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 March 5th, 2012, 01:16 AM #1 Senior Member   Joined: Jan 2012 Posts: 100 Thanks: 0 derivatives of exponential functions Hey I have been working on my assignment and I cannot get the correct answer: the question is: The average cost of producing q units of a product is given by $\bar{c}=\frac{880}{q} + 3500 \cdot \frac{e^(2q+6/880)}{q}$ What is the marginal cost if q=98 my keyboard is a little mesed up. it should be e^2q+6/880 The answer i get is 12.8400...but it is incorrect... I tried getting rid of the q by multiplying. Please help!
 March 5th, 2012, 03:22 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: derivatives of exponential functions $\bar{c}=\frac{880+3500e^{2q+\frac{3}{440}}}{q}=\fr ac{20$$44+175e^{2q+\frac{3}{440}}$$}{q}$ $\frac{d\bar{c}}{dq}=20$$\frac{q\(2\cdot175e^{2q+\f rac{3}{440}}$$-44-175e^{2q+\frac{3}{440}}(1)}{q^2}\)$ $\frac{d\bar{c}}{dq}=\frac{20}{q^2}$$175(2q-1)e^{2q+\frac{3}{440}}-44$$$ Multiplying through by q will also work: $\bar{c}q=880+3500e^{2q+\frac{3}{440}}$ Implicitly differentiate with respect to q: $\bar{c}+q\frac{d\bar{c}}{dq}=7000e^{2q+\frac{3}{44 0}}$ $\frac{d\bar{c}}{dq}=\frac{1}{q}$$7000e^{2q+\frac{3 }{440}}-\bar{c}$$$ $\frac{d\bar{c}}{dq}=\frac{1}{q}$$7000e^{2q+\frac{3 }{440}}-\frac{1}{q}\(880+3500e^{2q+\frac{3}{440}}$$\)$ $\frac{d\bar{c}}{dq}=\frac{1}{q^2}$$7000qe^{2q+\fra c{3}{440}}-880-3500e^{2q+\frac{3}{440}}$$\)$ $\frac{d\bar{c}}{dq}=\frac{20}{q^2}$$175(2q-1)e^{2q+\frac{3}{440}}-44$$$ So now we let q = 98: $\frac{d\bar{c}}{dq}\|_{q=98}=\frac{20}{98^2}$$175( 2\cdot98-1)e^{2\cdot98+\frac{3}{440}}-44$$\approx9.47\,\times\,10^{86}$ I suspect you did not properly express the exponent of e with bracketing symbols, i.e., e^((2q+6)/880) = e^((q+3)/440). I cannot stress enough how important it is to express the problem in such a manner that guessing what is meant is not necessary. However, you should now have a better idea of how to work this problem.
 March 5th, 2012, 08:23 AM #3 Senior Member   Joined: Jan 2012 Posts: 100 Thanks: 0 Re: derivatives of exponential functions Nvm, I worked it out and solved the problem!!

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### the average cost of producing q units of a product is given by

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