Mathematical Methods of Economics I,ECON 7025XBrooklyn College,Spring 2021, Instructor: Sagnik DasAssignment V Solutions1 Question Differentiate the following functions
(i)y=x5−x4+ 3x2Sol:y=x5−x4+ 3x2dydx=ddx(x5−x4+ 3×2)dydx=ddx(x5)−ddx(x4) +ddx(3×2) = 5×4−4×3+ 6x
(ii)y=43×3−54x4Sol:y=43×3−54x4dydx=ddx(43×3−54×4)dydx=ddx(43×3)−ddx(54×4) = 4×2−5×3
(iii)y=3×2+4x5Sol:y=3×2+4x5dydx=ddx(3×2+4×5)dydx=ddx(3×2) +ddx(4×5) =ddx(3x−2) +ddx4x−5=−6x−3−20x−6
(iv)y= 3×3−53x3Sol:y= 3×3−53x3dydx=ddx(3×3−53×3) =ddx(3×3)−ddx(53×3) = 9×2−5×2= 4×2
(v)y= 0.5×2+ 0.25×4−0.75x34Sol:y= 0.5×2+ 0.25×4−0.75x34dydx=ddx(0.5×2+ 0.25×4−0.75×34)dydx=ddx(0.5×2) +ddx(0.25×4)−ddx(0.75×34) =x+x3−916x−14
(vi)y= (x−3)(2×2+ 3)Sol:y= (x−3)(2×2+ 3)1
dydx=ddx(x−3)(2×2+ 3)dydx= (x−3)ddx(2×2+ 3) + (2×2+ 3)ddx(x−3)dydx= (x−3)4x+ (2×2+ 3)
(vii)y= (4×3+ 5×2−2x+ 3)(4×2+ 3x)Sol:y= (4×3+ 5×2−2x+ 3)(4×2+ 3x)dydx=ddx(4×3+ 5×2−2x+ 3)(4×2+ 3x)dydx= (4×3+ 5×2−2x+ 3)ddx(4×2+ 3x) + (4×2+ 3x)ddx(4×3+ 5×2−2x+ 3)dydx= (4×3+ 5×2−2x+ 3)(8x+ 3) + (4×2+ 3x)(12×2+ 10x−2)
(viii)y= (3×2+ 5x)(4×4+ 3×3+ 2×2+x)Sol:y= (3×2+ 5x)(4×4+ 3×3+ 2×2+x)dydx=ddx[(3×2+ 5x)(4×4+ 3×3+ 2×2+x)]dydx= (3×2+ 5x)ddx(4×4+ 3×3+ 2×2+x) + (4×4+ 3×3+ 2×2+x)ddx(3×2+ 5x)dydx= (3×2+ 5x)(16×3+ 9×2+ 4x+ 1) + (4×4+ 3×3+ 2×2+x)(6x+ 5)
(ix)y=(3×2+5x−2)(4×3+2x)Sol:y=(3×2+5x−2)(4×3+2x)dydx=ddx[(3×2+5x−2)(4×3+2x)]dydx=(4×3+2x)ddx(3×2+5x−2)−(3×2+5x−2)ddx(4×3+2x)(4×3+2x)2dydx=(4×3+2x)(6x+5)−(3×2+5x−2)(12×2+2)(4×3+2x)2(x)y=(x2+3x+1)(x+2)Sol:y=(x2+3x+1)(x+2)dydx=ddx[(x2+3x+1)(x+2)]dydx=(x+2)ddx(x2+3x+1)−(x2+3x+1)ddx(x+2)(x+2)2dydx=(x+2)(2x+3)−(x2+3x+1)(x+2)2(xi)f(x) = (x2−6×6)4Sol:f(x) = (x2−6×6)4dydx=ddx[(x2−6×6)4]dydx= 4(x2−6×6)3(2x−36×5)
(xii)f(x) =x5−5(x3−4×2)22
Sol:f(x) =x5−5(x3−4×2)2dydx=ddx[x5−5(x3−4×2)2]dydx= 5×4−10(x3−4×2)(3×2−8x)
(xiii)f(x) = (x3+ 8×5)4(x2−5x)Sol:f(x) = (x3+ 8×5)4(x2−5x)dydx=ddx[(x3+ 8×5)4(x2−5x)]dydx= (x3+ 8×5)4ddx(x2−5x) + (x2−5x)ddx[(x3+ 8×5)4]dydx= (x3+ 8×5)4(2x−5) + (x2−5x)4(x3+ 8×5)3(3×2+ 40×4)(xiv)f(x) = (x−23−5×34)5(6x−56+ 7×4)67
Sol:f(x) = (x−23−5×34)5(6x−56+ 7×4)67dydx=ddx[(x−23−5×34)5(6x−56+ 7×4)67]dydx= (x−23−5×34)5ddx(6x−56+ 7×4)67+ (6x−56+ 7×4)67ddx(x−23−5×34)5dydx= (x−23−5×34)5[67(6x−56+ 7×4)−17(−5x−116+ 28×3)] + (6x−56+ 7×4)67[5(x−23−5×34)4(−23x−53−154x−14)](xv)f(x) =(x3−2×4)3(4×3−5×2)5
Sol:f(x) =(x3−2×4)3(4×3−5×2)5dydx=ddx[(x3−2×4)3(4×3−5×2)5]dydx=(4×3−5×2)5ddx(x3−2×4)3−(x3−2×4)3ddx(4×3−5×2)5[(4×3−5×2)5]2dydx=(4×3−5×2)53(x3−2×4)2(3×2−8×3)−(x3−2×4)35(4×3−5×2)4(12×2−10x)[(4×3−5×2)5]23
