幾題a.maths Chain rule問題，唔該幫下手(20分)

(1)
dy/dx=(3x-2)[3(5x+4)(5)]+(5x+4)[2(3x-2)(3)]
=15(3x-2)(5x-4)+6(5x+4)(3x-2)
=3(3x-2)(5x-4)[5(3x-2)+2(5x+4)]
=3(3x-2)(5x-4)(25x-2)

(2)(我唔清楚係(5cosx/x) +1定係5cosx/(x +1)，所以我兩個做晒)
y=5cosx/(x +1)
dy/dx=5[(x+1)(-sinx)-cosx(2x)]/(x+1)
dy/dx=-5[(x+1)sinx+2xcosx]/(x+1)

y=(5cosx/x) +1
dy/dx=5[x(-sinx)-cosx(2x)]/x^4
=-5[xsinx+2cosx]/x

(3)
y=sinx/(1+cosx)
dy/dx=[(1+cosx)(cosx)-sinx(-sinx)]/(1+cosx)
=(cosx+cosx+sinx)/(1+cosx)
=(1+cosx)/(1+cosx)
=1/(1+cosx)

(4)
(x+1)e^(-x)
=(x+1)(-e^-x)+e^(-x)(2x)
=-e^(-x)(x-2x+1)
=-e^(-x)(x-1)

P.S. Question 4 should not be included in the syllabus as the syllabus stated that exponential function are not included in the syllabus

2008-07-14 16:33:51 補充：
Yahoo而家改左出唔到次方......
再打一次......
dy/dx=(3x-2)²[3(5x+4)²(5)]+(5x+4)³[2(3x-2)(3)]

=15(3x-2)²(5x-4)²+6(5x+4)³(3x-2)

=3(3x-2)(5x-4)²[5(3x-2)+2(5x+4)]

=3(3x-2)(5x-4)²(25x-2)

2008-07-14 16:36:33 補充：
(2)(我唔清楚係(5cosx/x²) +1定係5cosx/(x² +1)，所以我兩個做晒)

y=5cosx/(x² +1)

dy/dx=5[(x+1)(-sinx)-cosx(2x)]/(x²+1)²

dy/dx=-5[(x+1)sinx+2xcosx]/(x²+1)²



y=(5cosx/x²) +1

dy/dx=5[x²(-sinx)-cosx(2x)]/x^4

=-5[xsinx+2cosx]/x³

2008-07-14 16:37:18 補充：
y=sinx/(1+cosx)

dy/dx=[(1+cosx)(cosx)-sinx(-sinx)]/(1+cosx)²

=(cosx+cos²x+sin²x)/(1+cosx)²

=(1+cosx)/(1+cosx)²

=1/(1+cosx)

2008-07-14 16:37:53 補充：
(x²+1)e^(-x)

=(x²+1)(-e^-x)+e^(-x)(2x)

=-e^(-x)(x²-2x+1)

=-e^(-x)(x-1)²