# The derivative of natural logs

1. Feb 14, 2012

### bobsmith76

1. The problem statement, all variables and given/known data

I don't see why it's -LN it should be positive LN

2. Feb 14, 2012

### bobsmith76

never mind. it's the reciprocal rule in action. ln 1/x = -ln x

3. Feb 14, 2012

### HallsofIvy

Staff Emeritus
No, that's not the reason- there is no ln|1/1-y| involved. To integrate dy/(1- y), make the substitution u= 1- y. What is du? The - in the integral is because of the -y.

4. Feb 14, 2012

### bobsmith76

I don't follow you, if you use the substitution rule

u = 1 - y
du = -1

so then you have

du = -1dx

divide both sides by -1

du/-1, so when you integrate you would get 1/-y which is -LN |y| not -LN |1-y|

In any case, here is a similar problem:

both LN's should share the same sign. one should not be positive and the other negative.

5. Feb 14, 2012

### vela

Staff Emeritus
That's wrong. You can't have just one differential there.

How did x make an appearance here?

Show your work here. You seem to be jumping to conclusions and making mistakes.

6. Feb 14, 2012

### bobsmith76

the derivative of 1 - y is -1. what do you think it is?

whenever you take the derivative of a term I thought you always multiply it by dx

"Show your work here. You seem to be jumping to conclusions and making mistakes."

I did show my work. If you say I'm making a mistake then please point this out. It does me no good to say I'm making a mistake and not show how or why. Of course I know I'm wrong, I'm just trying to figure out how.

7. Feb 14, 2012

### SammyS

Staff Emeritus
If u = 1 - y, then the derivative of u (with respect to y) is -1. That is correct.

What you're being told (and you seem to not understand) is that if du/dy = -1 , then
du = -dy​

There's no x involved. no differential stands alone on one side of the equation w/o there being a differential on the other side.

8. Feb 15, 2012

### vela

Staff Emeritus
It's hard to show you where you're making mistakes if you don't clearly show what you did. We can't read your mind, and we can only guess what you did if you don't show every step.

From what you've written, this is what I have so far. You start with
$$\int \frac{dy}{1-y}$$ Now you let u=1-y, then got dx = du/-1. That's what I assume you meant when you wrote, "du/-1". Then you continued, "so when you integrate you would get 1/-y", by which I took to mean you claim
$$-\int \frac{du}{u} = \frac{1}{-y},$$ which is wrong. How did you get that? Then somehow turned that into "-LN |y|". How did you do that?

Or perhaps you didn't really integrate when you said you did. Instead, after doing the substitution, you arbitrarily turned all the u's back into y's and said
$$-\int \frac{du}{u} = -\int \frac{dy}{y} = -\ln |y|$$

9. Feb 15, 2012

### bobsmith76

The derivative of ln x = |1/x|

If we use u substitution then dy/(1-y) = dy/u

When we take the derivative of dy I think dy (I'm still not sure, I find the notation of derivatives very confusing) becomes 1, so we have

1/u or 1u-1

Now, I don't know what to do because if I integrate 1/u that will give ln u or ln |1-y|, not

-ln |1-y|

I still don't understand where the negative comes from.

10. Feb 16, 2012

### bobsmith76

I'm still lost on this one.

11. Feb 16, 2012

### I like Serena

With $u=1-y$, it follows that $y=1-u$.
Taking the derivative on both sides, we get $dy=-du$.
Substituting, we replace $(1-y)$ by $u$, and $dy$ by $-du$.

So:
$$\int {dy \over 1-y} = \int {-du \over u} = -\ln |u| + C$$

Now we can back substitute $u=1-y$, giving:
$$\int {dy \over 1-y} = -\ln |u| + C = -\ln |1-y| + C$$

12. Feb 16, 2012

### bobsmith76

thanks, now I got it.