Secant line with arbitrary difference | Derivatives introduction | AP Calculus AB | Khan Academy

A secant line intersects the curve ( y ) equal to the natural log of ( x ) at two points with ( x ) coordinates ( 2 ) and ( 2 + h ). What is the slope of the secant line?

Well, they're giving us two points on this line. It might not be immediately obvious, but they're giving us the points when ( x ) is equal to ( 2 ) and when ( x ) is equal to ( 2 + h ). What is ( y )? Well, they tell us ( y ) is equal to the natural log of ( x ), so in this case, it is going to be ( \ln(2) ).

And when ( x ) is equal to ( 2 + h ), what is ( y )? Well, ( y ) is always going to be the natural log of whatever ( x ) is, so it's going to be ( \ln(2 + h) ). These are two points that sit on the secant line. This happens to be where the secant line intersects our curve, but these are two points on the line.

If you know two points on the line, you will then be able to figure out the slope of that line. We can just remind ourselves that a slope is just change in ( y ) over change in ( x ). What is this going to be?

Well, if we view the second one as our end point, our change in ( y ) going from ( \ln(2) ) to ( \ln(2 + h) ), so our change in ( y ) is going to be our end point, ( \ln(2 + h) ), minus our starting point or our end ( y )-value minus our starting ( y )-value, ( \ln(2) ). Our change in ( x ) is going to be our ending ( x )-value, ( 2 + h ), minus our starting ( x )-value, ( 2 ).

Of course, these twos cancel out. If we look here, it looks like we have a choice that directly matches what we just wrote. This right over here is ( \frac{\ln(2 + h) - \ln(2)}{h} ).

Now, if you want to visualize this a little bit more, we could draw. We could draw a little bit. I'm going to clear this out so I have space to draw the graph just so you can visualize that this is a secant line.

Let me draw my ( y )-axis and let me draw my ( x )-axis. ( y = \ln(x) ) is going to look something like this. I'm obviously hand drawing it, so not a great drawing right over here.

When we have the point ( (2, \ln(2)) ), that would be, let's say it's over here. So if this is ( 2 ), then this right over here is ( \ln(2) ). So that's the point ( (2, \ln(2)) ).

Then we have some other point, just noting in the abstract ( 2 + h ). So it's ( 2 + something ), so let's say that is ( (2 + h, \ln(2 + h)) ).

The exercise that we just did is finding the slope of the line that connects these two, so the line will look something like that. The way that we did this is we figured out, okay, what is our change in ( y )?

We are going from ( y = \ln(2) ) to ( y = \ln(2 + h) ). So our change in ( y ) is ( \ln(2 + h) - \ln(2) ). Our change in ( x ) is going from ( 2 ) to ( 2 + h ).

So our change in ( x ) we just increased by ( h ); we're going from ( 2 ) to ( 2 + h ). Our change in ( x ) is equal to ( h ).

So the slope of the secant line, the slope of this secant line that intersects our graph in two points, is going to be change in ( y ) over change in ( x ), which is once again exactly what we have over there.