Differentiation

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DIFFERENTIATION Dr. T.K. Jain.

AFTERSCHO☺OL Centre for social entrepreneurship Bikaner M: 9414430763 [email protected] www.afterschool.tk, www.afterschoool.tk www.afterschoool.tk 04/30/09 www.afterschoool.tk

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DERIVATIVE OF X^N • N * X ^ (N-1)

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DERIVATIVE OF X^10 • = 10 X ^ 9

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DERIVATIVE OF 5X^3 – 2X^2 + 10 X -3 • 5*3 X^(3-1) -4X + 10 • =15X^2 – 4X +10

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DERIVATIVE OF (3X^2 -1)/(X+1) • • • •

[(X+1) * (6X ) - (3X^2 -1) (1) ] / (X+1)^2 (6x^2 + 6x - 3x^2 + 1) / ( x^2 + 2x + 1) =(3x^2 + 6x +1) / (X^2 + 2x +1) Answer.

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Differentiate e^x / x^2 • • • • •

= [(e^x * 2X) - (X^2 * e^x) ]/ (X^4) = (2x e^x - X^2e^x ) / X ^4 X (2e^x - Xe^x) / X^4 = (2e^x - Xe^x) / X^3 =e^x (2 - X) / X ^3 answer.

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Differentiate 2 ^(X^3 +X^2 +1) = 2^ (X^3 + X^2 + 1) * log2 * differentiation of (X^3 + X^2 + 1) = 2^ ( X ^3 + X ^2 + 1) * log 2 * ( 3x^2 + 2x) Answer.

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Differentiate (2x^3 + x^2 + 1) ^(3/2) • = 3/2 * ( 2x^3 + X^2 +1)^(1/2) * differentiation of (2x^3 + X^2 +1) • =3/2 * ( 2x^3 + X^2 +1)^(1/2) * (6x^2 + 2x) • Answer.

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Find differentiation of log (X^2 + 1) • = 1/ (x^2 + 1) * differentiation of (X^2 +1) • = 1/ (x^2 + 1) * 2X • Answer.

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Differentiate (log x)^2 • = 2 * log X * diffferentiation of log x • = 2 * log x * 1/x answer

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Find the total cost when output is 15000, C = (X^2 / 10000) + 500 • • • •

X = 15000 Thus C =[ (15000)^2 / 10000] + 500 C = 22500 + 500 Total cost = 23000 answer.

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Find the marginal cost when output is 15000, C = (X^2 / 10000) + 500 • Differentiate the function : • = 2X / 10000 X = 15000 = 30000/ 10000 = Rs. 3 answer.

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Find the average cost when output is 15000, C = (X^2 / 10000) + 500 • Average is obtained by dividing the total cost by the number of units. • The number of units is denoted by X. • Thus divide total cost by X • = [(X^2 / 10000) + 500 ] / x • =(x/10000) + (500/x) • = 1.5 + 500/15000 • = 1.53 answer. 04/30/09 www.afterschoool.tk

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During a period, a retail shop can sell X units at price P. What is the total revenue, if P = 20 - .03X and X = 50 ? • Total revenue is total units multiplied by price. • = (20 -.03X) X • =(20 – 1.5) * 50 • = 925 answer. 04/30/09 www.afterschoool.tk

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During a period, a retail shop can sell X units at price P. What is the marginal revenue, if P = 20 - .03X and X = 100 ? • • • • •

Differentiate total revenue Total revenue = P * X= (price * number of units) =(20 - .03x) X Differentiated = 20 - .06X (put x = 100) = 14 Answer

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During a period, a retail shop can sell X units at price P. What is the average revenue, if P = 20 - .03X and X = 100 ? • • • • •

Average revenue = total revenue / no. of units =(20X - .03X^2) / X = 20 - .03x = 20 – 3 = 17 answer.

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Find the maximum value of the function : f(x) = 2x^3 + 3x^2 -12x +1 • Maximum value is when first order differentiation is zero & second order differentiation is less than zero. • First differentiation : • 6X^2 + 6x -12 = 0 • X^2 +x – 2 = 0 • X^2 +2x –x -2 =0 • X (X +2) -1(X +2) •04/30/09 Thus X = 1 or -2. AFTERSCHO☺OL's MATERIAL 18 www.afterschoool.tk

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Solution…. • Second differentiation: • 12X + 6 • Here if we put the value of X = -2, we get value in negative. • f(x) = 2x^3 + 3x^2 -12x +1 • = 2 * ( -8) + 3 * 4 +24 +1 • =-16 +12 +24 +1 • =21 answer. 04/30/09 www.afterschoool.tk

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Find the minimum value of the function : f(x) = 2x^3 + 3x^2 -12x +1 • Minimum value is when first order differentiation is zero & second order differentiation is less than zero. • First differentiation : • 6X^2 + 6x -12 = 0 • X^2 +x – 2 = 0 • X^2 +2x –x -2 =0 • X (X +2) -1(X +2) 04/30/09 AFTERSCHO☺OL's MATERIAL 20 • Thus X = 1 or -2.FOR PGPSE PARTICIPANTS www.afterschoool.tk

Solution… • Second differentiation: • 12X + 6 • Here if we put the value of X = 1, we get value in positive, so X = 1 • f(x) = 2x^3 + 3x^2 -12x +1 • =-6 answer. ( minimum value).

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Differentials • Definition – (dy/dx = f (x)) – dy = f (x) Δx

• Use: – Δy ≈ dy

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A truck has a top speed of 75 miles per hour. When travelling @x miles per hour, consume diesel @ 1/200 ((1600/x) +x) liters per mile. Distance to be travelled is 200 KM and price of diesel is Rs. 16 per liter. What is the most economical speed of the truck? 04/30/09 www.afterschoool.tk

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Solution… • Minimum value is when first order differentiation is zero & second order differentiation is less than zero. • Cost of trip = use of diesel: 200 * 1/200 ((1600/x) +x) liters • Cost = qty * price = ((1600/x) +x) * 16 • Differentiation of ((1600/x) +x)*16 • = 16 ((-1600 / x^2) + 1) = 0 (by law) 04/30/09 www.afterschoool.tk

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• • • • • •

= 16 ((-1600 / x^2) + 1) = 0 =16((-1600 + X^2) / X^2) = 0 =-25600 + 16X^2 = 0 =16x^2 = 25600 X^2 = 1600 X = 40

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2 order differentiation… nd

• • • • • • • •

((1600/x) +x)*16 Where X = 40 Second differentiation: 16 ((-1600 / x^2) + 1) 16( 1600 *2 / X^3 ) =16 (3200/ 64000) <0 Thus this condition is fullfilled. Thus X = 40 Therefore the ideal speed is 40 KM per hour.

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About AFTERSCHO☺OL • PGPSE - World’s most comprehensive programme on social entrepreneurship – after class 12th • Flexible – fast changing to meet the requirements • Admission open throughout the year • Complete support from beginning to the end – from idea generation to making the project viable. 04/30/09 www.afterschoool.tk

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Branches of AFTERSCHO☺OL • PGPSE programme is open all over the world as free online programme. • Those who complete PSPSE have the freedom to start branches of AFTERSCHO☺OL • A few branches have already started one such branch is at KOTA (Rajasthan).

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Workshop on social entrepreneurship • We conduct workshop on social entrepreneurship – all over India and out of India also - in school, college, club, association or any such place - just send us a call and we will come to conduct the workshop on social entrepreeurship. • These workshops are great moments of learning, sharing, and commitments. 04/30/09 www.afterschoool.tk

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