數學挑戰題 (4) -- 微分 (中) (20分)

Question

微分 :
(1)
y = ( x - 3 ) / [ (x - 1) (x - 2) ]

(2)
y = √(2x + 3)

(3)
y = 1 / [ x + √(x^2 + 1)]

Answer 1

(1)
d[(x-1)(x-2)]/dx
=(x-1)d(x-2)/dx+(x-2)d(x-1)/dx
=2x-3
y=(x-3)/[(x-1)(x-2)]
dy/dx={(x-1)(x-2)d(x-3)/dx-(x-3)d[(x-1)(x-2)]/dx}/[(x-1)(x-2)]^2
=[(x-1)(x-2)-(x-3)(2x-3)]/[(x-1)(x-2)]^2
=-(x+7)(x-1)/[(x-1)(x-2)]^2
=-(x+7)/(x-2)^2

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

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