My Math Forum Derivatives: Exponential Function

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 October 23rd, 2013, 04:46 PM #1 Newbie   Joined: Oct 2013 Posts: 3 Thanks: 0 Derivatives: Exponential Function The average cost of producing q units of a product is given by C average=(860/q)+4100(e^((3q+3)/860)/q) Find the marginal cost if q=101 .
 October 24th, 2013, 01:40 AM #2 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: Derivatives: Exponential Function marginal cost = dC/dq = -860/q^2 + (4100/860)*(-3/q^2)*(e^((3q+3)/860)/q) Let substitute 101 for q.
 October 24th, 2013, 02:19 AM #3 Senior Member   Joined: Oct 2013 From: Sydney Australia Posts: 126 Thanks: 0 Re: Derivatives: Exponential Function marginal cost is the change in the total cost that arises when the quantity produced changes by one unit. Firstly (3q+3)/860)/q = (3q+3)/(860q) If f(q)=(3q+3)/(860q) then f'(q) = [(860q)(3) - 860(3q+3)] /[860^2 * q^2] using the quotient rule =[3q - (3q+3)] /[860 * q*q] = -3 / (860qq) C average= 860*q^(-1) + 4100e^[(3q+3)/860q] dC/dq = (-860 /q^2) + 4100 * [ -3 / (860qq) ] e^[(3q+3)/860q] When q=101 dC/dq = (-1/10201)[(860 + 14.3023255 e ^ (306 / 86860)] = -0.086 This doesn't sound right. It would ususally be a positive answer. It can be negative it just would rarely be.

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