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

Pre Calculus innovative 5 - Assignment Examplef(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 + x100) f (x) = 200 2x + 100x99 e (200X-X2 + x100) 2. uprise the differentials for the following shares 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 first 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 e quation 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 detachment 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 Let 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 work 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