Math 130 Winter 2010 Quotient/Product/Chain Rule Practice January 28, 2010


 Lambert Ramsey
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1 Math Winter 0 Quotient/Product/Chain Rule Practice January 8, 0 1. Find the derivative of the following functions: a) gx) = x + 4x) 5 + x x + 1. To do this problem we treat each part seperately. F the first piece we use the fact that if we have fx) n then the derivatives is nfx) n 1 g x). F the second piece we us the quotient rule with u, the numerat equal to x and v = x + 1. Thus u = 1 and v = 1. Putting this together we get: g x) = 5fx)) 4 f x) + vu uv = 5x + 4) 4 x + 1)1) x1) x + 4) + x + 1) = 5x + 4) 4 x + 4) + v 1 x + 1) b) fx) = x + x 4x 5 x ) To do this problem we will apply the product rule with: u = u = ) 1 x + x ) ) 1 x + x x + 1 ) v = 4x 5 x ) v = 0x 4 + 6x 4 Here we have used rewritten the cube root as a power of 1 and rewritten x as x. To compute u we used the generalized power rule f functions that look like gx) n and f v it only requires our standard power rule that we learned a while ago. Putting this all together we have: f x) = uv + uv ) 1 = x + x = x + x ) ) x + 1 x + 1 ) 4x 5 x ) + x + x ) 1 0x 4 + 6x 4) ) 4x 5 x ) + x + x ) 1 0x 4 + 6x 4) 1
2 c) hx) = x4 + x x. F this problem we might first rewrite the function as: hx) = x 4 ) 1 + x x To compute the derivative we then apply the generalized power rule which gives: h x) = 1 ) x 4 ) 1 + x d x dx [ x 4 ] + x x We are not done though because we still have to compute the d dx [ ]. Which means compute the derivative of what is inside the square brackets. To do the derivative of what is inside the square brackets we have to use the quotient rule. Here we have: u = x 4 + x v = x u = 4x + v = 1 Applying this we have: ) 1 x h 4 ) 1 + x x )4x + ) x 4 ) + x)1) x) = x x ) One could simplify this me but I will not do that here.
3 . We have not yet learned this but if fx) = x then the derivative is f x) = x ln). Use this fact and the chain rule find the following derivatives a) hx) = 4x F this problem we first think of hx) as the composition of two function hx) = fgx)). Here fx) = x and gx) = 4x. You are given that f x) = x ln) although we now know this fact from class. g x) = 4. If we apply the chain rule we have: b) gx) = x + x ) 5 h x) = f gx))g x) = gx) ln)g x) = 4x ln)4) = 4) 4x ln) F this problem we apply the generalized power rule. When we compute the derivative of the inside function we have to use the fact that is given in the instructions to compute the derivative of x. When we do this we obtain: g x) = 5 x + x ) 4 x ln) + x) c) kx) = 6x x) This problem is extremely similar to the first problem on this page except that gx) = 6x x here and thus g x) = 1x. So we have: k x) = f gx))g x) = gx) ln)g x) = 6x x) ln)1x ) = 1x ) 6x x) ln) d) lx) = xx x + 1 F this problem we must use the quotient rule with: u = x x v = x + 1 u = 1) x + x x ln) v = 1 = x + x x ln) Here to compute u I had to use the product rule and the derivative of x which was given in the instructions. Applying the quotient rule we have: l x) = vu uv v = x + 1) x + x x ln)) x x 1) x + 1)
4 . Suppose the cost in dollars of manufacturing x items is given by : and the demand equation is given by: C = 000x x = p. Here p represents the price of one item. In terms of the demand x, a) find an expression f the revenue R; The Revenue is equal to the price, p, times the demand, x. So we have: Rx) = x p, but we want a function of x only so we must eliminate p from this expression. To do so we use the demand equation that relates x and p. We need to solve this equation f p in terms of x so we first square both sides giving: x = p Then subtracting we have: Dividing by gives: p = x. p = 000 x. If we use this to substitute into the revenue expression we obtain: ) Rx) = x 000 x = 000x x b) find an expression f the profit P ; The profit is the Revenue minus the Cost. Using the answer from a) we have: P x) = Rx) CX) c) find an expression f the marginal profit. = 000x x 000x 5000 = 8000x x 5000 The marginal profit is the derivative of the profit function. We can find this using the power rules we developed. Namely, P x) = 8000 x P x) = 8000 x 4
5 d) determine the value of the marginal profit when 1 items are in demand. Give a practical interpretation of this quantity. Should the company sell me less than this number? Why? We just need to evaluate P 1) which is: P 1) = ) = 1768 dollars item This means that if one was to go from selling 1 to 1 items then the profit would decrease by approximately $1,768. Thus the company certainly would not want to increase production. In fact if they decrease production from 1 to 11 then the profit would increase by approximately $1,768 so they would want to decrease production. e) find the value of x that makes the marginal profit equal to 0. What is the practical interpretation of this number? Should the company sell me less than this number x? Why? We set P x) = 0 P x) = 8000 x = 0 Now we solve this expression f x., 8000 x = 0., 8000 = x 4000 = x ± 4000 = x Since we are only interested in positive values of x it represents the number of items produced and sold). Then we take x = items. Since the marginal profit at this number of items is 0 then increasing decreasing the number of items will not change the profit, thus you would want to stay at this level of production. There is slightly me to this because we have to make sure that we are indeed at a maximum profit and not a minimum profit, but in this case we are indeed at a maximum when x = items) 5
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