Page 1 Differentiation and Integration Calculus: Differentiation 7 71 Introduction 74 Applications of Differentiation 7.2 Differentiation 7.4.1 Profit function 73 Rules of Differentiation 7.422 Marginal Cost 7.3.1 Constant Rule 7.4.3 Marginal Revenue 7.3.2 Power Function Rule 7.44 Marginal Utility 7.3.3 Product Rule 7.4.5 Marginal Rate of Substitution 73.4 Quotient Rule 7.5 Turning points (maxima & minima) 7.3.5 Chain Rule 7.4.1 Application of Turning Point 7.346 Inverse Function Rule 76 Self practice questions 7.1 Introduction Differentiation is the method of determining the rate of change of a function. This is the basis of nomics where the rate of change of the total cost function is known as the marginal cost function. Marginal cost is the extra cost incurred when producing an extra unit of output. Similarly the rate of change of total revenue is the marginal revenue. Marginal revenue is the extra income earned from sale of extra unit of product. Microcconomics theory suggests that the optimum production level is when marginal revenue function and marginal cost function are equal, if the firm is attempting to maximize profits Behavioural theories of the firm suggest that production will be increased until the level marginal revenue is zeto (or given the acceptance minimum), Differentiation is said to find the slope of a function and since for non-linear functions the slope will change depending upon the value of x. The results of differentiating @ non-linear function will not be a single number. To find the slope at a particular point it is always useful to have the function rather than the slope at a point The ability to move between two different versions of the cost and revenue will allow modelling of several relationships in basic economics and accounataney 7.2 Differentiation Branch of mathematies that deal in minimization and maximization Can be expressed in terms of quantity, price, demand, revenue and cost D=f@) D=atbp Dif ferentiation Demand (D) Price (p)Page 2 Differentiation and Integration Differentiation is the rate of change if'y with respect to x 7.3 Rules of Differentiation 73.1 Constant Rule dy Gao Ify = 0,15, 100 find & Given that y = 15,then = = 0(15)x°-? 7.3.2 Power Function Rule Y is a function of x [y = f(x)] such that y = x” ay _ et] y= = ne] y= Fe Example: Differentiate y nax*™| y=f) Example: Dif ferentiate y = 4x? dy a4 73 Combined rule of (i) and (ii) Y isa function of x [y = f()] such that y = x" + k [2 = ne" + 0 = nx*] This isthe combination of constant rule and y = fC) 2x Illustrations Differentiate the following: 2x3 + 3x41 Solution i) ys2x*taxd1 wy - - - = 3(2)x5t + 1)x11 + Ox aPage 3 Differentiation and Integration ii) iti) #22) et43Q)et+ cart Baxtpe-S 7.3.3 Product Rule ‘Suppose y = mn Where m and n are functions of x m.n FO) dy dn, dm < +n This is referred to as product m ax dx th ag Thisis referred to as product rule Example: Differentiate y = (x? ~ 2)(x + 2) In the above equation the first part (x? ~ 2) represent m while the second part (x + 2) represent n Hence dy 2 ® 2 G2 —-n) ++ DED ay Fe 7 2) + 2x? + tx) x? + dae — 2 7.3.4 Quotient Rule « Suppose Y= Where v and w are functions of x u ya fe ay vit yt 2 ae this is referred to as Quotient Rule Example: Given y = =? find(Page 4 Differentiation and Integration dy _@+DG)-6?7-D) dx G+ 2)@ +2) dy 2x? 44x—3x2 +2 dx x +4 $4 dy _x?+4x42 de P4444 dy _@+1)@+2) dx @ +2042) dy _(x+1) dx” @¥2D) 7.3.5 Chain rule or Function of a function rule or Composite funetion rule ify is a fimetion of m, and m is a function of x, that is ify = f(m) and m = f(x) then y is called the composite function of x and the derivative of y with respect to x is equal to the derivative of y with respect to m, times the derivative of m with respect to xcthis is expressed symboli If Then dy dy dm Ge dm t dx (isis referred to as Chain Rule Example 1: y =m? and m = 2x? + 2x —1 find Solution: ‘This means that y = (m)? f(x) When m = 2x? + 2x — 1 then y = (2x? + 2x — 1)? x am Thus = 2m; = 4x +2 ‘Therefore, using the formulae for Chain rule & = & a di Pm ame car 42) Then subsitute form dy 2 y= 22H? + 2x1) + Ax +2) Example 2: Differentiate z = y!® where y = 2+ 3x Solution Given z = y"® where y = 2+ 3x Differentiate z with respect y then y with respect to x dz, dy 7 Wy and T= 3 Substitute the values in the formulaPage 5 Differentiation and Integration dz_dz dy gg Ten By de 7 1073 = 30? duty =2 43x dz ~ Hence 0(2 + 3x)" Example 3: Differentiate y = (3x? + 7)!° where y = 2+ 3x Solution Let y =u? where u= 3x +7 Differentiate y with respect u then w with respect to x du 2 = tou and S = 6x ay Oe Substitute the values in the formula BOO iow vox = °) =3x2 +7 = = ou! 6x = 6xC1OW) but w= 3x? + dy Hence & fence 0x(3x? +7)? Example 4: Differentiate y = x — x? where x Solution Differentiate y with respect x then xwith respect to z 6S ang de ene eB Substitute the values in the formula dy_dy dx_ sy(7h eae ae 8-1) 7.3.6 Inverse Funetion Rule Ify = f(), an inverse function x = g(y), that is 96) = f~*(y) exists only and only if each value of y yields one and only one value of x. If the inverse function exists, the inverse function rule states that the derivative of the invrse function is the reciprocal of of the derivative of the original function, that is, dx 1 1 a = gy Which can also be written as fF Q)= FO dx Example 1: Find the derivative for the inverse of the following function: i) yame dy dx 1 1 Dem then === axe MO yn de Example 2: Find the derivative forthe inverse of the following function: i) qQeur dq rue a aPage 6 Differentiation and Integration Example 3: Find the derivative for the inverse of the following function: iii) Can you solve for L = q? Task Differentiate the followin, ) y=xt8 ii) y=2xt 43x41 ii) y=vxe iy y= x Yo yea vi) yas eS vi) Y= vii) y= fs 7.4. Application of differentiation in economics and business i) It is applied in deriving the profit function: Profit (7) = Total Revenue (TR) — Total Cost (TC) That is = TR -TC But TR = (AR)x where x is quantity and AR is the average revenue And TC = (AC)x where x is quantity and AC is the average cost Example: The revenue function of an itemis AR = 1000 — 2x It is known that AR leads to TR; TR = (AR)x where x is quantity Then it follows that, the Total Revenue TR = (1000 - 2x)x ii) It is applied in deriving Marginal Revenue function Marginal revenue is the increase in revenue that results from the sale of one additional unit of output It is the corresponding change in total revenue or total sales when demand or sale increase by one unit This can be expressed geometrically as: Change in Revenue Marginal Re ————o— ar ginal Keven’ = Change in Quantity TR MR =—— where TR is Total Revenue and x is Demand or Sales ‘The above can be expressed by means differentiation as indicated: aR dTR ATR MR = “7 -which can also be written as MR = 7>-or 7Differentiation and Integration This means that Marginal Revenue (MR)is a derivative of Total Revenue (TR) Let us now find the marginal revenue of the previous example i.e. Find the marginal revenue when TR = (1000 ~ 2x) aTR _ d(1000x ~ 2x?) ce dx MR = 1000 — 4x It is applied in deriving Marginal Cost (MC) Marginal cost of production measures the change in the total cost of a good that arises from producing one additional unit of that good. Change in Total Cost Mar ginal Cost = Tio ge in Quantity arc MC == where TC is Total Cost and x is Quantity produced The above can be expressed by means differentiation as indicated! arc ; arc dTR MC = <= which ean also be written as MC = “Fe or This means that Marginal Cost (MC) is a derivative of Total Cost (TC) Example The average cost function ofan item is AC = 13x? ~ 210x +1150 Ibis known that AC lead to TC; TC = (AC)x where x is quantity produced ‘Then it follows that, the Total Cost TR = (13x? ~ 210x + 1150)x ATC _ d(13x? — 210x? + 1150%) “dx dx MC = 39x? — 420x +1150 MC = Using the above Total revenue function and Total Cost function, we can now derive the profit function as indicated: Recall that: TR = (1000x ~ 2x) and TC = (13x? - 210x? + 1150x) ‘Therefore TR—TC = (1000x ~ 2x?) ~ (13x3 — 210x? + 1150x) m= 1000x ~ 2x? ~ 13x3 + 210x? ~ 1150x m= 208x? ~ 150x — 13x? ‘Note: Profit is maximized when Marginal Revenue is Equal to Marginal cost Example: ‘A firm has analysed their operation conditions, prices and costs and have developed the following functions: Revenue, (R) = 400Q — 49? Cost, (C) = Q? + 100 +30 Where @ is the number of units sold ‘The firm's objective is to maximize profits and wishes to know: a) What quantity should be sold? b) At what price should it be sold? ©) What will be the amount of profit?ii) Differentiation and Integration Solution: a) What quantity should be sold? It is known that profit is maximized when MR = MC MR = MC = Maximum Profit dR MR= 2 = 400 -8Q dc a =2Q+10 At maximum profit the marginal revenue is equal marginal cost + 400—8Q = 20 +10 10Q 90 Q=39 'b) At what price should it be sold? ‘The quantity that should be sold to maximize profit is = 39 units Price (P) = Average Revenue (AR) = Revenue per unit sale Total Revenue = 4009 — 497 = ap = TR — 4000-402 ButP = AR =" > P= AR = 400~4Q but Q = 39 ‘Thus P = AR = 400 ~ 4(39) = 400 ~ 156 = 244 ‘The products will thus be sold at Sh.244 per unit MC ©) What will be the amount of profit? Profit (m) = Total Revenue (TR) ~ Total Cost (TC) That is w= TR —TC TR = 4009 - 49? TC = Q? +109 +30 =TR-TC (400@ — 4Q*) — (Q? + 109 + 30) 400(39) — 4(39?) — 39? — 10(39) — 30 15600 — 6080 ~ 1521 ~ 390 ~ 30 = 7579 It is applied to find elasticity of demand at a point on the demand curve ‘The basic definition of the price elasticity of demand is that it measures the responsiveness of the quantities sold (demand) to changes in prices Price elasticity of demand is thus given as Proportionate change in quantity proportionate change in prices She Phy ‘The above equation can be rewritten by doing away with the fraction in the denominatorPage 9 Differentiation and Integration Thus Ep sah AX fy wP yy = ELE a, Py apeg Hence Axep_ Ax p ?dpeq apg Remember the above formula gives elasticity over a range of output levels; but what is required is elasticity at a point. Now the change in Q divided by P is a slope, and the differentiation gives the slope of the function at a point ‘Thus, differentiate the demand function and divide by [price (i.e. average revenue) divided by quantity]. The price elasticity of demand is thus found. D=atbp Ap Di EB FE = Differentiation Price (P) ae s i.e. and if you ean remember inverse funetion rule, then the differential (price elasticity of demand) becomes ay Simo $5 = ay Demand (x) Example: if a firm’s demand function is p = AR = 1000 ~ 25x ‘And it is decided to produce 20 units, then the price would be: p = AR = 1000 ~ 25(20) = 1000-500 00 Now dx 1 1 ap —25, and so 28 dx The price elasticity of demand at this point will be: 1 Ey rag huey = 500 wheng = 20 =25 p, -2 , 500 » 225" 20 Ep=-1 What does the above price elasticity of demand mean? See the interpretations below: The interpretations of price elasticity of demand are as follows a. Price elasticity can either be positive (increasing the prices) or it can be negative (decreasing the prices)Page 10° Differentiation and Integration b. When price elasticity of demand is greater than one, demand is elastic. This can be interpreted as consumers being very sensitive to changes in price: a 1% increase in price will lead to a drop in quantity demanded of more than 1%, ©. When price of elasticity of demand is less than one, demand is inelastic. This can be interpreted as consumers being insensitive to changes in price: a 1% increase in price will lead to a drop in quantity demanded of less than 1%, d. When price of elasticity of demand is equal to one, demand is unitary. This ‘ean be interpreted as consumers responding proportionally to changes in rice: a 1% increase in price will lead to a drop in quantity demanded by the same 1%. Note that: ‘+ A-good with more close substitutes will likely have a higher elasticity. ‘+ The higher the percentage of a consumer's income used to pay for the product, the igher the elasticity tends to be. ‘+ For non-durable goods, the longer a price change holds, the higher the elasticity is likely to be. The more necessary a good i the lower the price elasticity of demand 7.3 Turning Points (Maxima and Minima Values) Ify is a function of x that is y = f(x) then we either get a straight line or a curve Linear Curvilinear ~ lea ‘Some functions have turning points of minima and maxima and these points have business applications of minimum cost and maximum output or profits or revenue 7 Maxima point i Maxima point Quantity Quantity Minima and maxima are points of zero gradient or zero slope. ‘The derivative of a function shows the slope or the rate of change of a function. This means that when a derivative is calculated and equated to zero, then it shows the turning point of the function,| Page 11 Differentiation and Integration Certain functions have more than one turning point hence is a minima or a maxima. is necessary to test whether the turning point “ « For maxima 2 = 0 and > <0 anima 2% ay For anima = 0 and £3>0 Note that © is referred to as the second derivative since © is the first derivative Example: Given that y = 2x? — 8x + 50 check whether it has a turning point and if it is a minima or maxima turing point, and which point is it (that is which coordinate?) First get the first derivative and equate it to zero: =4x-8=0 8 a@ Therefor, 4x Thus x = $= 42 ‘This means that the turning point is at x = 2 Secondly get the second derivative from the first derivative Remember that we have computed the first derivative and found 4 — 8 So, a = 4. You realize that the second derivative is positive hence the turning point is a Thirdly, substitute for x in the original equation (y = 2x? — 8x + 50) to find y Since x = 2, then y = 2(2)? ~ 8(2) +50 = 42 Thus y = 2x? — 8x +50 has a minimum turning point at (2,42) Example 2: Given y = 4x? +5 + 10 determine whether it has turning point hence find the turning point, Ify = 2x3 — 4x? + 15x +10 then the first derivative shall be: 2 8x +15 bi ine = 0 rence 8x+15=0 Bax? ut at turning point = 0 hence = x? = 8x + 18 = When we factorize x2 — 8x +15 = 0 we shall obtain (x — 3) = 0 and (x — 5) Thus x = 3 and x = 5 which means that the function has two turning points 2x-8 The second derivative © = Using the second derivative we ean now determine the turning points for x = 3 and x = § Atx = 3, 2x8 = 2(3) — 8 = 6 — 8 = —2 The value is negative hence this is maxima point Atx =5, 2x —8 = 2(5) — 8 = 10 —8 = 2 The value is positive hence this is minima point The last step is to find the coordinates of the turning point, as follows Atx = 3, y= 5x3 — dx? + 15x +10 = 5 (3)? — 4(3)? + 15(8) +10 = 28Page 12° Differentiation and Integration Ata =5, y=tx?— 4x? + 15x +10 = 4 (5)? — 4(5)? + 15(5) + 10 = 26.67 ‘Thus the function y = 4x? + 15x + 10 has the following two turning points: thas a maximum turning point at (3,28) and a minimum turning point at (5,26.67) 7.5.1 Application of turning points in Business Its applied in cost minimization and profit maximization or output maximization problems Example: Ifa firm has TR = 40x — 8x? and TC = 8 + 16x +x? Where x = thousands of units of production its profit function (rr) will be the difference between its total revenue and total cost m= TR-TC m= (40x — 8x?) — (8 + 16x +x") m= Ox ~ 8x? — 8 — 16x +x? 1 = ~8 + 24x — 72x? this is the profit function At turing point where the profit is maximum, the slope is zero. ‘Therefore, we find the first derivative of the profit function we shall get zero: Hence we can now differentiate profit (m) with respect to the quantity output (x) Given n = -8 + 24x — 722, then = 24 — 14x = 0 Hence x == 1.714 We can now compute the second derivative from the first derivative (24 — 14x) We get £3 = 14, this is negative which represent a maxima point Hence achieve maximum profit when x = 1.714 and since x is measured in thousands, this is 171 4units. We can now confidently say that the firm maximizes profits when 1714 units are produced. ‘What shall be the total profits? To get the total profits, you substitute x = 1.714 in the profit function m = -8 + 24x — 7x*. You will get m = 12.571 but since the values are in thousands, the profit will be Sh.12571 ‘What shall be the sales maximizing output and the maximum sales? 7.6 Self-practice questions i) Ifa firm has TR = 40x ~ 8x? and TC = 8 + 16x + x7 Determine: a. The profit maximizing output b. The maximum profit fi) Describe briefly how differential calculus and integral calculus are used to solve problems in business managementDifferentiation and Integration ) Kamau and Mutua Company limited has the following demand and total cost function for a particular item: Demand p = 5625 ~ 3q? Total cost TC = 500 + 846q Where p is the price of the item, q is the quantity of produced and sold, while TC is the total cost of the item, Required: a, Determine the price and quantity for maximum sales revenue and the ‘maximum revenue b. Determine the price and quantity of maximum profit and determine the ‘maximum profit iv) The demand q for a product manufactured by Nyangwachi industries change with price p in Kh. of the product and is given by: q = 30,000 ~ Sp The total cost of manufacturing q units is given by: C = 300g + 0.04q? + 175000 Required: a. Determine the number of units that should be manufactured and sold in order to maximize the profits for the industry 'b. What price should be charged to the product to maximize profits? Calculate the expected maximum profits 4. Explain briefly the application of: i) marginal cost, and break even analysis 100 —5q and the output produced is 8, vy) If the demand curve is given by Determine a. The slope of the curve b. The price elasticity of demand €. The interpretation of the price elasticity of demand obtained in (i) above