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

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Derivatives and slope 2.1 update day1

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