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# Derivatives and slope 2.1 update day1

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### Derivatives and slope 2.1 update day1

1. 1. Derivatives and Slopes:Goal: identify tangent lines to a graphUse limits to find slope Use limits to find derivatives<br />Tangent Line to a Graph:<br />The slope of the line tangent to a curve at a given point determines the rate at which the graph rises or falls at that given point.<br />http://www.ies.co.jp/math/java/calc/index.html<br />
2. 2. m=6<br />
3. 3. Slope and limit:Just looking at the graph can be difficult and inaccurate.<br />Secant line<br /> recall from geometry a secant is a line that passes through two points on a circle.<br />
4. 4. Is a secant line of circle C<br />
5. 5. Δy<br />Δx<br />Slope m= Δy/Δx, we can use this fact and the idea that f(x) = y<br />
6. 6. This point has the <br />Coordinates (x + Δx, f(x + Δx))<br />This point has the coordinates<br />(x, f(x))<br />
7. 7. (x + Δx, f(x+Δx))<br />(x, f(x))<br />As we bring this difference in closer making Δx approach 0, the secant line approaches the tangent line<br />
8. 8. 1<br />
9. 9. 2<br />
10. 10. 3<br />
11. 11. Going back to the original coordinates for the secant line:<br />(x + Δx, f(x + Δx)<br />f(x + Δx) – f(x)<br />(x, f(x)) Δx<br />Slope = m = f(x + Δx) – f(x) Difference Quotient<br />Δx<br />
12. 12. Definition of the Slope of a Graph:<br />The slope m of the graph of ƒ at the point<br /> (x, ƒ(x)) is equal to the slope of its tangent line at (x, ƒ(x)), and is given by<br /> msec =<br /> m = <br />Provided this limit exists.<br />
13. 13. Use the limit process to find slope.<br /> m =<br />
14. 14. Ex. ƒ(x)= x2 at the point (2, 4) find the slope of the graph at this point.<br />lim ƒ(2 + Δx) – ƒ(2)<br />Δx->0 Δx<br /> lim (2 + Δx)2 – (2)2<br />Δx->0 Δx<br /> lim 4 + 4 Δx + Δx2 -4<br />Δx->0 Δx<br />
15. 15. lim 4 Δx + Δx2<br />Δx->0 Δx<br /> lim Δx(4 + Δx) <br />Δx->0 Δx<br /> lim (4 + Δx) = 4<br />Δx->0<br />
16. 16. Find the slope of the graph y = 2x + 5 using the limit process<br />