Module 10  Derivative of a Function  
Introduction  Lesson 1  Lesson 2  Lesson 3  Self Test  
Lesson 10.1: The Derivative at a Point  
In this lesson you will use several different features of the TI83 to find and understand derivatives.
In Module 9 you saw that velocities correspond to slopes in the graph of position vs time. Average velocity corresponds to the slope of a
Average velocity is given by
, which is the slope of a secant line through the points Instantaneous velocity is given by , which is the slope of the tangent line to the curve at (a, f(a)). The slope of the tangent line to the graph of a function at a point is called the derivative of the function at that point. The formal definition of derivative is given below. Formal Definition of Derivative The derivative of a function f at x = a is provided the limit exists. Illustrating Secant Line Convergence For functions that have a tangent line, if the point (a, f(a)) on the curve is fixed, as h approaches zero the second point (a+h, f(a+h)) approaches the fixed point and the corresponding secant lines converge to the tangent line at that point. The procedure outlined below will find the value of the derivative of the function f(x) = 2x  x^{2} at the point (0.5, 0.75) using a method similar to the one you used to find instantaneous velocities.
The graph below illustrates f(x) = 2x  x^{2} in a [1, 3, 1] x [1, 2, 1] window, with three secant lines through the fixed point (0.5, 0.75) that approximate the tangent line at (0.5, 0.75). Finding Slopes of Secant Lines The first step in the procedure outlined above is to find the slopes of secant lines that will be used to estimate the derivative. To find the slopes you need to enter the function f(x) = 2x  x^{2} in the Y= editor.
The slope of the secant line through the points (0.5, f(0.5)) and (0.5 + h, f(0.5 + h)) can be found by evaluating the difference quotient We are interested in values of h that are small so that the two points are close together. The resulting secant line will approximate the tangent line. You may evaluate the difference quotient for h = 0.1 on the TI83 by using a twopart command. The first part of the command will store 0.1 in h and the second part of the command will evaluate the difference quotient. The two commands will be combined together with the colon symbol.
The slope of the secant line containing (0.5, f(0.5)) and (0.6, f(0.6)) is 0.9. Using Smaller Values of h As the point (0.5 + h, f(0.5 + h)) approaches the point (0.5, f(0.5)), h approaches 0 and the secant lines converge to the tangent line. To evaluate the difference quotient for smaller values of h, change the value of H in the last expression on the Home screen from 0.1 to 0.01 and evaluate the difference quotient.
The slope of the corresponding secant line is 0.99.
The slopes of the secant lines are 0.999 and 0.9999, respectively. 10.1.1 Predict the derivative at (0.5, f(0.5)). Click here for the answer. Lefthand Difference Quotients In the procedure above, righthand difference quotients were used. Lefthand difference quotients may be found by letting h be a negative number.
The slopes of the corresponding secant lines are 1.01 and 1.001. With the fixed point (0.5, 0.75), one secant line passes through (0.49, f(0.49)) and the other through (0.499, f(0.499)). Finding a Derivative at a Point As stated earlier, the derivative at x = 0.5 is defined to be the limit Before this limit can be evaluated, the expression must be expanded and simplified. Recall that the function of interest is f(x) = 2x  x^{2}.
Therefore, and the derivative of f(x) = 2x  x^{2} at x = 0.5 is 1. Using the Numeric Derivative Command You can also approximate the derivative of the function at a point by using the numeric derivative command nDeriv(, which is found in the Math menu. The syntax for finding a derivative at a point is nDeriv(expression,variable,value).
Drawing the Tangent Line
Because the point on the curve and the derivative at that point are both known, an equation for the tangent line may be found by using the
Graph f(x) = 2x  x^{2} and its tangent line at (0.5, 0.75).
The line appears to be tangent to the curve at x = 0.5. 

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