Math 251 Notes. Part I.

Transcription

1 Math 251 Notes. Part I. F. Patricia Meina May 6, 2013

2 Growth Moel.Consumer price inex. [Problem 20, page 172] The U.S. consumer price inex (CPI) measures the cost of living base on a value of 100 in the years The CPI for the years (see figure in the book) is moele by the function c(t) = 151e 0.026t, where t represents years after a. Is the average growth rate greater between or ? For the first perio, 1995 correspons to t = 0 an 2000 correspons to t = 5. The average growth rate in between t = 0 an t = 5 is given by c c(5) c(0) = = 4.19 t For the secon perio: the average growth between t = 10 an t = 15 is given by c c(15) c(10) = = t

3 The average growth rate is larger in the perio b. Is the growth rate greater in 2000 (t = 5) or 2005 (t = 10)? In orer to solve this exercises we nee to prove the following claim: Claim: Let A be a fixe real number an g(t) = e At. Then g (t) = Ae At. Proof. g(t + h) g(t) e A(t+h) e At lim = lim h 0 h h 0 h e At e Ah e At = lim h 0 h = lim h 0 e At (e Ah 1) h = e At e Ah 1 lim h 0 h ( = e At e Ah ) 1 lim A = Ae At 1 = Ae At. h 0 Ah

4 Now, we are reay to solve part b. c (t) = (151e 0.026t ) = 151(0.026)e 0.026t = 3.926e 0.026t So c (5) 4.47 an c (10) Growth rate is greater at the later ate.

5 Average an Marginal cost Associate with manufacturing process is a Cost function C(x) (cost of manufacturing x items of the prouct). The average cost is C(x) = C(x) x per item (cost of items alreay prouce). How about the cost of proucing aitional items? If we have prouce x items an we want to prouce x aitional items, the cost to prouce these aitional items is C(x + x) C(x) The average cost of proucing x aitional items is given by The marginal cost: C x = C(x + x) x C M(x) = lim x 0 x = C (x)

6 Average an marginal cost [ Problem 24, page 172] Consier the cost function C(x) = 0.04x x + 800, 1 x 100, a = 500. a. Fin the average cost an marginal cost functions. Average cost function Marginal cost C(x) = C(x) x = 0.04x x. M(x) = C (x) = 0.08x b. Determine the average an marginal costs when x = a. If x = 500 : C(500) = 0.04(500) = = $81.6/item

7 M(500) = C (500) = 0.08(500) = $60/item c. Interpret the values obtaine in b. The average cost per item when proucing 500 items is $ The cost of proucing the next item is $ 60.

8 Examples Theorem 3.14 [Textbook] Suppose y = f (u) an u = g(x) are ifferentiable functions. The composite function y = f (g(x)) is ifferentiable, an its erivative can be expresse in two equivalent ways: Version 1 Version 2 y x = y u u x x [f (g(x))] = f (g(x)) g (x)

9 Examples Guielines for using the chain rule [ page 176 ] Assume the ifferentiable function y = f (g(x)) is given. 1 Ientify the outer function f, the inner function g, an let u = g(x). 2 Replace g(x) by u to express y in terms of u : 3 Calculate the prouct 4 Replace u by g(x) in y y u to obtain x. u {}}{ y = f ( g(x)) = y = f (u) y u u x

10 Examples g( ) f ( ) x g(x) f (g(x)) g ( ) f ( ) g (x) f (g(x)) f (g(x)) g (x) Diagram create by F. Patricia Meina (2013)

11 Examples Use the chain rule to compute the erivative of the following composite functions. Example 1 Inner function: Outer function: y = (5x x) 20 g(x) = 5x x f (u) = u 20 erivative: f (u) = 20u 19 f (g(x)) = f (5x x) Multiply erivatives of inner an outer functions: x [(5x2 + 11x) 20 ] = f (5x x) (10x + 11) = 20(5x x) 19 (10x + 11)

12 Examples Example 2 Inner function: Outer function: y = 16x g(x) = 16x f (u) = u erivative: f (u) = 1 2 u f (g(x)) = f (16x 2 + 1) Multiply erivatives of inner an outer functions: x [ 16x 2 + 1] = f (16x 2 + 1) (32x) 1 = 2 16x (32x)

13 Examples Example 3 y = ( ) 2x x + 3 Inner function: Outer function: g(x) = ( ) 2x + 1 x + 3 f (u) = u 7 erivative: f (u) = 7u 6 ( ) 2x + 1 f (g(x)) = f x + 3 Multiply erivatives of inner an outer functions: [ ( ) ] 2x ( ) = f 2x + 1 ( ) 2x + 1 x x + 3 x + 3 x x + 3 ( ) ( ) 2x (x + 3) (2x + 1) 1 = 7 x + 3 (x + 3) 2 ( ) ( ) 2x = 7 x + 3 (x + 3) 2

14 Examples Example 4: Chain Rule for powers Suppose that g is ifferentiable for all x in its omain an n is an integer. Fin the erivative of y = (g(x)) n Inner function: Outer function: Evaluate at g(x) : g(x) f (u) = u n erivative: f (u) = nu n 1 f (g(x)) Multiply erivatives of inner an outer functions: x [(g(x))n ] = f (g(x)) g (x) = n(g(x)) n 1 g (x).

15 Examples Example 5 [ Composition of three functions] This is of the form: f (g(h(x)) First inner function: Secon inner function: Evaluate g at h(x) : Outer function: Evaluate f at g(h(x)) : y = cos 4 (7x 3 ) h(x) = 7x 3 g(z) = cos(z) erivative: g (z) = sin(z) g (h(x)) = g (7x 3 ) f (u) = u 4 erivative: f (u) = 4u 3 f (g(h(x))) = 4(g(h(x))) 3 Multiply erivatives first inner, secon inner, an outer functions...

16 Examples x [cos4 (7x 3 )] = f (g(h(x))) g (h(x)) h (x) = 4(g(h(x))) 3 g (7x 3 ) 21x 2 = 4(g(h(x))) 3 ( sin(7x 3 )) 21x 2 = 4(cos(7x 3 )) 3 ( sin(7x 3 )) 21x 2 = 4cos 3 (7x 3 )sin(7x 3 )21x 2.

17 Examples Example 6 [ Calculating erivatives at a point] Let p(x) = f (g(x)). Use the values in the following table to compute p (1) an p (2). Recall that for any a the chain rule gives x f (x) g(x) g (x) If a = 1 : p (a) = f (g(a))g (a) p (1) = f (g(1))g (1) = f (2)g (1) = 5 8 = 40

18 Examples If a = 2 : p (2) = f (g(2))g (2) = f (1)g (2) = 3 5 = 15

19 Example 1 a. Use implicit ifferentiation to fin the erivative y x, where y 2 + 3x = 2. b. Fin the slope of the curve at the point ( 1, 5). a. Recall that y = y(x) (x is an inepenent variable.) Write (y(x)) 2 + 3x = 2 Take erivative to both sies of the equation with respect to x: x [y(x)]2 + x (3x) = x (2) x [y(x)]2 = 2y(x)y (x) ( chain rule ) x (3x) = 3 an x (2) = 0

20 We are left with ( Recall: y (x) = y x ) Solve for y x : Divie both sies by 2y 2y(x)y (x) + 3 = 0 2y y x = 3 y x = 3 2y

21 b. To fin the slope of the curve at ( 1, 5), y x = = 3 ( 1, 5) 2 10 Note: You may have cases where you have to use both coorinates of the given point.

22 Example 2 [ Tangent lines ] a. Verify that the point ( π 2, π 4 ) lies on the curve cos(x y) + sin(y) = 2. b. Determine the equation of the line tangent to the curve at the given point. a. cos( π 2 π 4 ) + sin( π 4 ) = cos( π 4 ) + sin( π 4 ) = = = 2

23 b. We fin x (y): x [cos(x y)] + x [sin(y)] = x (2) sin(x y) (1 x (y) ) + cos(y) x (y) = 0 sin(x y) + sin(x y) (y) + cos(y) x x (y) = 0

24 So Equation of the line: Or, sin(x y) + (sin(x y) + cos(y)) x (y) = 0 (sin(x y) + cos(y)) (y) = sin(x y) x y x = ( π 2, π 4 ) x (y) = sin(x y) (sin(x y) + cos(y)) sin( π 2 π 4 ) (sin( π 2 π 4 ) + cos( π 4 )) = y π 4 = 1 2 y = 1 2 x. ( x π ) =

25 Logarithmic ifferentiation Inverse properties 1 e ln(x) = x for x > 0, an ln(e x ) = x for all real number x. 2 y = ln(x) if an only if x = e y. 3 For b > 0, b x = e ln(bx) = e xlnb. Theorem 3.17 [ Textbook] x (ln(x)) = 1 x x (ln x ) = 1 x for x > 0 for x 0

26 Logarithmic ifferentiation We will use implicit ifferentiation : Let y = ln(x) If x > 0 by one of the inverse properties, Differentiate both sies: x = e y. 1 = e y y x Solve for y x : y x = 1 e y = 1 x Hence, for x > 0 : x (ln(x)) = 1 x

27 Logarithmic ifferentiation Example 1 Fin x (ln(3x2 )) ( ) 1 x (ln(3x2 )) = 3x ( 2 1 = 3x 2 = 6x 3x 2 = 2 x x (3x2 ) ) 3 2x Notice that ln(3x 2 ) is efine for all x such that 3x 2 > 0. Solving this inequality we obtain that the result is vali for all x (,0) (0, ).

28 Logarithmic ifferentiation General logarithmic functions Let f (x) = b x (b > 0) If b > 0 with b 1 then y = log b (x) x = b y x (log b x) = 1 xln(b) x (log b x ) = 1 xln(b) for x > 0 for x 0

29 Logarithmic ifferentiation Example 2 Compute the erivative of f (x) = log 3 (5x + 2) The result is vali if 5x + 2 > 0. That is x > 5 2. By the chain rule, f (x) = General power rule 1 (5x + 2)ln3 (5x + 1 2) = (5x + 2)ln3 5 = 5 (5x + 2)ln3 For real numbers p an for x > 0, Differentiate both sies of x p = e pln(x)... x (xp ) = px p 1.

30 Logarithmic ifferentiation Let ( f (x) = ) 2x x Use logarithmic ifferentiation to evaluate f (x). Take the natural logarithm to both sies: [ ( ln[f (x)] = ln ) ] 2x x ( = 2xln ) x loga b = blog(a) Differentiate both sies using the prouct an chain rules: ( 1 f (x) f (x) = 2 ln ) 1 + 2x ( ) ( 1x ) x x

31 Logarithmic ifferentiation Multiply both sies by f (x) : [ Hence, f (x) = f (x) 2 ln ( ) 1 + 2x ( ) ( 1x ) ] x x ( f (x) = ) [ 2x ( 2 ln ) + 2x x x ( = ) [ 2x ( 2 ln ) 2 x x ( = ) 2x [ ( ln ) 1 ] x x x ( ) ( 1x ) ] x 1 ( x ) ( ) ] 1 x