Math 131. Test 2, Form 115, Friday, October 31 Name. Instructions. Do each of the following 8 questions. Question 1 is worth 10 pts, the remaining questions are worth 5 points each. Please show all appropriate work, and do your best! Express answers in exact form, unless asked to round. You may use a scientific non-graphing calculator. 1. State the requested derivative rules. Solution: (a) Constant Multiple rule: d [ch(x)] = ch0 (x) dx (where c is a constant) f 0 (x)h(x) − h0 (x)f (x) d f (x) = (b) Quotient rule: dx h(x) h2 (x) (c) Difference rule : (d) Power rule: d p [x ] = pxp−1 dx (e) Product rule: (f) Chain rule: d [h(x) − f (x)] = h0 (x) − f 0 (x) dx (where p is a rational number) d [h(x)f (x)] = h0 (x)f (x) + f 0 (x)h(x) dx d [f (h(x))] = f 0 (h(x)) · h0 (x) dx (g) d [sin x] = cos x dx (i) d [tan x] = sec2 x dx (h) (j) √ 2. Let f (x) = 18x + x3 d [cot x] = − csc2 x dx d [csc x] = − csc x cot x dx √ 10 x − πx . x9 (a) Rewrite f (x) using exponents to make its derivative easy to compute: (b) Find f 0 (x) √ √ 18x1/2 , and distribute the x3 and then use properties of exponents √ √ √ 1/2 3 10 x 3 f (x) = 18x + x − x (πx) = 18x1/2 + 10x−11/2 − πx4 x9 Solution: (a) First, to obtain 18x = (b) Now using basic rules for differentiation √ 18 −1/2 0 f (x) = x − 55x−13/2 − 4πx3 2 3. (a) State the limit definition for the derivative of a function f (x). (b) Let f (x) = 6x2 − 9x. Use the definition of derivative to find f 0 (x). Solution: (a) The derivative of f at x is f (x + h) − f (x) h→0 h f 0 (x) = lim provided the limit exists. (b)For f (x) = 6x2 − 9x, we compute f 0 (x) = f (x + ∆x) − f (x) ∆x→0 ∆x lim = 6(x + ∆x)2 − 9(x + ∆x) − (6x2 − 9x) ∆x→0 ∆x = 6(x2 + 2x∆x + (∆x)2 ) − 9x − 9∆x − 6x2 + 9x ∆x→0 ∆x = (∆x)(12x + 6∆x − 9) ∆x→0 ∆x = lim lim lim lim 12x + 6∆x − 9 = 12x − 9. ∆x→0 4. Let f (x) = 13 csc x + 3x3 . (a) Find f 0 (x) (b) Find f 00 (x) Solution: (a) f 0 (x) = −13 csc x cot x + 9x2 (b) Then f 00 (x) = −13(− csc x cot x)(cot x) − 13(csc x)(− cot2 x) + 18x1 = 13 csc x cot2 x + 13 csc3 x + 18x1 5. A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the front of the boat, which is 12 feet below the level of the pulley. If the rope is pulled through the pulley at a rate of 27 ft/min, at what rate will the boat be approaching the dock when 140 ft of rope is out? Round your final answer to four decimal places, and state the units of measure in your answer. Solution: Let x be the horizontal distance of the front of the boat from the dock and let r be the length of rope that is out. Then by the pythagorean theorem 122 + x2 = r2 . Differentiating this equation with respect to t we find 2x dx dr dx r dr = 2r and so = · dt dt dt x dt dr From the information given in the question, we know = −27 ft/min (the negative is because dt dx the rope is being pulled in). We are asked to find when r = 140 feet. With this information, dt from the pythagorean equation above, we know p x = 1402 − 122 ≈ 139.4848 and therefore, dx 140 =√ · (−27) ft/min. ≈ −27.0997 ft/min. dt 1402 − 122 The negative sign indicates that the boat is approaching the dock. Thus, the boat is approaching the dock at a rate of approximately 27.0997 feet per minute. 6. Let h(x) = 3x2 − 6 6x + 3 3 . Find h0 (x), do not simplify your answer. Solution: First write g(x) = 3x2 − 6 . Then 6x + 3 g 0 (x) = = = 6x1 (6x + 3) − (3x2 − 6)(6) (6x + 3)2 36x2 + 18x1 − 18x2 + 36 (6x + 3)2 2 18x + 18x1 + 36 (6x + 3)2 Therefore, h0 (x) = 3(g(x))2 g 0 (x) 2 2 18x2 + 18x1 + 36 3x − 6 = 3 6x + 3 (6x + 3)2 3(3x2 − 6)2 (18x2 + 18x1 + 36) = (6x + 3)4 √ √ y = 7 and y(25) = 4. Find y 0 (25) by implicit differentiation. √ √ (b) Find the equation of the tangent line to the graph of x + y = 7 at the point (25, 4). Write your answer in slope intercept form. 7. (a) Suppose x+ Solution: (a) Differentiating √ x+ √ y = 7 implicitly we obtain 1 −1/2 1 −1/2 0 x + y y =0 2 2 or √ y dy x−1/2 = − −1/2 = − √ dx x y √ 2 4 Thus y (25) = − √ = − . 5 25 0 (b) Using point slope form with m = −2/5 and the point (25, 4), we write y − 4 = − 52 (x − 25) and so 2 (2)(25) 2 y =− x+ +4 or y = − x + 14 5 5 5 8. For what values of x is the tangent line of the graph of f (x) = x3 − 1x2 − 15x + 2 horizontal? 3 Solution: First, f 0 (x) = x2 − 2x − 15. There is a horizontal tangent when f 0 (x) = 0 and so we solve x2 − 2x − 15 = 0, thus x2 − 2x − 15 = (x − 5)(x + 3) = 0. Thus x = 5 and x = −3 are the x-coordinates where the graph has a horizontal tangent line.

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