Lecture Notes 1-7

# Lecture Notes 1-7 - MIT OpenCourseWare http/ocw.mit.edu...

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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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Lecture 1 18.01 Fall 2006 Unit 1: Derivatives A. What is a derivative? Geometric interpretation Physical interpretation Important for any measurement (economics, political science, finance, physics, etc.) B. How to differentiate any function you know. d For example: e x arctan x . We will discuss what a derivative is today. Figuring out how to dx differentiate any function is the subject of the first two weeks of this course. Lecture 1: Derivatives, Slope, Velocity, and Rate of Change Geometric Viewpoint on Derivatives Tangent line Secant line f(x) P Q x 0 x 0 + Δ x y Figure 1: A function with secant and tangent lines The derivative is the slope of the line tangent to the graph of f ( x ). But what is a tangent line, exactly? 1
Lecture 1 18.01 Fall 2006 It is NOT just a line that meets the graph at one point. It is the limit of the secant line (a line drawn between two points on the graph) as the distance between the two points goes to zero. Geometric definition of the derivative: Limit of slopes of secant lines PQ as Q P ( P fixed). The slope of PQ : P Q (x 0 +∆x , f(x 0 +∆x )) (x 0 , f(x 0 )) Δ x Δ f Secant Line Figure 2: Geometric definition of the derivative lim Δ f = lim f ( x 0 + Δ x ) f ( x 0 ) = f ( x 0 ) Δ x 0 Δ x Δ x 0 Δ x �� �� “difference quotient” “derivative of f at x 0 1 Example 1. f ( x ) = x One thing to keep in mind when working with derivatives: it may be tempting to plug in Δ x = 0 Δ f 0 right away. If you do this, however, you will always end up with = . You will always need to Δ x 0 do some cancellation to get at the answer. Δ f 1 1 1 x 0 ( x 0 + Δ x ) 1 = x 0 x x 0 = = Δ x = 1 Δ x Δ x Δ x ( x 0 + Δ x ) x 0 Δ x ( x 0 + Δ x ) x 0 ( x 0 + Δ x ) x 0 Taking the limit as Δ x 0, lim 1 = 1 Δ x 0 ( x 0 + Δ x ) x 0 x 2 0 2

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Lecture 1 18.01 Fall 2006 y x x 0 Figure 3: Graph of x 1 Hence, f ( x 0 ) = 2 1 x 0 Notice that f ( x 0 ) is negative as is the slope of the tangent line on the graph above. Finding the tangent line. Write the equation for the tangent line at the point ( x 0 , y 0 ) using the equation for a line, which you all learned in high school algebra: y y 0 = f ( x 0 )( x x 0 ) Plug in y 0 = f ( x 0 ) = 1 and f ( x 0 ) = 2 1 to get: x 0 x 0 y x 1 0 = x 2 0 1 ( x x 0 ) 3
Lecture 1 18.01 Fall 2006 y x x 0 Figure 4: Graph of x 1 Just for fun, let’s compute the area of the triangle that the tangent line forms with the x- and y-axes (see the shaded region in Fig. 4). First calculate the x-intercept of this tangent line. The x-intercept is where y = 0. Plug y = 0 into the equation for this tangent line to get: 0 1 = 2 1 ( x x 0 ) x 0 x 0 1 1 1 = x + 2 x 0 x 0 x 0 1 2 x = 2 x 0 x 0 2 x = x 2 0 ( ) = 2 x 0 x 0 So, the x-intercept of this tangent line is at x = 2 x 0 .

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