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**Finding the Derivative of a Function at a Number Algebraically**

As we can see, the formula is really the same as the formula So, Furthermore, if an equation of the tangent line at (a, f(a)) can be written as: y – f(a) = mtan(x – a) it can be also written as: y – f(a) = f (a)(x – a) How to find the derivative of a function f at a, i.e., f (a)? Example: If f(x) = x2 – 2x + 3, find f (1). Step 0: Find f(a) if it’s not given. Step 0: f (1) = 12 – 2(1) + 3 = 2 Step 1: Step 1: Whatever you’ve obtained in step 1 is f (a). f (1) = 0 Note: In almost every case, when you try to find , it will be a [0/0] indeterminate form. Examples: Find the derivative at the indicated x-value or at the indicated point. 1. f(x) = x2 + 3 at x = –1 2. f(x) = 4x – 5 at (2, 3) [Note: f(2) = 3] f(1) = (1)2 + 3 = 4 mtan = f (a)

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**Finding an Equation of the Tangent Line Algebraically**

How to find an equation of the tangent line of a function f(x) at a given x-value a or a given point (a, f(a))? Step 0: Find f(a) if it’s not given. Step 1: Find the slope of the tangent line using the formula Step 2: Use point-slope form of a line y – y1 = m(x – x1) by substituting a into x1, f(a) into y1 and f (a) into m. Examples: Find an equation of the tangent line of the given function at the x-value or at the given point. 1. f(x) = 2x2 – 3x – 2 at x = 2 2. f(x) = x3 + 1 at (1, 2) Equation of the tangent line is: y – 0 = 5(x – 2) y = 5x – 10 Equation of the tangent line is: y – 2 = 3(x – 1) y – 2 = 3x – 3 y = 3x – 1

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**Graphical Interpretation of the Derivative of a Function at a Number**

Determining whether the derivative of f at a number should be positive, negative or zero when a graph is given: When we are given a graph, to determine whether a function value f(a) is positive, negative, or zero, all we need to do is to see whether the point (a, f(a)) is above, below or on the x-axis, respectively. For example, f(–3) is positive because the point (–3, f(–3)) is above the x-axis, f(3) is negative because the point (3, f(3)) is below the x-axis, and f(0) is zero because the point (0, f(0)) is on the x-axis. On the other hand, to determine the derivative of f at a number a, i.e., f (a), is positive, negative, or zero, we need to ask ourselves this question: If we can construct a tangent line at (a, f(a)), what is the slope of this tangent line? For example, if we construct the tangent line at x = –5, what is the slope—is it positive, negative, or zero? Since the tangent line is going up—the slope is obviously positive! Therefore, f (–5) is positive. Determine whether the following derivatives are positive (+), negative (–) or zero (0): f (–4) = (+) f (–3) = 0 f (–2) = (–) f (–1) = (–) f (0) = (–) f (1) = (–) f (2) = 0 f (3) = (+) f (4) = (+)

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**By constructing the tangent line, of course!**

Estimating the Derivative of a Function at a Number In the last slide we learned how to determine whether the derivative at a certain number is positive, negative, or zero, i.e., whether the slope of the tangent line is positive, negative, or zero. The question now is: Can we find out what the exact value of the derivative is (if it’s not zero)? The answer is yes and no. Yes, if we know the equation of the function (with or without the graph). On the other hand, if we don’t know the equation of the function but only know the graph, we can still estimate the value of the derivative to the best extent we can. How? By constructing the tangent line, of course! For example, after we construct the tangent line at x = –5, we can find its slope by picking two points on this tangent line. You might ask: Which two points? In theory, any two points. However, in practice, we want to pick the two points which have integral values. Unfortunately, this is not always possible. On the other hand, the two points are never difficult to pick since we can pick any two points—we only need to estimate their coordinates if they are not integers. For example, to find f (–5), i.e., the slope of the tangent line at x = –5, we can pick the two points indicated, which are (–4, 4.5) and (–5, 0.2). Hence we have: f (–5) = (4.5 – 0.2)/(4 – (5)) = 3.3/1 = 3.3 Now, use any two points on the tangent line at x = 1 to find f (1): The two points are (1, 1) and (2, 3). f (1) = (1 – (3))/(1–2) = 4/3 1.3 Disclaimer: Here, we say we are only estimating f (a) using the slope of the tangent line at x = a—we are not claiming we are finding the exact value of f (a). That’s because the accuracy of the calculation can be affected by the following factors: 1. The points on the tangent line may be an estimation and not the exact value. 2. The tangent line we draw may be a little off from the actual tangent line.

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**Sketching the Derivative of a Function—Just a Rough Sketch Anyway**

The derivative of a function is also itself a function (as we will see later). Therefore, we can sketch the derivative function too. We can do a precise sketch if we know the equation of the function. If we are only given the graph, we can only draw it to the best extent as we can. How? By connecting the dots, of course. This is what we mean: For each x (preferably with integral value), we calculate or estimate f (x) by the method described on the previous page. For example, we have estimated f (–5) to be 4.3 and f (1) to be 1.3 and we know f (–3) and f (2) are both 0. So only couple more derivatives we need to calculate before we can connect the dots. f (–1) = (4, 3.5) f (–2) = (5, 1.5) f (–4) = (3.5 – 1.5)/(4 – (5)) = 2 f (4) = x f (x) –5 4.3 –4 2 –3 –2 –1.3 –1 1 3 4 f (0) = f (3) = Figure 4 The graph of y = f(x) with some tangent lines Figure 5 The graph of y = f (x)

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**Sketching and Determining the Derivative Functions**

As we can see, the derivative of a function itself is a function! The question now becomes: Is there a way we can sketch the derivative of a function without estimating the derivatives of the function at various points? Yes and no—it depends on how complicated the graph of the function is. Of course, it will be more difficult (if not impossible) to try to sketch the derivative of a function with a complicated graph (i.e., one with many twists and turns) than one with a simple graph (i.e., one without many (and perhaps without any) twists and turns). The following are some examples of functions without any twists and turns. They are lines! Examples: f(x) = 4 f (x) = 0 f(x) = 2x – 3 f (x) = 2 f(x) = –3x f (x) = 3 On the other hand, very often you will be given the graphs of two functions (on the same set of axes) and asked to identify the function f(x) and its derivative function f(x). There are many ways to distinguish a function and its derivative, but one of the easiest and fastest ways is: where can we draw a horizontal tangent line? Examples: In each of the graph on the right, it shows the graph of a function and its derivative. Identify the one that is the function and the one that is the derivative of the function. [Note: For the curve you can draw a tangent line at a number c, if the other curve has an x-intercept at c, then that curve is f(x) and the other curve is f(x). y x y x y = f(x) y = f(x) y = f(x) y = f(x)

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**The Derivative of a Function—Definition and Application**

Recall that the derivative of a function at a number a, by considering (a, f(a)) as the fixed point and (x, f(x)) as a point that varies, is So, how do we find the derivative of a function at any number x, i.e., the derivative of a function? Same idea, only now we let (x, f(x)) to be the fixed point, and let the point (x + h, f(x + h)) varies. If we want the point (x + h, f(x + h)) to approach (x, f(x)), we simply let h approaches 0. Therefore, we will have as the definition of the derivative of a function. (a, f(a)) (x, f(x)) x a x (x, f(x)) (x+h, f(x+h)) x+h x x+h Use the limit definition of derivative to find the derivatives of the following functions: 1. f(x) = x 2. f(x) = x2 3. f(x) = x3 Conclusion: If f(x) = x, f (x) = 1 If f(x) = x2, f (x) = 2x If f(x) = x3, f (x) = 3x2 A slightly different notation:

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