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### Course: Class 12 math (India) > Unit 6

Lesson 6: Proofs for the derivatives of eˣ and ln(x)# Proof: d/dx(ln x) = 1/x

Taking the derivative of ln x. Created by Sal Khan.

## Want to join the conversation?

- 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?(18 votes)
- 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.(33 votes)

- @7:52Sal 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?(13 votes)
- 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.(13 votes)

- at8:00Sal 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?(12 votes)
- 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.(11 votes)

- 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.(2 votes)
- 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.(12 votes)

- Why can i put ln in front of the limit ? at8:00

Thanks(6 votes)- why can i put ln infront of ln? Is is just because that u is not in 1/x and limit cant influe it?(1 vote)

- Wouldn't it be the limit as xu approaches 0--not as u approaches 0??(4 votes)
- 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?(5 votes)
- 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.(4 votes)

- at3:10how 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)(5 votes)- 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.(3 votes)

- why we use natural logarithm instead of common or another logarithm in calculus. please make a video mentioning the reason behind it .(3 votes)
- 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 ](3 votes)

- 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?(2 votes)
- 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.(4 votes)

## Video transcript

I'm now going to show you what
I think are probably the two coolest derivatives
in all of calculus. And I'll reserve that. None of the other ones have
occurred to me right now. But these are definitely to
me some of the neatest. So let's figure out
what the derivative of the natural log is. And just as a review,
what is the natural log? Well the natural log of
something is the exact same thing as saying logarithm
base e of that something. That's just a review. So let's take the
derivative of this. I think I'm going to need
a lot of space for this. I'm going to try to do it
as neatly as possible. So the derivative of the
natural log of x equals-- well let's just take the definition
of a derivative, right? We just take the slope at some
point and find the limit as we take the difference between
the two points to 0. So let's take the limit as
delta x approaches 0 of f of x plus delta x. So I'm going to take the
limit of this whole thing. The natural log ln of x plus
delta x-- right, that's like one point that I'm going to
take evaluate the function-- minus the ln of x. All of that over delta x. And if you remember from the
derivative videos, this is just the slope, and I'm just taking
the limit as I find the slope between a smaller and
a smaller distance. Hopefully you remember that. So let's see if we can do some
logarithm properties to simplify this a little bit. Hopefully you remember-- and if
you don't, review the logarithm properties-- but remember that
log of a minus log of b is equal to log of a over b, and
that comes out of the fact that logarithm expressions are
essentially exponents, so they follow the exponent rules. And if that doesn't make
sense to you, you should review those as well. But let's apply this logarithm
property to this equation. So let me rewrite the whole
thing, and I'm going to keep switching colors to keep it
from getting monotonous. So we have the limit as delta x
approaches 0 of this big thing. Let's see. So log of a minus b equals log
a over b, so this top, the numerator, will equal the
natural log of x plus delta x over x. Right? a b a/b, all of
that over delta x. And so that equals the limit as
delta x approaches 0-- I think it's time to switch colors
again-- delta x approaches 0 of-- well let me just write
this 1 over delta x out in front. So this is 1 over delta x,
and we're going to take the limit of everything. ln x divided by x is 1
plus delta x over x. Fair enough. Now I'm going to throw out
another logarithm property, and hopefully you remember that--
and let me put the properties separate so you know it's not
part of the proof-- that a log b is equal to log
of b to the a. And that comes from when you
take something to an exponent, and then to another exponent
you just have to multiply those two exponents. I don't want to confuse
you, but hopefully you should remember this. So how does apply here? Well this would be a log b. So this expression is the
same thing as the limit. The limit as delta x approaches
0 of the natural log of 1 plus delta x over x to the
1 over delta x power. And remember all this
is the natural log of this entire thing. And then we're going
to take the limit as delta x approaches 0. If you've watched the compound
interest problems and you know the definition of e, I think
this will start to look familiar. But let me make a substitution
that might clean things up a little bit. Let me make the substitution,
let me call it n-- no, no, no, let me call u-- is equal
to delta x over x. And then if that's true
then we can multiply both sides by x and we get
xu is equal to delta x. Or we would also know
that 1 over delta x is equal to 1 over xu. These are all equivalent. So let's make the substitution. So if we're taking the limit is
delta x approaches 0, in this expression if delta x
approaches 0, what does u approach? u approaches 0. So delta x approaching 0 is the
same exact thing as taking the limit as u approaches 0. So we can write this as the
limit as u approaches 0 of the natural log of 1 plus-- well we
did the substitution, delta x over x is now u-- to the 1
over delta x, and that same substitution told us that's the
same thing as one over xu. Remember we're taking the
natural log of everything. And we know this is an exponent
property, which I'll now do in a different color. We know that a to the bc
is equal to a to the b to the c power. So that tells us that this me
is equal to the limit as u approaches 0 of the natural log
of 1 plus u to the 1/u, because this is one over xu, right? 1/u, and then all of
that to the 1/x. And how did I do that? Just from this exponent
property, right? If I were to simplify this, I
would have 1/x times 1/u, and that's where I get
this 1 over xu. Well then we can just do this
logarithm property in reverse. If I have b to the a I can
put that a out front. So I could take this 1/x
and put it in front of the natural log. So now what do I have? We're almost there. We have the limit
as u approaches 0. Take that 1/x, put it in front
of the natural log sign. 1/x times the natural log
of 1 plus u to the 1/u. Fair enough. When we're taking the limit as
u approaches 0, x, this term doesn't involve it at all. So we could take this out in
front, because the limit doesn't affect this term. And then we're essentially
saying what happens to this expression as the limit
as u approaches 0. So this thing is equivalent to
1/x times the natural log of the limit as u approaches
0 of 1 plus u to the 1/u. And by now hopefully you
would recognize that this is the definition. This limit comes to e, if
you remember anything from compound interest. You might remember it as the
limit-- as n approaches infinity of 1 plus
1 over n to the n. But these things
are equivalent. If you just took the
substitution u is equal to 1/n, you would get this. You would just get this. So this expression right here
is e That expression is e. So we're getting close. So this whole thing is
equivalent to 1/x times the natural log, and this
we know, this is one of the ways to get to e. So the limit as u approaches
0 of 1 plus u to the 1/u. That is e. And what is the natural log? Well it's the log base e. So you know this is equal to
1/x times the log base e of e. So that's saying e
to what power is e. Well e to the first
power is e, right? This is equal to 1. So 1 times 1/x is equal to 1/x. There we have it. The derivative of the natural
log of x is equal to 1/x, which I find kind of neat, because
all of the other exponents lead to another exponent. But all of a sudden in the mix
here you have the natural log and the derivative of that
is equal to x to the negative 1 or 1/x. Fascinating.