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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)
• @ 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?
(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)
• at 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?
(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 ? at
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)
• at 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)
(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.