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# Nachylenie siecznej do wykresu funkcji, przechodzącej przez dowolny punkt

## Transkrypcja filmu video

A curve has the equation y equals the natural log of x, and passes through the points P equals e comma 1. And Q is equal to x natural log of x. Write an expression in x that gives the slope of the secant joining P and Q. So I think I'm going to need my little scratch pad for this one right over here. So this is the same question over again. Now let's just try to visualize this curve right over here. So let me draw my axes. So let's say this is my y-axis. This is my y-axis, and then this is my x-axis. This is my x-axis right over here. That's my x-axis. And natural log of x-- so let's think about it a little bit. The natural log of 0, e to what power is equal to 0? Well, that's going to be 1. So you're going to have the point 1, 0 on this. So that's 1 right over there. And then the natural log of smaller and smaller numbers as we approach 0 is going to get more and more negative, all the way going down to negative infinity. So this curve is going to look something like this. It's going to look something like that. And we know it also has the point e comma 1 on it. So it has a point, the natural log of e. So if that's 1, that'll be about 2. That'll be about 3. So that is e right over here, roughly e, right over here. And that's the point e comma 1. So that's the point e comma 1 right over there. And I'll just mark that as P there to give us a label for it. So that is P. And we want to find the secant line between P and an arbitrary Q for any given x. The y value is going to be natural log of x. So let's say that this is our Q right over here. So let's say this right over here is our-- actually, I want to make it clear that it is a, I'll do it our here. I'm being a little indecisive. So let's say that is our Q right over there. And that's the point x natural log, natural log of x. And we want to find the slope of the secant line joining these two points. So this, the slope of this line, I want to try to make it so it doesn't look tangent, so it's secant. So you see it cuts through the curve. So that's the secant line right over there. It intersects the curve at P and at Q. So I want to find the slope of that secant line. Well, to find the slope of the secant line, I just need to find the change in y and the change in x between these two points. So what's the change in-- so let's be clear here. This is the point, this is when x is equal to-- well, it's just a kind of arbitrary x. And this right over here is the point lnx. So what is our change in x? Our change in x-- this value right over here, our change in x-- is just going to be x minus e is equal to x minus e. And what is our change in y? Our change in y is going to be lnx is going to be the natural log of x minus 1, minus 1. That's this distance right over here. So the slope of this line, the line that contains both of these points, the slope-- I could write m for slope-- is going to be our change in y over our change in x, which is equal to lnx minus 1 over x minus e. And now we can just input that and make sure that we got that right. So let me try to remember it-- lnx minus 1 over x minus e. So let me input that. So we have the natural log of x minus 1 over-- and it interprets it nicely for us right below, so we make sure that we're doing over x minus e. And now let us check our answer. And we got it right.