a-maths 既微分(計左好耐都計唔到)

1. x^(3/2) + y^(2/3) = a^(2/3) [AT THIS POINT, WHEN YOU DIFFERENTIATE, IGNORE a]

(3/2)x^(1/2) + [(2/3)y^(-1/3)][dy/dx] = 0

(dy/dx) = - [(3/2)x^(1/2)]/[(2/3)y^(-1/3)]
= - [(9/4)][x^(1/2)][y^(1/3)]

2. x = 2y + sin (y)

1 = 2(dy/dx) + [cos (y)][dy/dx]

[dy/dx][2 + cos (y)] = 1

(dy/dx) = 1/[2 + cos (y)] or [2 + cos (y)]^(-1) .....(*)

(d2y/dx2) = - [ 2 + cos (y)]^(-2) [- sin (y)] [dy/dx]
= [sin (y)][2+ cos (y)]^(-2) [dy/dx]

from (*),

(d2y/dx2) = [sin (y)][2+ cos (y)]^(-2) [2 + cos (y)]^(-1)
= [sin (y)][2 + cos (y)]^(-3)

3.

dy/dx= 2tanx(sec^2x)

d^2y/dx^2=2[2tanxsecx(secxtanx)+sec^4x]
=2[2sec^2x(tan^2x)+sec^4x]
=2[2(1+tan^2x)(tan^2x)+(1+tan^2x)^2] *(sec^2x=1+tan^2x)
=2[2(1+y)y+(1+y)^2] *(y=tan^2x)
=2(2y+2y^2+1+2y+y^2)
=2(3y^2+4y+1)
=2(1+y)(1+3y)


4.

y=x+√(x^2-1)
y=x+(x^2-1)^1/2

dy/dx=1+[1/2(x^2-1)^-1/2][2x]
=1+x(x^2-1)^-1/2

d^2y/dx^2=x{[-1/2(x^2-1)^-3/2][2x]}+(x^2-1)^-1/2
d^2y/dx^2=-x^2(x^2-1)^-3/2+(x^2-1)^-1/2
(x^2-1)d^2y/dx^2=-x^2(x^2-1)^-1/2+(x^2-1)1/2 *(全式乘x^2-1)
(x^2-1)d^2y/dx^2+x^2(x^2-1)^-1/2 -(x^2-1)1/2=0
(x^2-1)d^2y/dx^2+x(dy/dx-1)- (x^2-1)1/2=0 *(dy/dx-1=x(x^2-1)^-1/2)
(x^2-1)d^2y/dx^2+xdy/dx-x- (x^2-1)1/2=0
(x^2-1)d^2y/dx^2+xdy/dx-[x+(x^2-1)1/2]=0
(x^2-1)d^2y/dx^2+xdy/dx-y=0 *(y=x+(x^2-1)^1/2)