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MathReview

# MathReview - Rule 1 ln(xy = ln(x ln y 2 ln(i = ln(x — ln...

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Unformatted text preview: Rule 1) ln(xy) = ln(x) + ln( y) 2) ln(i) = ln(x) — ln( y) y 3) A ln(x) = ln(xA) Calculus Review Log Rules Example ln(5 X 3) = ln(5) + ln(3) ln(g) = ln(5) — ln(3) 5111(3) = ln(35 ) = ln(243) Exponent Rules Rule Example 1) xaxb =xa+b x5x3 = x8 2) (xa)b =xab (x5)3 _x15 0 5 3) £— : xa_b ..x_ = x2 LI: Introduction & Calculus Review Derivative Rules Type V Function Derivative Power Function y = Ax” il—‘l: = nAx "'1 dy Exam le =5x3 ——=15x2 P y dx Log Function y = A ln(x) i): = A dx x Example )1 = 7 ln(x) Q = 1 dx x d ProductRuIe y=f(x)g(x) i=f(x)g'(x)+g(x)f'(x) Example y=J§(x3 +4) %=\/;(3x2)+(x3 +4)?“ Note:f(x)=\/;, g(x)=(x3 +4) Quotient Rule y = f(x) £13 =w g(x) dx g(x) l (x3 + 4)* x'“2 — J;(3x2) Example y = 2/; dy _ 2 (x + 4) dx (x3 + 4)2 . d d a' Chum Rule: y=f(g(x)) j=éﬁ=f'(g)g'(x) 7 dy 14x6 1 =1 — = Examp e y n(2x ) dx 2x7 Note: f (g) =ln(g), g(x) = 2x7 Exponent Rule: Special Case Example I .' Example 2: Partial Derivatives: Example 1: Example 2: L1: Introduction & Calculus Review x Q 2"” Order Partial Derivatives: Example 1: Example 2: y =a =a" lna yzaﬂx) g=af(x)fv(x)lna yzeﬂx) iii—d:=e/(X)fv(x)lne=ef(x)fv(x) y =5“ % =5x33x21n5 _ (x2+3x+5) dy (xz+3x+5) y —e a =e (2x + 3) a 3 y=f(x1’x2’x3""’xk) l=i 2f] 8x, 8x1 2 x— y =3x15+x12-4x22— 2x1x2 3—: = fl =15x14+ 2xl — 2x2 l y =e(7x,3x§ +3x1+5x2) 88—3} =f2 = e(7x5x§+3x1+5x2)(14x13x2 + 5) x2 62 y=f(x,,x2,x3,...,xk) y =fij axiaxj =60 3 + 2 y =3xl5 + x12 — 4x22 — 2x1x2 f” x1 flz :- 2 (Mix; +3x| +5x2) y=e 3 2 X 3 Z f22 = e‘7“‘2+3x'+'5 2)14x13 + (14x13x2 + 5)e<7"1*z+3"1 +5x1)(14x,3x2 + 5) 3 Z a n. 3 Z ﬂ] = e(7x, x2 +3x,+5)c2)42)‘:12x2 +4x15x2 + 5)e(7x,x2 +3x|+5xz)(21x12x2 + LI .‘ Introduction & Calculus Review Young’s Theorem 32 y 32 y = 0r .. = :. axidxj ijaxi f” f" Total Differential Univariate Case Function: y = f (x) Total Differential: dy = gidxz fxdx x Multivariate Case Function: y=f(x],x2,x3,...,xk) Total Differential: dy= fla'x1 + f2de +---+ fndxn Implicit Functions and Their Derivatives Given F (y, x1,x2,x3,..., xk) = 0 , the total differential of F is: chzﬁz+Fldxl +F2dx2 +---+F,,dxn =0 . . d . . In order to calculate the derivative 21% , holding the other varlables constant, one 1 need merely manipulate the total differential to show: dy _ Fx. Ex: _ _ F, Examgle Suppose you have a production possibility frontier of the form 5x2 + 3 y2 = 768. This equation could be written implicity as F (x, y) = 5x2 + 3 y2 — 768 = 0 . To derive the slope of the production possibility frontier one can use the implicit function theorem as follows: dy F)r _10x__5x dx IVE—3 y Interpretation: At the point (9,1 l) the slope is —45/33, or approximately -1.36. This means that if x decreased 1 unit from that point, y must increase by 1.36 in order to stay on the PPF. Univariate Optimization Objective Function First Order Condition F.O.C. Necessary Condition Second Order Condition S.O.C. Sufﬁcient Condition Example Objective Function First Order Condition Second Order Condition Optim L1: Introduction & Calculus Review ization Rules y=f(x) -:y—x=f'(x)=0 d2 d); =f"(x) < 0 Maximum d2 d); =f"(x) > 0 Minimum y=x4 +108x~5 dy dx dzy de 4x3 +108 = 0 , satisﬁed whenx = —3 =12x2 =12(—3)2 2108 > 0 minimum L1: Introduction & Calculus Review Optimization Rules Multivariate Optimization Objective Function y = f (xl , x2 , x3 ,..., xk) First Order Conditions (9—)) za—y =a—y =...= a—y = 0 3x, 3x2 8x3 Bxk Second Order Conditions Depends upon the sign (+/—) of the determinant of the Hessian matrix See Chapter 11 of A.Chiang, Fundamental Methods of Mathematical Economics 3rd edition. Maximum: xx, <0 and n > 2 (Two Variable Case) f f” f f” f” Minimum: fmfyy > 0 and fxxfw > f; Example Objective function: f(x,y) = 60+8x+2y— \$2 — 0.5g2 First order conditions: f\$=8—2m=0, fy=2—y=0, => \$*=4,y*=2 Second order conditions: fm: = —2, fyy = —1: fmy = 0 Evaluated at the optimal values: form = —2, W = —1, fzy = 0 Hence, 2 . fan < 0: fyy < 0) fxxfyy > f i a maxrmum LI: Introduction & Calculus Review The Envelope Theorem The envelope theorem is concerned with how the optimal value of a function changes when a parameter of the function changes. The theorem says that ﬁnding how the optimal value changes with a parameter change is as easy as ﬁnding how the function changes with a parameter change (when you evaluate that change at the optimal value). For example, suppose the objective function is: y = f (x, a) where x = variable OL = parameter If we optimize f(x, (x) by choosing x we will have the optimal value of x being a function of a -- this happens through solving the ﬁrst order condition: F.O.C. ﬂ=§£~=o :> x" =g(a) Substituting x* into the objective function gives us the indirect objective function: )7 = f(x‘(a),06) If we want to know how y* changes as 0t changes, we would need to take the following derivative: 2J1E+BL aa " ax aa aa . i . . ' Note that at the optlmal value of y , — = O by the ﬁrst order condltions. Hence, ax when %y— is evaluated at x* the above equation collapses to: a i _ 3f(x,a) 3a 8a The next reasonable question is, “What does this have to do with an envelope?” The answer to that has to do with slopes and tangent lines. As you know, the derivative is the slope or tangent to a ﬁmction. If you mapped the function y = f(x , on) and then B i . . . * varied a, you would ﬁnd that —ay—would prov1de the tangent lines to the ﬁmction y at making an effective “envelope” of the ﬁlnction. See the diagram on p. 39 of the Nicholson text. L1: Introduction & Calculus Review An Application of the Envelope Theorem Pro 1t Maximization: p =price of product q = quantity of product = xle xi 2 input into production wi =price of input i Proﬁt Function: 7r(x1,x2,p,wl,w2)= TR—TC = pxlx2 —x,wl —x2w2 In this problem we maximize proﬁts by choosing input levels, (x1, x2 ), given the cost of the inputs (w1 , w2 ). The input costs are parameters in the above equation. Method 1: If you want to know how the optimal value of proﬁts changes as the input price of good 1 changes, you could invoke the envelope theorem to calculate: 87f . __ = _x1 8wl Method 2: Or you could go through the entire optimization problem as follows: an" * wl ———=px2—w1=0 x2:— 3x1 P F.O.C. :> j31:1’xi_w2:0 xlz’wi 8x2 P It is clear that x‘ = x‘ (w, p) such that the optimal proﬁt function is written as follows: if i 75* (xi (wap)=x: (W: P): Wis wzap)= pxl x2 "' xiw1_ xng and the derivative of 15‘ with respect to w, would be written as: an? . 3f . dx‘ i dx‘ Ext \$29951 )“2—+va ‘—]"x1"W1—l‘wz —2 3w, V ' 8wl 3w1 aw! x‘ I W t =—p‘+O—x1— ——2=—xl [7 P Note that methods 1 and 2 provide the same solution but 1 was a lot easier. ...
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MathReview - Rule 1 ln(xy = ln(x ln y 2 ln(i = ln(x — ln...

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