Pre Calculus Mod 5 Assignment Example | Topics and Well Written Essays - 1250 words

Pre Calculus Mod 5 - Assignment Example f(X) =ex3 f '(x) = 3x2 ex3 d. f(X) =2X2 e (1-X2) r (x) = 2x2, r' (x) = 4x s(x) = e (1-X2) s' (x) = -2 e (1-X2) Applying product rule, f '(x) = 2X2-2 e (1-X2) + 4x e (1-X2) = -8x2 e (1-X2) + 4x e (1-X2) = (4x -8x2) e (1-X2) e. f(X) =5X e (12-2x) Let r(x) = 5x, r' (x) = 5 and s(x) = e (12-2x), s' (x) = -2 e (12-2x) f '(x) = 5x (-2 e (12-2x)) + 5 e (12-2x) f '(x) = -10x e (12-2x) + 5 e (12-2x) = (-10 + 5) e (12-2x) f. f(X) =100e(x8 + x4) f '(x) = 8x7 + 4x3e(x8 + x4) g. f(X) = e (200X-X2 + x^100) f '(x) = 200 – 2x + 100x^99 e (200X-X2 + x^100) 2. Find the derivatives for the following functions: a. f(X) = ln250X b. f(X) = ln (20X-20) c. f(X) = ln (1- X2) d. f(X) = ln (5X + X-1) e. f(X) = Xln (12- 2X) f. f(X) = 2Xln(X3 + X4) g. f(X) = ln (200X - X2 + X100) Solutions The derivative of the function y = ln x is obtained by d/dx (ln x) = 1/x. d/dx logex = 1/x, suppose y = ln x, then dy/dx = 1/x a. f(X) = ln250X log ab = log a + log b Therefore, the equation can be rewritten as f (x) = ln 250 + ln x d/dx ln 250 = 0 (derivative of a constant) d/dx (ln x) = 1/x Hence dy/dx = 1/x. b. f(X) = ln(20X-20) If y = ln u and u is some function of x, then dy/dx = u'/u If y = ln f(x), then dy/dx = f ' (x)/ f(x) Let u = 20x – 20 u' = 20 dy/dx = 1. u'/u = 20/(20x – 20) c. f(X) = ln (1- X2) Let u = (1- X2) Then u' = -2x dy/dx = 1. ... ln (12-2x) f ' (x) = 2x/ (12 – 2x) + ln (12-2x). f. f(X) = 2X ln(X3 + X4) Let r(x) = 2x, therefore, r' (x) = 2 Similarly, if s(x) = ln (X3 + X4), s'(x) = (3x2 + 4x3)/ (X3 + X4) Therefore, f ' (x) = 2x ((3x2 + 4x3)/ (X3 + X4)) + 2 ln (X3 + X4) g. f(X) = ln(200X - X2 + X100) u = ln (200X - X2 + X100) u' = 200 -2x + 100x99 f ' (x) = dy/dx = u'/u = 200 -2x + 100x99/ (200X - X2 + X100) 3. Find the indefinite integrals for the following functions a. f(X) = e6X = ? e6X dx = 1/6e6X + C b. f(X) = e (5X-5) = 5/2 x2-5x e (5X-5) c. f(X) = 5eX = ? 5eX dx = 5 ? eX dx = 5eX + C d. f(X) = 1/ (1 + X) = ln ?1 + x? + C e. f(X) = 5/X = 5 integral [1/x] dx = 5 ln ?x?+ C 4. Find the definite integrals for the following functions a. f(X) = e2X over the interval [2, 4] =Integral 42 [ e2x ] dx = [ 1/2 e2 ( 4) + C ] - [ 1/2 e2 ( 2 ) + C ] = 1/2 [ e8 - e4 ] b. f(X) = 2eX over the interval [0, 2] =Integral 20 [2eX] dx = [e2 + C] - [e0 + C] = e2 – e0 d. f(X) = 2/ (2 + X) over the interval [2, 5] Le t u = 2 + x, when x = 2, u = 2 + 2 = 4 and when x = 5, u = 2 + 5 = 7 = ln [?2 + x?] 52 = ln (7) – ln (2) e. f(X) = 10/X over the interval [3, 10] =dx = 10 integral [ 10 / x ] dx = 10 [ ln | x | ] + C, so Integral103 [ 10/ x ] dx = [ 10 ln | 10 | + C ] - [ 10 ln | 3 | + C ] = 10 ln 10 – 10 ln3 = 10 [ln10 – ln 3] Part2: Application of Calculus in Business Decision-Making Calculus is extensively used in making business decisions, which are critical for the success and survival of every business enterprise. Derivatives have wide applications in the business world. Derivatives are used to measure rate of change of a function in relation to the changes in variables (inputs) under focus. At some given value of an input, the derivative tells us the linear estimate of the function, which is close to