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1 Differentiability Revisited Recall that the function f is differentiable at a if exists and is finite. f (a) = lim x a f (x) f (a) x a Another way to say this is that the function f (x) f (a) F a (x) = x a f (a) if x = a if x = a is. That is, lim F a(x) x a Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 2/26 The Wrong Product Rule Let P(x) = f (x)g(x). The Product Rule gives the rule for finding P (x). Is it true that P (x) = f (x)g (x)? In Example 37 we found the derivative of the function P(x) = x ( x 2 3 ) to be by multiplying through by the x before taking the derivative. But we can write P(x) = f (x)g(x) where f (x) and g(x). Then, f (x) and g (x). So that f (x)g (x) Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 3/26

2 Derivation of the Product Rule For the function P(x) = f (x)g(x) suppose that f and g are differentiable at a. Let F a and G a be the functions given in the alternate description of differentiability. Then F a and G a are continuous at a with and if lim F a(x) x a lim G a(x) x a F a (x) G a (x) Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 4/26 Derivation of the Product Rule Continued We now use Formula 1 to find P (a) as P (a) Now, using the expressions for f (x) and g(x) in terms of F a (x) and G a (X), gives P(x) = f (x)g(x) = [f (a) + F a (x) (x a)] [g(a) + G a (x) (x a)] since P(a). So we have P(x) P(a) Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 5/26

3 The Product Rule The Final Form So that P(x) P(a) x a Thus, finally we have P (a) Replacing the a by x we have the Product Rule P(x) = f (x)g(x) P (x) or using operator notation d [f (x)g(x)] Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 6/26 Extending the Product Rule The reason for writing the Product Rule in the given form is that the extension to products with more factors is easy. For a product with three factors we have d [f (x)g(x)h(x)] You can think of the Product Rule as simply applying the derivative to each factor in the product individually leaving the other factors alone, and repeating for every factor. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 7/26

4 Example 39 Verifying the Product Rule Verify that the Product Rule gives the correct derivative for the function P(x) = x ( x 2 3 ) from Example 37. Solution: Write the function as P(x) = f (x)g(x) where f (x) g(x). Then using the Product Rule gives and P (x) = 3x 2 3 This is the same result obtained in Example 37. This example illustrates that fact that it is not always desirable to use rules such at the Product Rule. Here it is easier to simply multiply out and then take the derivative. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 8/26 Visualizing the Product Rule Using Formula 3 for the derivative of the function P(x) = f (x)g(x) as a function of x gives P (x) = Now let f (x + h) and g(x + h). Then f and g so that using Formula 3 again gives f f (x + h) f (x) (x) = lim h 0 h g g(x + h) g(x) (x) = lim h 0 h Further note that since f and g are differentiable, they are continuous. So that and. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 9/26

5 Visualizing the Product Rule Continued In the limit for P (x) we need to simplify the expression f (x + h)g(x + h) f (x)g(x) = g(x) g f To do this consider the rectangle with sides f (x) + f and g(x) + g, divided into four regions as shown. f (x) Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 10/26 Visualizing the Product Rule Continued The product g(x) g (f (x) + f ) (g(x) + g) represents the total area of the rectangle, which is the sum of the areas the four regions. Thus, from the diagram f f (x) (f (x) + f ) (g(x) + g) = And f (x + h)g(x + h) f (x)g(x) The remaining three terms represent the change in the area of the rectangle when x changes by h. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 11/26

6 Visualizing the Product Rule Continued Then the derivative is P (x) Note that the third term above comes from the small rectangle in the upper corner. Its contribution becomes negligible as h 0. This is similar to what we saw when we found the rate of change of the area of the square with respect to its side length in Example 26. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 12/26 Example 40 Using the Product Rule Differentiate each function (a) f (x) = ( x 2 + x + 1 ) ( x 2 x + 1 ) (b) y = x 2 e x (c) P(x) = ( x 3 + 3x 2 + 4x + 5 ) 2 Solution (a): We could multiply the function out, or use the Product Rule. Using the Product Rule gives f (x) Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 13/26

7 Solution: Example 40(b) Solution (b): Applying the Product Rule gives dy Solution (c): Applying the Product Rule with f (x) = g(x) gives P (x) In most cases it is not necessary, or desirable, to multiply out an expression like this. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 14/26 Example 41 Rate of Change Using the Product Rule In Example 33 we considered the beat capacity of the heart, C = f (r), giving the number of millilitres of blood pumped by the heart per beat as a function of the heart rate r in beats per minute. In that example we used a table of values to estimate the rate of change of the beat capacity with respect to the heart rate when the heart rate is 70 beats per minute. We found that C(70) = 4.2 ml/beat and C (70) (ml/beat) / (beat/min). Now consider the cardiac flow rate of the heart which is the number of millilitres of blood pumped by the heart per minute. It is given by F(r) = rc(r). (a) (b) Find the value of F(70) and verify that it has the units you expect. Estimate the value of F (70). Give the units of your answer and explain what your answer means in practical terms. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 15/26

8 Solution: Example 41(a) Using the values given F(70) The units are as you would expect, since the flow rate gives. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 16/26 Solution: Example 41(b) First use the Product Rule to find an expression for F (r). Hence F (r) F (70) = 2.45 From the expression above it is apparent that the units of F (70) are the same as the units of C(70), which are ml/beat This value, and the units, mean that. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 17/26

9 The Reciprocal Rule Let g(x) be differentiable and define the reciprocal function as f (x) = 1 g(x) Find f (x) in terms of g(x) and g (x). First note that f (x)g(x) = 1, so that d [f (x)g(x)] But using the Product Rule gives d [f (x)g(x)] = Then solving for f (x) gives f (x)g(x) Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 18/26 The Reciprocal Rule Continued The Reciprocal Rule is d ( 1 ) g(x) We will see that the Reciprocal Rule is a special case of a more general rule soon. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 19/26

10 The Power Rule for Negative Integer Powers Recall that we only proved the Power Rule for non-negative integer powers. We can now prove it for negative integer powers as well. For a positive integer n let f (x) = x n This is a reciprocal function with g(x) ( ) d 1 x n. So the derivative is So that d ( x n ) which is the Power Rule for the case of a negative integer power. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 20/26 The Quotient Rule Let Q(x) = f (x) g(x) Find Q (x). We could go back to the limit definition of the derivative as we did for the Product Rule. Instead, we will apply the Product and Reciprocal Rules to the function This gives Q(x) Q (x) Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 21/26

11 The Quotient Rule Continued Thus, the Quotient Rule is ( ) d f (x) g(x) Most computer algebra systems such as Maple do not use the Quotient Rule in this form. They take the derivative a quotient using the Product Rule as in the second line in the development above. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 22/26 Example 42 Verifying the Quotient Rule In Example 40(b) we used the Product Rule to find the derivative of y = x 2 e x. Use the Quotient Rule to find the derivative of this function and verify that you get the same answer as you got using the Product Rule. Solution: Write the function as Then applying the Quotient Rule gives y Q (x) This is the same result that we got using the Product Rule. Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 23/26

12 Example 43 Applying the Quotient Rule Let (a) Find f (x). f (x) = x 2 x (b) Determine the intervals in which f (x) > 0 and f (x) < 0. (c) Sketch the graphs of f and f and explain how the intervals found in part (b) relate to the graph of f. Solution (a): Using the quotient rule gives f (x) Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 24/26 Solution: Example 43(b) Since we must solve the inequalities f (x) = 2x (x 2 + 1) 2 2x (x 2 + 1) 2 and 2x (x 2 + 1) 2 But the denominator is always, so these inequalities become and Hence f (x) > 0 on the interval f (x) < 0 on the interval Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 25/26

13 Solution: Example 43(c) The graph of this function is the same as the graph in Example 31(b). The graph of f looks like this, and the graph of f looks like this. Using the same notation as in Example 37(c) we indicate intervals found in part (b) like this. So the graph of f is sloping downward on the interval (, 0) and is sloping upward on the interval (0, ). Clint Lee Math 112 Lecture 11: Differentiation Product & Quotient Rules 26/26

Limits at Infinity. Horizontal Asymptotes. Definition (Limits at Infinity) Horizontal Asymptotes

Limits at Infinity. Horizontal Asymptotes. Definition (Limits at Infinity) Horizontal Asymptotes Limits at Infinity If a function f has a domain that is unbounded, that is, one of the endpoints of its domain is ±, we can determine the long term behavior of the function using a it at infinity. Definition