Chapter

1. (a) /(x) = 9x4 + 6x2 + 1. so f'(x) = 36x3 + 12x. (b) Using [5.1]: /'(x) = 2(3x2 + l)2"]6x = 36x3 + 12x.

3. (a) \(l+ x)-"2 (b)§*2(*3 + ir,/2 (c) -§(x - I)"2 (tzy) !/2 (d) —66*(1 - x1)11 (e) 3x2JT^--i-

(f) 1(1 +*)"2/3(l -x)1/5 - ¿(1 + x)l/3(l -x)"4/5

5. (a) 1 + f(x) (b) 2f(x)f'(x) - 1 (c) A[f(x)Yf'(x)

(d)2xf(x)+x2f\x) + 3{f(x)]2f'(x) (t) f(x) + xf(x) (f) /JL

(g) [2xf(x) - x2f'(x)]/[f(x)]2 (h) [2xf(x)f'(x) - 3(f(x))2]/x4 (i) ¿{fix) + [f(x)Y + x}-2^{f'ix) +3[fix)]2fix) + 1}

9. fix) = mix - a)m~] ■ ix - b)n + nix - a)m ■ ix - b)n~\ Factoring this expression yields fix) = [mix -b) + nix - a)](x - a)m_I • ix - b)n~] = 0 at = ina + mb)/in + m). Clearly, a < ina + mb)/in + m) < b iff a(n+m) < na+mb < bin+m). Here ain+m) < na+mb iff ia — b)m < 0, which is true because a < b and m is positive. Moreover, na+mb < bin+m) iff 0 < nib — a), which is also true.

11. Use the product rule and [5.1]. m = n = 1 yields the product rule for derivatives, m = —n = 1 yields the quotient rule.

1. (a) dy/du = 20ul and du/dx = 2x so dy/dx = idy/du) ■ idu/dx) = 40u3x = 40x(1 + X2)3

(b) dy/dx = (1 - 6u5)idu/dx) = [l - 6(1/* + l)5] ( - l/x2)

3. (a) / = —5(x2 + * + 1)"6(2jc + 1) (b) >•' = {[x + ix + x1'2)1'2]_1/2[l + {ix +*1/2)"1/2(1 (c) y' =axa~{ipx+q)b+xabpipx+q)h~]

(c) x/x = icc+ ß)iaaa~lä + ßbß-lb)/iaa + bß) (d) x/x = aä/a + ßb/b

11. (a) hix) = f{gix)). where g(x) = 1 + * + x2 and fiu) = w1/2. so h'ix) = (1/2)(1 +* + x2)-'/2(l +2x). (b) hix) = f ig ix)), where gix) = xm + 28 and fiu) = u~]. so h'ix) = ~ixm + 28)-2100*99.

13. (a) >•' = 5(jc4)4 - 4*3 = 20*19 (b) >•' = 3(1 - x)2(-1) = -3 + 6* - 3x2

15. dR/dt = idR/dS)idS/dK)idK/dt) = aSa~]ßyKy~lAptp~]

1- (a) Differentiation w.r.t. x yields 1 • y -f xy' = 0. so y' = -y/x. Because y = l/x. this gives >•' = -l/x2. (b) >•' = (1 - 3y)/(l - 3x). Because y = (x - 2)/(l - 3jc), this gives y' = -5/(1 - 3x)2. (c) y' = 5x4/6y5. Because y = x5/6, this gives y' = (5/6)x~'/6. dv 2u + v

3. — = —--. Hence, dv/du = 0 when v = —2u. Inserting this value into du 3v~ — u the original equation implies that dv/du = 0 at (u. v) = (1/8. —1/4). 5. dQ/dP = -19/P3/2

7. Y = f(Y) +1 + A — g(Y). Differentiating w.r.t. I using the chain rule yields dY/dl = f'(Y)(dY/dI) + 1 - g'(Y)(dY /d I). Solving for dY/dl gives dY/dl = 1 / [ 1 — f'(Y) + g'(K)]. Imports should increase when income increases, so g'(y) > 0. We find that dY/dl > 0.

9. Differentiation w.r.t. x yields g'[f(x))f'{x) = 1, so f'{x) = 1 /g'(f(x)) (provided that g'(/'(*)) #0).

1. If f(x) = y/lTx, then f'(x) = 1/2VTT^, so /(0) = 1 and /'(0) = 1/2. Hence, [5.5] gives VTTI % 1 + - 0) = 1 + \x. See Fig. 18.

3. (a) 1/(1 -hjc) =»1 — jc (b) (1+*)5 ^ \+5x (c) (1 -x)]/* % 1 -x/4

5. (a) 4/TT = (1 + 1/10)1/3 % 1 + (1/3)(1/10) % 1.033 (b) ^33 = 2(1 + 1/32)1/5 % 2(1 + 1/160) = 2.0125 (c) ^9 = 2(1 + 1/8)1/3 % 2(1 + 1/24) % 2.083 (d) (1.02)25 = (1 + 1/50)25 % 1 + 1/2 = 1.5 (e) JTi = (36 + 1),/2 = 6(1 + 1/36)1/2 % 6(1 + 1/72) % 6.083 (f) (26.95)I/3 = (27 - 5/100)1/3 = 3(1 —0.05/27)1/3 as 3 - 0.05/27 % 2.998

7. V(r) = {4/3)ixr3. The linear approximation is: V(2 + 0.03) - V(2) % 0.03 • V'(2) = 0.48tt. Actual increase: V(2.03) - V(2) = 0.487236;r.

1. (a) (l+x)5 % l+5x+10.r2 (b) AKa * A+aA(K-l)+{a(a-l)A(K-l)2 (c) (l + §£ + ±e2)1/2 % 1 + §£ - ¿e2 (d) (1 - x)-1 % 1 + x + x2

3. Implicit differentiation yields: [*] 3x2y + x3y' + 1 = iy_1/2y'. Inserting x = 0 and y = 1 gives 1 = (^)l~1/2y', so y' = 2. Differentiating [*] once

FIGURE 18 '

FIGURE 18 '

900 Answers to Odd-Numbered Problems more w.r.t.jc yields 6xy + 3x2y'+ 3x2y'+x3y" = - \ \-3/2(y')2 + v Inserting x = 0. y = 1. and y' = 2 gives v" = 2. Hence. y(x) % 1 + 2x —

5. Use [5.10] with f(x) = (1 +x)n and x = p/100. Then f'{x) = «(i +x)»-i and fix) = n(n — 1)(1 + x)"~2. The approximation follows.

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