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This is the third video on the derivative.
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And what we are working with is the function f of x equals x squared minus 3.
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That was a parabola opening upward.
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And we were trying to find the slope of the tangent line at the point 1, negative 2.
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And we were using the definition for the slope of the tangent line,
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which is m equals the limit is h approaches 0 of f of c plus h minus f of c over h.
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Remember this formula came from the formula for slope?
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And c is the x value in the ordered pair.
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So when I had it went through this process and we ended up with a slope of 2,
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because the limit as h approaches 0 was 2.
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So remember our function was f of x equals x squared minus 3, the parabola opening upward.
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And using the definition for the slope of the tangent line,
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it was the formula, the limit as h approaches 0, etc.
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We found that at the point 1, negative 2, the slope of that tangent line was positive 2.
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So if we wanted to, we could use the point slope formula to find the equation of the tangent line.
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So let's just do that.
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Remember the point slope formula is y minus y1 equals m times x minus x1.
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And if you did want to use that formula then where we would have y minus y1 is negative 2.
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The slope is 2.
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So we put that m for m.
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And x1 is positive 1.
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That goes from the order pair. So we have y plus 2 equals 2x minus 2,
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or y equals 2x minus 4.
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So now we're going to go back up to our original picture of this graph and see if it makes sense that
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that looks like that would be the equation of the tangent line.
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This is where we started. We considered the function f of x equals x squared minus 3.
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And then we were just estimating where the tangent line might be at the point 1, negative 2.
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Remember we said the tangent line down here, the red line.
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It looked like the slope at 0, negative 3.
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The slope of that tangent line was 0.
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And then we were trying to find out what the slope of this green line is.
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So we were trying to draw a tangent line.
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And so the question does it look like it is what we just came up with?
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We came up with y equals 2x minus 4 for that tangent line.
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And actually that's what it looks like.
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It looks to me like it did go through a y intercepted n84 and a slope of 2.
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That was approximation. We could have been off a little bit, but for my picture we sort of get the idea.
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So the slope of the tangent line was 2.
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And the actual equation if we wanted to use the slope point formula.
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Let's look back at the definition of tangent line with slope m.
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We found it took this limit as h approaches 0 of f of c plus h minus f of c over h.
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And if it existed it equaled m, then we say that the line passing through that order pair c of f of c
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with that slope m, whatever you get when you apply the limit, that is going to be the slope of the tangent line to the graph.
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Now what's interesting is we're going to now go on to the formal definition of a derivative.
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It's going to look exactly like that.
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The difference is instead of you seeing this letter c, we're going to have an x.
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So c was a particular value in an ordered pair.
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So here's the formula you've been waiting for.
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The derivative of a function is the definition.
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f prime of x, that's the way we read this, f prime of x.
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It's equal to the limit as h approaches 0 of this difference quotient.
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Remember this had to do with the formula for slope.
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f of x plus h minus f of x, that was the difference of the y values.
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And at the bottom we originally would have x plus h minus x.
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So we ended with just the h in the denominator.
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So that is the derivative.
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So if you have any ordered pair, x, y, you can plug in the x value into this f prime of x, take the limit,
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and you'll be able to figure out the slope of the tangent line to the curve at that point on the curve.
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So let's do it for the problem we've been working on.
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We've been working on f of x equals x squared minus 3.
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And just notice we're going to have to take a limit as h approaches 0.
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And in that limit we're going to also have to compute what f of x plus h is.
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So to make it easier, let's just compute f of x plus h.
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So we're going to plug in x plus h for x.
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And you're going to square the binomial.
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Remember it's x plus h times x plus h.
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So that's x squared plus 2xh plus h squared minus 3.
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So we know f of x, we know f of x plus h.
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Now we're going to plug that into the formula to get the derivative.
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So we've got f prime of x is equal to...
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Alright, it says we're going to plug in...
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Oops, I forgot, it's got to be the limit.
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Don't forget to write the limit as h approaches 0.
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And in the numerator we're going to have f of x plus h,
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which is this x squared plus 2xh plus h squared minus 3.
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So f of x appears f of x, which is x squared minus 3.
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And all of that is going to be over h.
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Alright, we have to keep going.
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Let's write that is the same thing.
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Remember you have to keep writing the limit as h approaches 0.
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And let's simplify the numerator.
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So I have x squared plus 2xh plus h squared minus 3 minus x squared.
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Plus 3 when I distribute the minus sign.
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And in the numerator, let's see, the x squared minus x squared cancels,
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the negative 3 plus 3, I end up with just 2xh plus h squared over h.
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This should look familiar.
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It's almost like what we just did a few minutes ago.
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I'm going to go down the page still.
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Because the limit is h approaches 0, let's see.
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I'm just going to show the factoring here.
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h times 2x plus h all over h.
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See how the h is canceled?
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So I really end up with just 2x plus h.
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Which if you plug in, now you could plug in 0 for h.
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That's the first time you don't have to write the limit.
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So you have 2x plus 0.
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And the answer is 2x.
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See the difference this time, I got 2x.
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Alright, so what does this mean?
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We had this function.
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f of x equals x squared minus 3.
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And we now just came out with f prime of x was equal to 2x.
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So what that means is that any point on the curve,
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if you find the slope, I could plug in the x value
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and it'll tell me the slope.
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Interesting, didn't you think?
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So let's look back at our graph.
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Here is our graph.
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Here is our original function.
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f of x equals x squared minus 3.
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What it says is f prime of x equals 2x.
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So at any order pair, we should be able to tell the slope of the line.
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If we know the ordered pair.
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So here's an order pair.
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This was our order pair, 1 negative 2.
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And what happens?
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When we plug in 1 for x and the f prime of x, you get 2.
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So that's why the slope of this line was 2.
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How about this order pair down here?
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It was 0 negative 3.
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And here the x value is 0.
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If you plug in f prime of 0,
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you are going to get a slope of 0, which is what it looks like.
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Now let's say we were going to do this point over here.
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This was negative 3, 6.
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And I want to know the slope of this line.
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That looks pretty steep.
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And it also looks like a negative slope.
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If you plug in negative 3 for x, you're going to get a slope of negative 6.
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So this tangent line, wherever it is,
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it's going to be very steep.
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Hard to even see it.
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It has a slope of negative 6.
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And that's all about the interesting part of finding the derivative.