Presentation on theme: "Calculus 2.1 Introduction to Differentiation"— Presentation transcript:
1Calculus 2.1 Introduction to Differentiation Mrs. Kessler
2Rate of ChangeTo begin our study of rates and changes we must realize that the rate of change is the change in y divided by the change in x.We are assuming two pointsWith a straight line, this has been the slope, m.withHowever, with a curve, this is a little more complicated.
3Examples of tangents of curves Since the curve is increasing and decreasing at different rates, we are looking for the instantaneous rate of change at a particular point.This means we need the slope of the tangent line at the point P.
4Approximating the slope of the tangent line with the secant line x + x
14Find equation of the line tangent to and parallel to y = 2x -7
15That’s what we are looking for! Find its equation. Find equation of the line tangent to and parallel to y = 2x -7Original functionThat’s what we are looking for! Find its equation.Original function plus line
16Find equation of the line tangent to and parallel to y = 2x -7
17is the slope of the tangent line at any point x. Continued:is the slope of the tangent line at any point x.We want a tangent line parallel to y = 2x – 7We can see m = 2 in y = 2x – 7and from the calculus, m = .Now set the slopes equal and solve for x.Now sub back into the original function, and find y.y = 1/4
18Now find the equation y – y1 =m ( x – x1) At the point (17/16, ¼), the tangent line is parallel toNow find the equationy – y1 =m ( x – x1)
20y = x2. The two tangent lines intersect at (1, -3). Find their equations.(x1, y1)(1, -3)Let the red dot be the point (x1, ,y1). By the delta process the derivative is 2x.At the point (x1 ,y1), the derivative which is the slope = 2x1.Now use the slope formula setting the two =.
21The tangent points are ( 3, 9) and (-1, 1) If x =3, f(x) = y = 9 y = x2. The two tangent lines intersect at (1, -3).Find their equations.(x1, y1)Cross multiply and set equal to 0. Note: y1=(x1)2(1, -3)The tangent points are( 3, 9) and (-1, 1)If x =3, f(x) = y = 9If x = -1, f(x) = y = 1
22Now find the equation of the line. y = x2. The two tangent lines intersect at (1, -3).Find their equations.The tangent points are( 3, 9) and (-1, 1)Now find the equation of the line.y - y1= m ( x – x1)(x1, y1)(1, -3)For first equation use ( 3, 9) and (1, -3)For second equation use ( -1, 1) and (1, -3)y = 6x – and y = - 2x -1 Graph and check.
24Alternative form of the derivative provided the limit exists.exist and are equal.
25Alternative form of the derivative at x = 4Since the limit from left is not equal to the limit from the right, there is no limit, which means there is no derivative at x = 4 even though the function is continuous there.
28Using the TI 83/84 to draw the tangent line Enter the function into the y -editor.Graph the function:y = x sin(x)Suppose I want the tangent at x = π/3.Press 2nd Draw 5Type π/3Note: Decimal place was set a 3 fixed places.