Derivative of ln(x) | Advanced derivatives | AP Calculus AB | Khan Academy

So in this video, we're going to think about what the derivative with respect to X of the natural log of X is. I'm going to go straight to the punch line: it is equal to 1 over X. In a future video, I'm actually going to prove this. It's a little bit involved, but in this one, we're just going to appreciate that this seems like it is actually true.

Right here is the graph of y is equal to the natural log of X. Just to feel good about the statement, let's take the slope. Let's try to approximate what the slope of the tangent line is at different points. So let's say right over here when X is equal to 1. What does the slope of the tangent line look like? Well, it looks like here, the slope looks like it is equal, pretty close to being equal to 1, which is consistent with this statement. If X is equal to 1, 1 over 1 is still 1, and that seems like what we see right over there.

What about when X is equal to 2? Well, this point right over here is the natural log of 2. But more interestingly, what's the slope here? Well, it looks like—let's see—if I try to draw a tangent line, the slope of the tangent line looks pretty close to 1/2. Well, once again, that is 1 over X; 1 over 2 is 1/2.

Let's keep doing this. If I go right over here when X is equal to 4, this point is 4 comma natural log of 4. But the slope of the tangent line here looks pretty close to 1/4, and if you accept this, it is exactly 1/4. You could even go to values less than 1. Right over here when X is equal to 1/2, 1 over 1/2, the slope should be 2, and it does indeed—let me do it in a slightly different color—it does indeed look like the slope is 2 over there.

So once again, you take the natural log, you take the derivative with respect to X of the natural log of X, it is 1 over X. Hopefully, you get a sense that that is actually true. Here, in a future video, we will actually prove it.