The derivative of natural logs

[bobsmith76]

Homework Statement

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

 

[bobsmith76]

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

 

[HallsofIvy]

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.

 

[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.

 

[vela]

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

u = 1 - y
du = -1

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

so then you have

du = -1dx

How did x make an appearance here?

divide both sides by -1

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

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

 

[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.

 

[SammyS]

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.

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.

 

[vela]

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.

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
$$-\int \frac{du}{u} = -\int \frac{dy}{y} = -\ln |y|$$

 

[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.

 

[bobsmith76]

I'm still lost on this one.

 

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

 

[bobsmith76]

thanks, now I got it.