Derivatives of Exponential and Logarithmic

[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.)
http://www.synthdriven.com/images/deletable/calc01.jpg
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.)
http://www.synthdriven.com/images/deletable/calc02.jpg
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.)
http://www.synthdriven.com/images/deletable/calc03.jpg
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?

 

[cepheid]

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:

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

 

[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?

 

[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.

 

[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 x^x 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.
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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 e^2x (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?

 

[d_leet]

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.

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

 

[cepheid]

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

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.

 

[HallsofIvy]

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 e^4x, then take the limit.

 

[verd]

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 x^x 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 e^2x (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?

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??

 

[VietDao29]

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?

This is correct.
Define f(x) := a^x.
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) := e^x is increasing.
If you graph that function, you will see that e^x tends to 0, as x tends to negative infinity, and e^x 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:
$$\lim_{x \rightarrow - \infty} e ^ {-x} \rightarrow \infty$$
$$\lim_{x \rightarrow + \infty} e ^ {-x} = 0$$

And in that case, it would be $$\infty - 1$$

I wouldn't say $$\infty - 1$$ really, since $$\infty$$ is not a number. In fact $$\infty - 1$$ is still $$\infty$$

... 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...?

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 e^2x. Because at first, you get the indeterminate form $$\frac{\infty}{\infty}$$ if you divide both numerator, and denominator by e^2x (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.
To answer your last question:

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

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? :)

 

[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?

 

[VietDao29]

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 e^4x (which tends to 0, as x tends to negative infinity.)
If you divide everything by e^2x, you'll get e^-4x, but if you divide everything by e^-2x, you will get e^4x.
Look again at the above post for why you should divide by e^-2x, and not e^2x.

As for Number 1, I understand that anything less than 0 DNE, correct?

What do you mean by that?
A limit does not exists if it tends to infinity (positive, or negative).
In your question:
$$\lim_{x \rightarrow 0 ^ -} \ln (\sin (-x))$$. 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?

 

[verd]

Haha! I see my mistake! Thank you!