# youtube to mp3 of “Calculus – Derivatives 2 Taking derivatives Differential Calculus”

In the last presentation, I hopefully gave you a little bit of an intuition of what a derivative is. It’s really just a way to find the slope at a given point along the curve.

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## Calculus – Derivatives 2 Taking derivatives Differential Calculus

Now we’ll actually apply it to some functions. So let’s say I had the function f of x. f of x is equal to x squared. And I want to know what is the slope of this curve.

What is the slope at x is equal to– let’s say at x equals 3. What is the slope of x? Let’s draw out what I’m asking. Coordinate axis. x-coordinate, that’s the y-coordinate.

And then if I were to draw– let me pick a different color. So we want to say what is the slope when x is equal to 3. This is x equals 3.

And of course when x equals 3, f of x is equal to 9. We know that, right? So what we do is we take a point, maybe a little bit further along the curve. Let’s say this point right here is 3 plus h. And I keep it abstract as h because as you know we’re going to take the limit as h approaches 0.

And at this point right here is what? It’s 3 plus h squared, right? Because the function is f of x is equal to x squared. So this point right here is 3 plus h, 3 plus h squared. Because we just take the 3 plus h and put it into x squared and we get 3 plus h squared.

And this point here is of course 3, 9. What we want to do is we want to find the slope between these two point. I really have to find a better tool.

This one keeps freezing, I think it’s too CPU intensive. But anyway. So we want to find the slope between these two points.

So what’s the slope? so it’s a change in y, so it’s 3 plus h squared minus this y minus 9 over the change in x. Well that’s 3 plus h minus 3. So if we simplify this top part or we multiply it out, what’s 3 plus h squared? That’s 9 plus 6h plus h squared, and then get the minus 9, and all of that is over– well this 3 and this minus 3 cancel out, so all you’re left is with h. And even if we simplify this, this 9 minus 9, they cancel out.

So let me go up here. We’re left with– this pen keeps freezing– it’s 6h plus h squared over h. And now we would simplify this, right, because we can divide the top and the bottom, that numerator and the denominator by h. And you get 6 plus h squared.

So that’s the slope between these two points. It’s 6 plus h squared. So if we want to find the instantaneous slope at the point x equals 3, f of x is equal to 9, or the point 3,9, we just have to find the limit as h approaches 0 here. So we’ll just take the limit as h approaches 0.

Well this is an easy limit problem, right? What’s the limit of 6 plus h squared as h approaches 0? Well it equals 6. So we now know that the slope of this curve at the point x equals 3 is 6. So if you actually did a traditional rise over run, the slope, this change in y over change in x is 6.

So we have the instantaneous slope at exactly the point x is equal to 3. So that’s useful. You know if this was a graph of someone’s position, we would then know kind of the instantaneous velocity, which is– well I won’t go into that. I’ll do a separate module on physics.

But this was useful, but let’s see if we can do more generalized version where we don’t have to know ahead of time what point we want to find the slope at. If we can get a generalized formula for the slope at any point along the graph f of x is equal to x squared. So let me clear this. So we’re going to stick with f of x is equal to x squared.

And we know that the slope at any point of this is just going to be the limit as h approaches 0 of f of x plus h minus f of access. All of that over h. This part right here, this is just the slope formula that you learned years ago.

It’s just change in y over change in x. And all we’re doing is we’re seeing what happens as the change in x gets smaller and smaller and smaller as it actually approaches 0. And that’s why we can get the instantaneous change at that point in the curve.

So let’s apply this definition of a derivative to this function. And actually if you want to know the notation, I think this is the notation Lagrange came up with. This is equal to f prime of x.

Don’t take my word on it on Lagrange. You might want to look it up on Wikipedia. But this [UNINTELLIGIBLE] derivative of f of x is f prime of x. Let’s apply it to x squared.

So we’re going to say the limit as h approaches 0 of f of x plus h. Well, f of x plus h is just– this pen driving me crazy– x plus h squared. I just took the x plus h and put it into f of x.

Minus f of x–well that’s just x squared– over h. And this is equal to the limit as h approaches 0. Just multiply this out of.

x squared plus 2xh plus h squared minus x squared– running out of space– all of that over h. Let’s simplify this. This x squared cancels out with this minus x squared. And then we can divide the numerator and the denominator by h, and we’re left with the limit as h approaches 0.

Numerator and denominator by h of 2x plus h. Well this is easy. This goes to 0, this is just equal to 2x. So there we have it.

The limit as h approaches 0 is equal to 2x. And this is equal to f prime of x, so the derivative of f of x, which is the denoted by f prime of x is equal to 2x. Well what does it tell us? What have we done for ourselves? Well now I can give you any point along the curve.

Let’s say we want to know the slope at the point 16, right. When at the point 16,256. That’s a point along f of x equals x squared. It’s just 16 and then 16 squared.

What’s the slope at that point? Well we now know the slope is 2 times 16. So the slope is equal to 32. Whatever the x value is you just put into this f prime of x function or the derivative function, and you’ll get the slope at that point.

I think that’s pretty neat and I’ll show you how in future presentations how we can apply this to physics and optimization problems and a whole other set of things. And I’m also going to show you how to find the derivatives for a whole set of other functions. I’ll see in the next presentation.