Current time:0:00Total duration:7:35

0 energy points

# Antiderivative of x^-1

Can you find a function whose derivative is 1/x? Created by Sal Khan.

Video transcript

What I wanna do in this video is think about the anti-derivative of one over x,
or another way of thinking about it, another way of
writing is the anti-derivative x to the negative
one power. And we already know if we somehow tried to apply that anti power rule, that inverse
power rule over here we would get something that's
not defined, we would get x to the zero over zero. Doesn't make any sense. And you might have been saying okay, well
I know what to do in this case. When we first learned about derivatives,
we know that the derivative, let me just put this in
yellow, the derivative with respect to x of the
natural log of x is equal to one over x. So why can't we just say that the
anti-derivative of this right over here is equal to the natural log of x
plus c. And, this isn't necessarily wrong, the
problem here is that it's not broad enough. And, when I say it's not broad enough, is
that the domain over here, for our original function that we're taking the
anti-derivative of, is all real numbers except for x equals 0. So over here x cannot be equal to zero. While the domain over here is only
positive numbers. So, over here, x, so for this expression,
x has to be greater than zero. So, it would be nice if we could come up
with an anti-derivative that has the same domain as the function
that we're taking the anti-derivative of. So, it would be nice if we could find an
anti-derivative that is defined everywhere that our original function is,
so pretty much everywhere except for x
equaling zero. So, how can we rearrange this a little
bit, so that could be defined for negative
values as well? Well one, one possibility is to think
about the natural log of the absolute value of X, the natural log of the
absolute value of X. So I will put a little question mark here
because we don't know really know what the derivative of
this thing is going to be. I am not going to rigorously prove it here
but I'll, I will give you kind of the
conceptual understanding. So to understand it, let's plot, let's
plot the natural log of x. And I had done this ahead of time. So that right over there is roughly what
the graph of the natural log of x looks like. So what would the natural log of the absolute value of x, is going to look
like? Well for positive xs, for positive xs it's
gonna look just like this. For positive Xs you take the absolute
value of it, It's just the same thing as taking that
original value. So it's gonna look just like that for
positive xs. But now this is also gonna be defined for
negative xs. If you're taking the, the absolute value
of negative one, that evaluates to just one, so it's just the
natural log of one. So you're gonna be right there. As you get closer and closer and closer to
zero from the negative side, you're just gonna
take the absolute value. So it's essentially going to be exactly
this curve for natural log of x. But the left side of the natural log of
the absolute value of x is going to be its mirror image if you were
to reflect around the y axis. It's gonna look something like this. It's gonna look something like this. So what's nice about this function is you
see it's defined everywhere. It's defined everywhere except for, except
for, I'm trying to draw it symmetrically as possible, is defined
everywhere except for x equals zero. So, if you combine this pink part and, and
this part on the right, if you combine both of
these, you combine both of these, you get, you get y
is equal to the natural log of the absolute
value of x. Now, let's think about its derivative. Well, we already know what the derivative
of the natural log of x is. And for positive values of x. So let me write this down. For x is greater than zero, we get the
natural log of the absolute value of x is equal to the
natural log of x, let me write this. Is equal to the natural log of, is equal
to the natural log of x. And we would also know, since these two
are equal for x is greater than zero, for x is greater
than zero, the derivative. The derivative of the natural log of the
absolute value of x, is going to be equal to the derivative, is
going to be equal to the derivative of the natural log of x, the
natural log of x, which is equal to, which is equal to 1 over x, for
x greater than 0. So let's plot that. Let's plot that. I'll do that in gradients, equal to 1 over
x. So 1 over x, we've seen it before. It looks something like. It looks something like this. So let me, my best attempt to draw it as
both vertical and horizontal isotopes. So it looks something like this, it looks
something like this. So this right over here is 1 over x for x
is greater than 0. So this is 1 over x when x is greater than
0. So all it's saying here and you can see it pretty clearly, is the slope, the slope
right over here. The slope of the tangent line is 1 and so
you see that when you look at the derivative slope right over here, the derivative should be equal 1
here. When you get close to zero you have a
very, very, steep positive slope here. And so you see you have a very high value
for its derivative. And then as you can move away from zero,
it's still steep, it's still steep, but becomes less and less and less steep,
all the way until you get to one. And then, and then it gets, and then it keeps getting less, and less and less
steep. But it never quite gets to an absolutely
flat slope. And that's what we see its derivative
doing. Now what is the natural log of absolute
value of x doing right over here? When we are out here, when we're out here
our slope is very close to 0. It's symmetric. The slope here is essentially the negative
of the slope here. I could do it maybe clearer showing it
right over here. Whatever the slope is right over here, whatever the slope is right over there,
it's the exact negative of whatever the slope
is at a symmetric point on the other side. So for the other side the slope is right
over here, over here it's going to be the negative of that, so it's going to be
right, it's going to be right over there. And then the slope just gets more and more
and more negative. Right over here the slope. Over here, the slope is a positive one. Over here, it's going to be a negative
one. So, right over here, our slope is a
negative one. And then as we get closer and closer to
zero, it's just gonna get more and more and more
negative. So, the derivative of the natural log of
the absolute value of x, where x is less than zero, looks
something like this. Looks like this. And you see, you see, and it's once again
it's not a ultra rigorous proof. But what you see is, is that the
derivative of the natural log of the absolute value of x is equal to 1 over x
for all xs not equaling 0. So what you're seeing and hopefully you get the, you can visualize that the
derivative. Let me write it this way. The derivative, the derivative of the
natural log of the absolute value of x is indeed equal
to one over x, for, for all, for all x does not equal
zero. So this is a much more satisfying
anti-derivative for one over x. It has the exact same, has the exact same
domain. So when we think about what the
anti-derivative is for one over x. And I didn't do a kind of rigorous proof
here. I didn't use the definition of the
derivative and all of that. But I kinda gave you a visual
understanding hopefully of it. We would say it's the natural log of the
absolute value of x plus c, and now we have an anti-derivative that has the
same domain as the function that we're taking the
anti-derivative of.