Question

微积分求导

怎么求导？隐函数求导

详细点

Answer 1

Answer 2

用对数求导法，两边取自然对数得
lny=lnx-(lnx)^2+1/3ln(2-x)-2/3ln(2+x)
两边求导得
y'/y=1/x-2lnx/x-1/3*1/(2-x)-2/3*1/(2+x)
y'=[1/x-2lnx/x-1/3*1/(2-x)-2/3*1/(2+x)]*y

Answer 3

y = (x/x^lnx) [( 2-x)/(2+x)^2]^(1/3)
lny = lnx - (lnx)^2 + (1/3)[ ln(2-x) - 2ln(2+x) ]
(1/y) y' = 1/x - 2lnx/x + (1/3)[ -1/(2-x) - 2/(2+x) ]
y'={1/x - 2lnx/x + (1/3)[ -1/(2-x) - 2/(2+x) ]} y
= {1/x - 2lnx/x + (1/3)[ -1/(2-x) - 2/(2+x) ]} (x/x^lnx) [( 2-x)/(2+x)^2]^(1/3)

Answer 4