Question

求下列函数的二阶导数

Answer 1

(4)y=xarccotx
则，y'=arccotx+x*[-1/(1+x²)]=arccotx-[x/(1+x²)]
则，y''=[-1/(1+x²)]-[(1+x²)-x*2x]/(1+x²)²
=-2/(1+x²)²
(6)y=lntanx
则，y'=(1/tanx)*sec²x=(cosx/sinx)*(1/cos²x)=1/(sinxcosx)=2/sin2x=2csc2x
则，y''=2*(-csc2x*cot2x*2=-4csc2x*cot2x
(8)y=2x(e^x²)
则，y'=2(e^x²)+2x*(e^x²)*(2x)=2(e^x²)+4x²*(e^x²)
=(4x²+2)*(e^x²)
则，y''=8x*(e^x²)+(4x²+2)*(e^x²)*2x=8x*(e^x²)+(8x³+4x)*(e^x²)
=(8x³+12x)*(e^x²)

Answer 2

1. y=e^(3x-1)

   y'=(3x-1)'·e^(3x-1)=3e^(3x-1)

   y''=3·(3x-1)'·e^(3x-1)=9e^(3x-1)

2. y=√(x²-1)

   y'=(x²-1)'/2√(x²-1)=x·(x²-1)^⁻½

   y''=(x²-1)^⁻½-½x·(x²-1)'·(x²-1)^⁻(3/2)

   =(x²-1)^⁻½-x²·(x²-1)^⁻(3/2)