We first review various rules for easy differentiation of common functions: The same procedure works for a larger number of terms.

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1 1 Math 182 Lecture Notes 1. Review of Differentiation To differentiate a function y = f(x) is to find its derivative f '(x). Another standard notation for the derivative is Dx(f(x)). Recall the meanings of the derivative: Suppose y = f(x). (1) The line tangent to the graph where x = a has slope m = f '(a). (2) The instantaneous rate of change of y with respect to x is f '(x). (3) The definition of the derivative is f '(x) = f(x+h) - f(x) lim h h We first review various rules for easy differentiation of common functions: Theorem. If n is a constant and A is constant then Dx (Ax n ) = A n x n-1 Example. If f(x) = 4 x 3 then f '(x) = 12 x 2. Theorem. Dx [ f(x) + g(x)] = Dx (f(x)) + D(g(x)) and Dx [ f(x) - g(x)] = Dx (f(x)) - D(g(x)) This means, to differentiate the result of adding or subtracting functions, you just add or subtract the derivatives: Example. If f(x) = 3 x x 5 then f '(x) = 21 x x 4. The same procedure works for a larger number of terms. Example. Find Dx( 2 x x 3-2 x + 5) Answer. 12 x x 2-2 Note how we used the special cases where the exponent is 1 or 0: Dx(Ax) = A, Dx(A) = 0. The same procedures work if the coefficients or the exponents are more complicated: Example. Find f '(x) if f(x) = 3 (x) + 5 x -3/2. Rewrite the function that there are only exponents, no roots. f(x) = 3 x 1/2 + 4 x -3/2 f '(x) = 3 ( 1/2) x 1/ (-3/2) x -3/2-1 = (3/2) x -1/2-6 x -5/2 Example. Find g'(x) if g(x) = 2π x 2 - (3) x + π Answer. 4 π x - 3

2 2 If y = f(x) then the derivative evaluated at a is f '(a). Example. If f(x) = 3 x 3-7, find f '(2). f '(x) = 9 x 2-0 = 9 x 2 Hence f '(2) = 9 (2) 2 = 36. Theorem. (Quotient Rule) Dx [ f(x) / g(x) ] = g(x) Dx[f(x)] - f(x) Dx[g(x)]. [g(x)] 2 Alternatively, [ f(x) / g(x) ] ' = g(x) f '(x) - f(x) g '(x) [g(x)] 2 Example. Find the derivative of 3 x x - 1 (5 x - 1) Dx(3x 2 + 2) - (3 x 2 + 2) Dx( 5 x - 1) (5 x - 1) 2 (5 x - 1) (6 x) - (3 x 2 + 2) (5) = (5 x - 1) 2 (30 x 2-6 x) - (15 x ) = (5 x - 1) 2 30 x 2-6 x - 15 x 2-10 = (5 x - 1) 2 15 x 2-6 x - 10 = (5 x - 1) 2 Theorem (Power Chain Rule) For any constants n and A, Dx[ A[f(x)] n ] = A n [f(x)] n-1 Dx[f(x)] Example. Find the derivative of (2 x 3 + 3) 7 7 (2 x 3 + 3) 7-1 Dx(2 x 3 + 3) = 7 (2 x 3 + 3) 6 (6 x 2 ) = 42 x 2 (2 x 3 + 3) 6

3 3 Example. If f(t) = 7 (3 t 4 - t) 5, find f '(t). f '(t) = 7 (5) (3 t 4 - t) 4 Dt(3 t 4 - t) = 35 (3 t 4 - t) 4 (12 t 3-1) Theorem (Product Rule) Dx [ f(x) g(x) ] = f(x) Dx[g(x)] + g(x) Dx[f(x)]. Using the prime notation, this says [f(x) g(x)] ' = f(x) g'(x) + g(x) f '(x). Example. Find the derivative of (x 2 + 1) 4 (2x 3 ) (x 2 + 1) 4 Dx(2x 3 ) + (2x 3 ) Dx[(x 2 + 1) 4 ] = (x 2 + 1) 4 (6x 2 ) + (2x 3 ) 4 (x 2 + 1) 3 Dx[(x 2 + 1)] = (x 2 + 1) 4 (6x 2 ) + (2x 3 ) 4 (x 2 + 1) 3 (2x) = (x 2 + 1) 4 (6x 2 ) + (16x 4 ) (x 2 + 1) 3 Example. Suppose y = 2 x 3. (a) Find the slope of the line tangent to the curve where x = 1. (b) Find the equation of the line tangent to the curve where x = 1. (a) Here f(x) = 2 x 3, so f '(x) = 6 x 2. Hence the slope of the tangent line is f '(1) = 6 (1) 2 = 6. (b) The equation of the line with slope m that goes through the point where x = x0 and y = y0 is y = y0 + m (x - x0). In this case, x0 = 1, y0 = f(x0) = f(1) = 2, and m = f '(x0) = f '(1) = 6. Hence y = (x - 1) y = x - 6 y = 6x - 4 Example. The area of a colony of mold t hours after the establishment of the colony is A(t) = 3.5 t t mm 2 (a) Find the area when t = 4 hours. (b) Find the instantaneous rate of change of the area with respect to time when t = 4 hours. (c) Find the equation of the tangent line when t = 4. (a) A(4) = 3.5 (4) (4) mm 2 = 81.3 mm 2 (b) A'(t) = 7 t Hence A'(4) = 7(4) = 34.2 mm 2 /hour (We know the units of f '(x) are given as in the definition of derivative as the units of f(x) divided by the units of h. In this case the units are those of A(t) divided by the units of t, hence mm 2 divided by hours.) (c) The usual formula for a tangent line is y = y0 + m(x-x0). Here y corresponds to A and x corresponds to t, so

4 4 A = A0 + m (t - t0). We have t0 = 4, A0 = A(t0) = A(4) = 81.3, and m = A'(4) = Thus A = (t-4) A = 34.2 t mm 2 2. Increasing and decreasing functions (See Schaum 17) A function f is increasing on a set A if, for any u and v in A, whenever u<v, then f(u) < f(v). This means that as you follow the graph moving toward the right, the graph also moves upward and hence increases The function f(x) = x 2-2 x for 1 x 4. It is increasing. Similarly, a function f is decreasing on a set A if, for any u and v in A, whenever u<v, then f(u) > f(v). As you follow the graph moving toward the right, the graph moves downward, or decreases The function f(x) = 4 - x 2 is decreasing for 0 x 2.

5 5 Most functions are increasing some places and decreasing other places: This example is increasing for 0<x<0.5, 1.5<x<2.5, and 3.5<x<4.5. It is decreasing for 0.5<x<1.5, 2.5 < x < 3.5, and 4.5 < 5. There is a relative maximum at x = a if f(x) f(a) for all x near a. Thus if there is a relative maximum at x = a, then x = a is the top of a hill. There is a relative minimum at x = a if f(x) f(a) for all x near a. Thus if there is a relative minimum at x = a, then x = a is the bottom of a valley. In the first example, there is a relative maximum at x = 0.5, at x = 2.5, and at x = 4.5. There is a relative minimum at x = 1.5, at x = This example is increasing for 0<x<1.25 and 3.25 < x < 5. It is decreasing for 1.25 < x < 3.25 and 5<x. There is a relative maximum at x = 1.25 and x = 5. There is a relative minimum at x = Note that at either a relative maximum or a relative minimum in the picture, f '(x) = 0. It is important to be able to tell when a function is increasing and decreasing.

6 6 Theorem. If f '(x) > 0 for a<x<b then f is increasing for a<x<b. If f '(x) < 0 for a<x<b then f is decreasing for a<x<b. Idea. If f '(x) > 0 then the tangent line has positive slope, so it is moving upward to the right. If f '(x) < 0 then the tangent line has negative slope, so it is moving downward to the right. Procedure for finding when f (x) is increasing, when it is decreasing. (1) Solve f '(x) = 0, and tell when f '(x) has no value. (2) List these values in order. These numbers are called the critical numbers. (3) At every point other than those values, f '(x) is either positive (+) or negative (-). Tell which. (4) Every interval in which f '(x) > 0 is an interval where f is increasing. (5) Every interval in which f '(x) < 0 is an interval where f is decreasing. Example. If f(x) = 3-4 x, tell where f is increasing and tell where f is decreasing. Note f '(x) = - 4 Hence f is decreasing everywhere, and f is increasing nowhere. Example. Let f(x) = x 2-6 x + 5. Tell where f is increasing and where f is decreasing. Note f '(x) = 2 x - 6. The critical numbers are where 2 x- 6 is undefined or 0: 2 x - 6 = 0 means x = 3. Hence x = 3 is the only critical number. When x < 3, 2x-6 < 0; when x >3, 2x-6 >0. We thus obtain this diagram: (-) 3 (+) Hence f is increasing for x > 3; f is decreasing for x < 3. Example. Let f(x) = 2 x 3-3 x 2-36 x + 4. Find where f is increasing, and where f is decreasing. f '(x) = 6 x 2-6 x f '(x) always has a value. We solve f ' (x) = 0. 6 x 2-6 x - 36 = 0 6( x 2 - x - 6) = 0 6 (x - 3) (x + 2) = 0. The roots are 3 and -2. List them in order -2 3 Outside these points we figure out the sign: For x < -2, f ' (x) = (+) (-) (-) = +. For -2 < x < 3, f '(x) = (+) (-) (+) = -. For 3<x, f '(x) = (+) (+) (+) = +. Thus we obtain the chart: (+) -2 (-) 3 (+) So f is increasing for x < - 2 and for 3 < x since that is where f '(x) >0. Also f is decreasing for -2 < x < 3 since that is where f '(x) < 0. It is common to write intervals using a different notation: (a,b) means a<x<b. In this form the answer would be f is increasing on (-, -2) and (3, ). f is decreasing on (-2, 3).

7 7 Example. Let f(x) = (16 - x 2 ). Find where f is increasing, and where f is decreasing. f(x) = (16 - x 2 ) 1/2 f '(x) = (1/2) (16 - x 2 ) -(1/2) (-2x) = - x /(16 - x 2 ) (1/2) f '(x) makes sense only 16 - x 2 > 0 hence only for -4 < x < 4. f ' (x) = 0 when x = 0. Hence the critical numbers are: There are no values to the left of -4 or to the right of 4. We find -4 (+) 0 (-) 4 f is increasing for -4< x < 0. f is decreasing for 0 < x < 4. Use your calculator to graph the function. f(x). Example. Let f(x) = 3/ (x-1). Tell where f is increasing and where f is decreasing. f '(x) = (x-1) (0) = (x-1) 2 (x-1) 2 The only critical number is x = 1, since there is no value at x = 1. The chart is (-) 1 (-) Thus f is never increasing. Moreover f is decreasing for x<1 or x > 1. In fact, note that there is no point when x = Relative maxima and minima Theorem. (First Derivative Test) Suppose f '(a) = 0. (1) There is a relative maximum at x = a provided f '(x) > 0 to the left of a (but near a) and f '(x) < 0 to the right of a (but near a). (2) There is a relative minimum at x = a provided f '(x) < 0 to the left of a (but near a) and f '(x) > 0 to the right of a (but near a). Again, this is obvious from the picture. If f is increasing to the left of a and decreasing to the right of a, then there is a relative maximum at x = a. If f is decreasing to the left of a and increasing to the right of a, then there is a relative minimum at x = a. If f is decreasing to the left of a and also decreasing to the right of a, then there is neither a relative maximum nor a relative minimum at x = a. If f is increasing to the left of a and also increasing to the right of a, then there is neither a relative maximum nor a relative minimum at x = a. (3) If f '(a) = 0, f '(x) > 0 to the left of a (but near a) and f '(x) > 0 to the right of a (but near a), then there is neither a relative maximum nor a relative minimum. (Instead, f is increasing near a.) (4) If f '(a) = 0, f '(x) < 0 to the left of a (but near a) and f '(x) < 0 to the right of a (but near a), then there is neither a relative maximum nor a relative minimum. (Instead, f is decreasing near a.) Example. Let f(x) = x x + 5.

8 8 (a) Tell where f is increasing. (b) Tell where f is decreasing. (c) Locate all relative maxima. (d) Locate all relative minima. f '(x) = 2x + 2. Hence f ' (x) = 0 when x = -1. We find the following chart for f ': (-) -1 (+) (a) f is increasing for x > -1. (b) f is decreasing for x < -1. (c) there are no relative maxima. (d) there is a relative minimum at x = -1. A shorter way to ask you to find all the relative maxima and all the relative minima is to say "classify all the critical numbers." This means, for each critical number, tell whether it corresponds to a relative maximum, a relative minimum, or neither. Example. Let f(x) = 4 x +5 - x 2. (a) Tell where f is increasing. (b) Tell where f is decreasing. (c) Locate and classify all critical numbers. Solution: f '(x) = 4-2 x Hence f '(x) = 0 when 4-2 x = 0 or x = 2. We obtain the chart for f ' as follows: (+) 2 (-) (a) f is increasing for x < 2. (b) f is decreasing for x > 2. (c) x = 2 is a relative maximum Example. Let P(t) = 2 t 3-3 t 2-36 t + 4. Find the locations of all relative maxima and the locations of all relative minima. P'(t) = 6 t 2-6 t - 36 = 6 (t 2 - t - 6) = 6 (t-3) (t + 2). Solve P'(t) = 0 to find that the critical numbers are 3 and -2. We obtain the chart (+) -2 (-) 3 (+) To the left of 3, f '(t) < 0 while to the right of 3, f '(t) > 0. By the First Derivative Test there is a relative minimum at t = 3. To the left of -2, f '(t) > 0 while to the right of -2, f '(t) < 0. Hence by the First Derivative Test there is a relative maximum at t = -2. Write the answer as: t = -2: relative maximum t = 3: relative minimum Example. Let f(x) = 6 x x 4-80 x Tell where f is increasing, where f is decreasing. Classify each critical number. Solution f '(x) = 30 x x x 2.

9 9 Solve f '(x) = x x x 2 = 0 30 (x x 3-8 x 2 ) = 0 30 x 2 (x x - 8 ) = 0 30 x 2 (x + 4) (x - 2) = 0. Thus roots are x = 0, x= -4, and x = 2. Put them in order: Find the sign of f '(x): (+) -4 (-) 0 (-) 2 (+) Thus f is increasing for x<-4 and x > 2. f is decreasing for -4<x<0 and 0<x<2; we just write that f is decreasing for -4 < x < 2. There is a relative maximum at x = -4. There is a relative minimum at x = 2. Note that f '(0) = 0 so 0 is a critical number yet there is neither a relative maximum nor a relative minimum at x = 0. Thus the classification of critical numbers is -4 relative maximum 0 neither 2 relative minimum Example. A population of bacteria grows so that after t hours there are P(t) = 2t 3-21 t t + 3 grams of bacteria. (a) Tell when the population is increasing. (b) Tell when the population is decreasing. (c) Tell when there is a relative maximum. (d) Tell when there is a relative minimum. (e) Tell the initial population. (f) Tell the population at the relative maximum. (g) Tell the population at the relative minimum. P'(t) = 6 t 2-42 t Hence P'(t) = 0 when 6 t 2-42 t + 60 = 0 6 (t 2-7 t + 10) = 0 6 (t - 5) (t - 2) = 0. Thus P'(t) = 0 when t = 2 or t = 5. For 0<t<2, the sign of P'(t) is (+) (-) (-) = +, so P is increasing. For 2<t<5, the sign of P'(t) is (+) (-) (+) = -, so P is decreasing. For 5<t the sign of P'(t) is (+) (+) (+) = + so P is increasing. Hence (a) P is increasing when 0<t<2 and when 5<t. (b) P is decreasing when 2<t<5. (c) There is a relative maximum when t = 2. (d) There is a relative minimum when t = 5. (e) The initial population is P(0) = 3 grams. (f) The population at the relative maximum is P(2) = 55 grams. (g) The population at the relative minimum is P(5) = 28 grams.

10 10 What to do if the expression does not factor: Sometimes it is hard to factor f '(x). In these cases, to find the sign of f '(x), one can test any point in the interval. We expect that f '(x) can only change sign at a place where it is zero (or undefined). Since we will have found all such places on the chart, everywhere else it cannot change sign. Example. Let f(x) = -2.2 x x x (a) Tell where f is increasing. (b) Tell where f is decreasing. (c) Locate and classify all critical numbers. f '(x) = -6.6 x x We want to solve f '(x) = -6.6 x x = 0. We can't factor, so instead we use the quadratic formula to obtain x = or x = The chart for f '(x) is then For x < we substitute a value into f ' to find whether f '(x) is positive or negative. Pick any such value (and avoid the endpoint). I'll choose -2. Then from the calculator f '(-2) = -6.6 (-2) (-2) = Hence for x < we have f '(x) < 0. For < x < we pick any value. I'll choose x = 0. Then f '(0) = 69.1 > 0. Hence for < x < we have f '(x) > 0. For x > we pick any value. I'll choose x = 8. Then f '(8) = -6.6 (8) (8) = Hence for x > we have f '(x) < 0. The chart for f '(x) is then (-) (+) (-) The answer is then (a) f is increasing for < x < (b) f is decreasing for x < or x > (c) when x = there is a relative minimum; when x = there is a relative maximum