{x \rightarrow 0 ^ -} \ln (\sin (-x))$$. 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:
$$\lim_{x \rightarrow 0 ^ +} \ln (x) \rightarrow - \infty$$
Just graph it, and you'll see. From what I write above, what's:
$$\lim_{x \rightarrow 0 ^ -} \ln (\sin (-x)) = ?$$ do you think?

Last edited: Feb 2, 2006
13. Feb 2, 2006

### verd

Haha! I see my mistake! Thank you!