Calculus: Derivatives 2 More intuition of what a derivative is. Using the derivative to find the slope at any point along f(x)=x^2
Calculus: Derivatives 2
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- 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.
- 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.
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