Question

函数y=ln(1-x)除以(1+x),则y''(0)

Answer 1

y = (1+x)⁻¹ln(1-x)

y' = (-1)(1+x)⁻²ln(1-x) + (1+x)⁻¹ (1-x)⁻¹(-1)

= -(1+x)⁻²ln(1-x) - (1-x²)⁻¹

y'' = -(-2)(1+x)⁻³ln(1-x) - (1+x)⁻²(1-x)⁻¹(-1) - (-1)(1-x²)⁻²(-2x)

= 2(1+x)⁻³ln(1-x) + (1+x)⁻²(1-x)⁻¹ -2x(1-x²)⁻²

x = 0,y'' = 2(1+0)⁻³ln(1-0) + (1+0)⁻²(1-0)⁻¹ -2*0)(1-0²)⁻²

= 0 + 1+ 0

= 1