Rate of Change Word Problems, Derivatives and Product and Quotient Rule (15 Problems)
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71. The local game commission decides to stock a lake with bass. To do this 200 bass are introduced into the lake. The
population of the bass is approximated by P(t) = 20 (10 + 7t)/(1 + 0.02 t) where t is time in months. Compute P(t) and P'(t) and interpret each.
57. The monthly sales of a new computer are given by q(t) = 30t ? 0.5 t^2 hundred units per month t months after it hits the market, where 0 < t < 7. Compute q(3) and q'(3) and interpret each.
58. Suppose that the computer described in Exercise 51 has a retail price, in dollars, given by P(t) = 2200 ? 34t^2 in
t months after it hits the market, where 0<t<7. Compute p(3) and p'(3) and interpret each.
59. For the computer described in Exercises 57 and 58:
(a) Determine the revenue function R(t). Do not simplify.
(b) Determine the rate of change of revenue R'(t).
(c) Compute R(3) and R'(3) and interpret each.
Please see the attached file for the fully formatted problems.
A few problems on Derivatives of products and Quotients attached in the file to be done. Problems to be done are highlighted.
They again are 9,11,13,15, 25, 27, 29, 35, 41, 43, 45, 57, 58, 59, 71.
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Solution Summary
Fifteen rate of change word problems, derivatives and product and quotient rule problems are solved. The solution is detailed and well presented.
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Please see the attached file for the complete solution.
9 this is of form u(x)v(x) now d/dx(uv) = du/dx v + u dv/dx This is the chain rule. All of these product functions can be treated this way u'(x) = 3: v'(x) = 2
11) Is the same format here u(x)=5x+3 v(x) = 3x^3 +2x^2 +1
u'(x)=5:v'(x)=9x^2 +4x Then y=u(x)v(x) and y'(x)= u'(x)v(x) + u(x)v'(x) =5(3x^3 +2x^2 +1) +(5x+3)(9x^2 +4x)
13) Still the same format u(x)=(3x^2-2x+1) v(x)=2x^2 + 5x -7 u'(x) = 6x-2 v'(x)=4x + 5
g(x) = u(x)v(x) g'(x) = u'(x)v(x) + u(x)v'(x) = (6x-2)( 2x^2 + 5x -7) +(3x^2-2x+1)(4x +5)
15 still the same format u(x) = v(x) = 3x-4 u'(x) = 2(1/2)x^(1/2-1) +4 = x^(-1/2) +4 v'(x)= 3
so y = u(x)v(x) and y'(x) = (x^(-1/2) + 4)(3x-4) +( )(3)
25. now these are quotients of form u(x)/v(x) or u(x)v^(-1)(x) d/dx[u(x)/v(x)]
= u'(x)/v(x) + u(x) (-1) v'(x)/v^2(x) = [u'(x)v(x) - u(x)v'(x)] x v(x)^(-2)
so u(x) = 4x -3 v(x) = 2x +1 u'(x)=4 v'(x)=2
y'(x) = [4(2x+1) -2(4x-3)]/(2x+1)^2
27 This is again of the same quotient format u(x) = 3x^2 -5x +1 v(x) = 5x^2 + 3x + 2
u'(x) = 6x -5 v'(x) = 10x +3
f'(x) = [(6x-5)( 5x^2 + 3x + 2 )-( 3x^2 -5x +1)( 10x +3)]/( 5x^2 + 3x + 2)^2
29 This is again of the same quotient format u(x) = x^(1/3)-5 v(x) = 6x-1 u'(x) = 1/3 x ^(-2/3) v'(x) ...
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