Business And Financial Mathematics-differential

Derivative Sub Topics I. Definition of Derivative II. Properties of Derivative III Application Derivative IV Problems

I. Definition of Derivative Suppose f(x) is a function, then derivative of function f(x) is: F’(x) =

Lim Δx→0

F(x+ Δx)-f(x) Δx

Lim Δx→0 Lim Δx→0 Lim Δx→0 Lim Δx→0

(x+ Δx)2-x2 Δx x2+ 2xΔx +Δx2 - x2 Δx 2xΔx +Δx2 Δx 2x +Δx

Example. 2 f(x)=x f’(x) = = = = = 2x

II. Properties. 1. If f(x)=c , where c=constant then f’(x)=0 n n-1 2. For function f(x)=a x , then f’(x)= a n x 3. If h(x)= a f(x) where a=constant then h’(x) = a f’(x) 4. If h(x)= f(x) + g(x) then h’(x) = f’(x) + g’(x) 5. If h(x)= f(x) g(x) then h’(x) = f’(x)g(x) + g’(x)f(x) 6. If h(x)= f(x)/g(x) where g(x ) unequal 0 then 2

h’(x) = {f’(x)g(x) - g’(x)f(x)}/ {g(x) } 7. Composite Function Suppose h(x)=f (g(x)) then h’(x)= f ‘(g(x)) g’(x) III. Application of Derivative

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3.a Find gradient or slope of a tangent line (Tangent line) A tangent line of curve y=f(x) at x=a is f’(a). Example. Find tangent line equation of curve y =X2+ 1 at point (3,10) Answer. f’(x) = 2x Gradient of tangent line = f’(a) = 2·3=6 tangent line equation : y=6 x + c. through (3, 10) 10=6 ·3+ c c=-8. Tangent line equation : y=6 x -8 35 30 25 20 15 10 5 0 -2

-1

-5 0

1

2

3

4

5

6

-10 -15 -20

Example. Find tangent line equation of curve y = 2X2- X+ 4, through (3,17). Answer. f’(x) = 4x -1 Gradien of tangent line m=4x -1, at touch point (pada titik singgung) : Xs= (m+1)/4 and Ys = 2{(m+1)/4}2 -(m+1)/4 +4 Since the tangent line through (3,17) then m can be written as follows: m= ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔

(Ys -17)/(Xs-3) m={2{(m+1)/4}2 -(m+1)/4 +4 -17}/ {(m+1)/4 -3} m{(m+1)/4 -3}={(m2+2m+1)/8 –(m+1)/4 -13) m2/4+m/4-3m= m2/8+m/4+1/8 –m/4-1/4 -13 m2/4 + m/4 - 3m - m2/8 - m/4- 1/8 +m/4 + 1/4 + 13 =0 m2/8 - 22m/8 + 105/8 =0 m2 - 22m + 105 =0 (m-7)(m-15)=0 40

⇔ m=7 or m=15 So, there are two tangent lines of curve y = 2X2- X+ 4. and through (3,17) those are: Y=7X +c and y=15X+C . Because they are through (3,17) then those tangent line are : (1) Y=7X -4 and (2) Y=15X-28 First touch point can be found at the intersection between curve y = 2X2- X+ 4. with Y=7X -4 , it is (2,10) and second touch point can be found at at the intersection between curve y = 2X2- X+ 4 with Y=15X -28 , it is (4,32) Graph 60 50 40 Y

30 20

3; 17,0

10 0 -2

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0

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8

X

3.b Demand Elasticity and Supply Elasticity. Related the change in quantity of demand or quantity of supply if the price change. Demand elasticity is ratio between relative change in demand with price relative change. Supply elasticity is ratio between relative change in supply with price relative change. Elasticity = ∆Q/Q : ∆P/P =∆Q/∆P · P/Q If ∆Q → 0 dan ∆P → 0, then Elasticity : η =dQ/dP · P/Q Demand Elasticity: η =-dQ/dP · P/Q(sign -, in order to make η positive. (Source : Johannes).

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Example. Demand Function : P=12-2Q Find demand elasticity at price (P)=6. dP/dQ= -2 dQ/DP=-1/2 At P=6, Q=3 So: η =-(-1/2) · 6/3 =1

4 Problem. 1. Find equation of tangent line of curve y =-X4+ 6 at (1,5) 2. Find equation of tangent line of curve y =X3+ 4 which is parallel with y= 12 x +3 . 3. Given Demand Function : P=12-Q2 Find demand elasticity at price (P)=8. 4. Given demand function is P=18- 0.25 Q2. Find the value of price where demand elasticity is 1.75. 5. Given supply function is P=0.5 Q2+4 . a. Find supply elasticity at P=6 b. Find the value of price where supply elasticity is 0.75 6. Given supply function is P=0.25 Q2+6 . a. Find supply elasticity at P=7 b. Find the value of price where supply elasticity is 1.25 7. Given demand function is P= - Q2-2Q+15 . a. Find supply elasticity at P=12 b. Find the value of price where demand elasticity is 7/12

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