Name: Proof: d/dx(ln x) = 1/x
Uploaded: 2011-02-20T16:32:21Z
Description: Taking the derivative of ln x

Question 1

Some sources give this identity as d/dx(ln[abs x])=1/x.  Is this correct, and if so, why is the derivative only value for the log of the absolute value of x?

Accepted Answer

Here's a rigorous proof:
Suppose x > 0.
Then ln|x| = ln x and d/dx ln x = 1/x
Suppose x < 0.
Then ln|x| = ln(-x) and d/dx ln(-x) = -1*1/(-x) (chain rule) = 1/x
QED

Anyway what's useful about d/dx ln|x| is that it allows you to take the indefinite integral of 1/x for all non-zero values of x, which is later on in the playlist.

Question 2

@7:52 Sal says that 1/x is taken out of the limit problem because you're taking the limit as u-> 0 but, you can only take something out of the equation if it's a constant.... wouldn't the x just make it a multivariable problem instead?

Accepted Answer

There is a subtle confusion in your question. Let me follow your line of thought first. To take the 1/x out of the limit expression, he could have done one of two things: 
1) After substituting u, kept limit as deltaX -> 0. The substitutions are still valid, the limit of u as deltaX->0 is still zero. Pull 1/x out of the limit, and THEN make the change to lim u->0. 
2) He could have unpacked every u back into terms of x, extracted 1/x, repacked it all back into u, and proceeded. 

Remember, x is going to stay x before and after you take that limit. If it's unaffected, it's unaffected. 

Now, if we had started with lim u->0, and were told u was a function of x and Dx, it could be unjustified to pull 1/x out of the limit, because we might not know the effect. 
However, in this example, the substitution is known. the limit as u->0 is the same thing as saying the limit as Dx->0. He is expecting you to make this reference in your head, for simplicity. 

The confusion is about variables and constants. let Dx be any size. Now, change it, and don't touch anything else. This is exactly the same thing as saying, "hold all other variables constant". And that is what this limit expression does. 

In sum: This is a multivariable equation, its two variables are x and delta-x. The limit expression relates the continuous range of (constant) values x=a, f(a) by way of the additional variable Delta-x. We take the limit as delta vanishes, leaving us with a continuous relation dy/dx. Holding x constant over a continuous range, we compute the complete derivative expression at once. 

Does that help at all? I want to emphasize that you are correct: we must guarantee that x varies independently from Dx, or it would be impossible to remove (an expression in x) outside the limit. The demonstration that Dx can be given as a fully independent variable is tedious, but available.

Question 3

at 8:00 Sal shows that lim u->0 of ln (Z)=  ln (lim u->0 of Z). Is this a property of limits that I don't remember?

Accepted Answer

It's the property of limits having to do with continuous functions. If f is continuous, then lim_{x->c}f(g(x)) = f(lim_{x->c}g(x)). basically you can move the limit inside a continuous function.

Question 4

At 8.45 Sal said that if you just took the u substitution U = 1/n  you would get the common expression for e.  My question is can you do that. I mean can you just make any value equal to another value? For this proof Sal already made U = deltaX/X. I don't know my mathematics very well and that's probably why I don't understand this but I'm still going to say that substitution stuff seems quite sneaky. So - when using substitution you always have to back substitute - if you do that just before Sal makes that expression = to e the original deltaX --> 0 becomes deltaX/x ---> 0    I just back substituted deltaX/x in for U. If you do that and take the lim as deltaX/x--->0 you get (1/x)(1)^(x/deltax) which would make this problem undefined if deltax goes to zero or even U.  Now the real important questions, can Sal make that expression equal to e before evaluating the back sub. If he were to back sub the value of U and evaluate the limit it would be undefined. Thoughts?

Accepted Answer

Yeah, Sal is kind of sneaky in the way that there are random variables that "pop up" and changes the whole final answer. This is why some of his videos on calculus are confusing. I think that what you are doing is possible, but there already is about a thousand proofs of using this delta x format for different variables. You could try this, but it would get limited to zero, making no difference to the overall result. There was this problem that you could sub in a term for a set of variables which was very useful for me, since it could be applied to other problems, but itʻs hard to sub in a term with a basic proof.

Question 5

at 3:10 how did we get exactly  1/delta x * ln(1+ delta x/x)??
i'd like to see more detailed way of getting 1/delta x * ln(1+ delta x/x)

Accepted Answer

a/b is the same thing as 1/b * a. In this case, the "a" is a complicated looking thing, but the rule still works. the "b" is Δx.

Question 6

why we use natural logarithm instead of common or another logarithm in calculus. please make a video mentioning the reason behind it .

Accepted Answer

The math is much, much easier if you use the natural logarithm. That's the reason. Looking at the definition of e^x gives us:
e^x = lim h→0 (1+h)^(x/h)

It naturally follows that its inverse function is:
ln x = lim h→0 [x^(h) - 1] /h

If you apply either of these to the definition of a derivative, the math is quite easy. So, since the derivative of e^x is so simple, we wouldn't want to take the derivative of any other exponential base if we can avoid it. Likewise, the natural log has a very simple derivative, so we wouldn't want to take the derivative using any other base if we can avoid it.
 
However, if needs be, we certainly can take the derivative of a log or exponential of other bases. They're just harder.
d/dx (a^x) = ln (a) ∙ a^(x) 
d/dx logₐ x = 1/[x ln a ]

Question 7

In the end, I don't understand why the limit of something approaching infinity (definition of e) would be the same as the limit of basically the same thing as it approaches zero.

Accepted Answer

I think you misunderstood what he said.  The two main ways (they are mathematically the same thing) of defining e are:
e = lim n→ 0  (1+n)^(1/n)
Now, let us define k as k = 1/n
Let us replace n with 1/k (remember to convert what the limit is approaching)

e = lim k → inifinity  (1 + 1/k)^(k)
This is the other standard definition of e.

In both cases, you are taking 1 adding something extremely close to 0 and then taking that sum to a power near infinity.  So, they are mathematically the same thing.

Question 8

Are we literally stretching out, bending the rules a little bit or going around the fact that you cannot have 0 in the denominator in this proof or am I just seeing things? Seems like he invented a variable U just to avoid division by zero.  Which is undefined. From what I can read, what he's actually doing 1/delta(x)* x/x and then storing one the x of the nominator into the x/delta(x) and calling it 1/u. This is a pretty neat proof. Quite tricky and inventive though. In hindsight, I've realized that as delta x approaches zero, so does u, so 1/u is still 1/0. So why use u? For simplicity sake?

Accepted Answer

A limit is NOT the same thing as evaluating the function at the limiting value. Instead, it is a means of measuring what the function is getting close to as the variable gets closer and closer to the limiting value. Since the limit is NOT being evaluated at the exact point where there would be division by zero, then there is not a problem.

Course: Class 12 math (India) > Unit 6

Proof: d/dx(ln x) = 1/x

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