1 Implicit Differentiation

Compute and apply derivatives that require implicit differentiation.

GivenIn(xy)+5y2=10x3+15,, computedxdy..

x3. sinx Using logarithmic differentiation, find the derivative of y = 3x3 +7 a.) dy 3. sinx 15x4 = cotx - O dx 3x3 + 7 X 3x5 + 7 b.) dy 3 15x 4 - - cotx - O dx X 3x3 + 7 c.) dy 3 15x 4 + cotx O dx X 3x5 +7 d.) dy x3 . sinx 3 15x 4 - + cotx - O dx 3x3 + 7 X 3x5 + 7Find the derivative of f(x) = (4x3+ 1) - sin - 1(x3). a.) O f'(x) = 12x2 . sin - 1(x3) + 4x3+1 1-X 6 b.) O f'(x) =12x2 . sin - 1(x3) + 12x3 + 3x2 V1-X" C.) O f'(x) = 12x2. sin - 1(x3) _ 24x8 +6x5 1-x6 d.) 36x 4 O f' ( x ) = 1-x6a.) dV cm 3 = 288- O dt sec b.) dV cm 3 = 512 O dt sec c.) dV cm 3 = 864 O dt sec d.) dV cm = 192 O dt sec3 feet b 12 feet 3 feet I h -a.) dh 1 O ft = dt 18 min b.) dh 1 ft E O dt 9 min c.) dh ft = 36 O dt min d.) dh 1 ft O dt 6 miny 20 18 16 14 12 10 8 6 4 2 -12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12 X -2 -4a.) 16 is a global maximum of f(x) at x = 2. O -1 is a global minimum of f(x) at x = 1. b.) 16 is a global maximum of f(x) at x = 2. O -1 is a global and local minimum f(x) at x = 1. c.) 16 is a global maximum of f(x) at x = 2. O -1 is a global minimum f(x) at x = 1. 7 is a local maximum of f(x) at x = - 1. d.) 16 is a global maximum of f(x) at x = 2. O -1 is a global and local minimum of f(x) at x = 1. 7 is a local maximum of f(x) at x = - 1.Find all the critical numbers of f(x) = x3 + 3x2. O a.) X = - 2 ( b.) x = - 2 and x = 0 O c.) x = 0 and X = - W / N ( d.) x = 0 and x = - 3Find all the critical numbers of f(x) =x +3x2 -1, then determine the local minimum and maximum points by using a graph. a.) Critical number: x = 0 O (-2,3) is a local maximum, and (0, - 1) is a local minimum. b.) Critical number: x = - 2 O (-2,3) is a local maximum. c.) Critical numbers: x = - 2 and x = 0 O (-2,3) is a local maximum and (0, - 1) is a local minimum. d.) Critical numbers: X = - W/ N and x = 0 O 59 27 is a local maximum, and (0, - 1) is a local minimum.Find the global maximum and minimum points of the function f(x) = - 5x 3 +X +4 on the interval [ - 1,3]. O a.) The global minimum is -2 at x = - 1. The global maximum is 4 at x = 0. O b.) The global minimum is -2 at x = - 1. The global maximum is - 2:/9 +4 at x = 3. O c.) The global minimum is - 219 +4 at x = 2. The global maximum is 4 at X = 0. O d.) The global minimum is - 314 +4 at x = 2. The global maximum is 4 at X = 0.Determine if the requirements for Rolle's theorem are met by the function f(x) = - 2x +54x +5 on the interval [- 3, 6]. If so, find the values of c in ( - 3, 6) guaranteed by the theorem. a.) f(x) is a polynomial so it is continuous on [ - 3, 6] and differentiable on (- 3, 6). When evaluated, f(-3) = - 103 and f(6) = - 103. Therefore, f(a) = f(b) and the conditions of Rolle's theorem are met. O The value guaranteed by Rolle's theorem is c = 54. b.) f(x) is a polynomial so it is continuous on [ - 3, 6] and differentiable on ( - 3, 6). When evaluated, f(- 3) = - 103 and f(6) = - 103. Therefore, f(a) = f(b) and the conditions of Rolle's theorem are met. O The value guaranteed by Rolle's theorem is c = 3. c.) f(x) is a polynomial so it is continuous on [ - 3, 6] and differentiable on (- 3, 6). When evaluated, O f(- 3) = 383 and f(6) = - 1399. Therefore, f(a) # f(b) and the conditions of Rolle's Theorem are not met. d.) f(x) is a polynomial so it is continuous on [ - 3, 6] and differentiable on (- 3, 6). When evaluated, f(-3) = - 103 and f(3) = - 103. Therefore, f(a) = f(b) and the conditions of Rolle's theorem are met. O Since we want all values on the interval ( - 3, 6), the values guaranteed by Rolle's theorem are c = - 3 and c = 3.Determine if the conditions of the mean value theorem are met by the function f(x) = x- - 2x -3 on [ - 4, 3]. If so, X+3 find the values of c in ( - 4, 3) guaranteed by the theorem. a.) f(x) is continuous on [ - 4, 3] and differentiable on the interval ( - 4, 3). O The values guaranteed by the mean value theorem are c = - 3+ 15 . b.) f(x) is continuous on [ - 4, 3] and differentiable on the interval ( - 4, 3). O The values guaranteed by the mean value theorem are c = - 3+ \\ 15 . c.) f(x) is continuous on [ - 4, 3] and differentiable on the interval ( - 4, 3). O The values guaranteed by the mean value theorem are c = - 6 and c = 0. d.) The function is NOT continuous on [ - 4, 3], and therefore, the mean value theorem does not apply for O this function on the given interval.e 5x - 1 Evaluate the following limit: lim x + 0 3x a.) e 5x -1 lim =0 O X + 0 3x b.) e 5x - lim = O X +0 3x C.) 5x e lim = O 3 X + 0 3x d.) 5x e -1 lim = OO O X + 0 3xEvaluate the following limit: lim X - 1+ Inx X - 1 a.) O lim X- 1+ Inx X - b.) lim = DO O x - 1\\ Inx X - C.) O lim x - 1 Inx X - d.) O lim x - 1\\Inx X - 1 2Evaluate the following limit: lim (2x + 1)5cotx x + 0+ O a.) lim (2x + 1)Scotx = 10 x - 0+ O b.) lim (2x + 1)Scotx = 10 x - 0+ O c.) lim (2x + 1)Scotx = 5 x - 0+ O d.) lim (2x + 1)Scotx =5 x + 0+25 Use summation formulas to evaluate the sum: _ (2k- -5k-1). k =1 a.) 25 O (2k2-5k-1)=5175 K =1 b.) 25 O E (2k2-5k-1) =5199 C.) 25 O E (2k2-5k-1) =9400 K=1 d.) 25 E (2k2-5k-1) = 9424 O K =1Given In(xy) + 5yz = 10x +15, compute dy dx a.) dy 30x3y+y = O dx x+ 10xy- b.) dy _ 30x3y-y+15 O dx x+ 10xy- c.) dy _ 30x y-y O dx x + 10xy- d.) dy = 30x y O dx 1+ 10xy2