发布时间:2019-08-01 12:51:47
y = ln[ ( 3x + 1 )( 1 - x ) ] = ln( 3x + 1 ) + ln( 1 - x )
∴ y' = 3[ ln( 3x + 1 ) ]' - [ ln( 1 - x ) ]' = 3/( 3x + 1 ) - 1/( 1 - x ) = 3/( 3x + 1 ) + 1/( x - 1 );
y'' = -3^2/( 3x + 1 )^2 - 1/( x - 1 )^2;
y''' = 2 * 3^3/( 3x + 1 )^3 + 2/( x - 1 )^3 = 2[ 3^3/( 3x + 1 )^3 + 1/( x - 1 )^3 ];
y'''' = -2 * 3 * 3^4/( 3x + 1 )^4 - 2 * 3/( 1 - x )^4 = -2 * 3 * [ 3^4/( 3x + 1 )^4 - 1/( 1 - x )^4 ];
y^(n) = (-1)^(n-1) * (n-1)! * [ 3^n/( 3x + 1 )^n - 1/( 1 - x )^n ];
也没有更简单的方法,网友寂园晓月已给出解答,但稍微乱了一些。现补充清晰解答如下: