第3章 3.3

(1)

$y = x \cos x$

$y' = \cos x - x \sin x$

$y'' = -\sin x - \sin x - x \cos x = -2 \sin x - x \cos x$

$y''' = -2 \cos x - \cos x + x \sin x = -3 \cos x + x \sin x$

(2)

$y = x^3 e^x$

$y' = 3x^2 e^x + x^3 e^x$

$y'' = 6xe^x + 3x^2 e^x + 3x^2 e^x + x^3 e^x = 6xe^x + 6x^2 e^x + x^3 e^x$

$y''' = 6e^x + 6xe^x + 12xe^x + 6x^2 e^x + 3x^2 e^x + x^3 e^x = 6e^x + 18xe^x + 9x^2 e^x + x^3 e^x$

$= (x^3 + 9x^2 + 18x + 6)e^x$

(3)

$y = x^2 \log x$

$y' = 2x \log x + x^2 \cdot \dfrac{1}{x} = 2x \log x + x$

$y'' = 2 \log x + 2x \cdot \dfrac{1}{x} + 1 = 2 \log x + 2 + 1 = 2 \log x + 3$

$y''' = \dfrac{2}{x}$

(4)

$y = e^{-x} \sin x$

$y' = -e^{-x} \sin x + e^{-x} \cos x$

$y'' = e^{-x} \sin x - e^{-x} \cos x - e^{-x} \cos x - e^{-x} \sin x = -2e^{-x} \cos x$

$y''' = -2(-e^{-x} \cos x - e^{-x} \sin x) = 2e^{-x}(\sin x + \cos x)$

(5)

$y = \dfrac{x^3}{1-x}$

$y' = \dfrac{3x^2(1-x)+x^3}{(1-x)^2} = \dfrac{3x^2-2x^3}{(1-x)^2}$

$y'' = \dfrac{(6x-6x^2)(1-x)^2+(3x^2-2x^3)\cdot 2(1-x)}{(1-x)^4} = \dfrac{6x-6x^2-6x^2+6x^3+6x^2-4x^3}{(1-x)^3}$

$= \dfrac{6x-6x^2+2x^3}{(1-x)^3}$

$y''' = \dfrac{(6-12x+6x^2)(1-x)^3+(6x-6x^2+2x^3)\cdot 3(1-x)^2}{(1-x)^6}$

$= \dfrac{6-12x+6x^2-6x+12x^2-6x^3+18x-18x^2+6x^3}{(1-x)^4} = \dfrac{6}{(1-x)^4}$

(6)

$y = \tan^{-1} x$

$y' = \dfrac{1}{1+x^2}$

$y'' = \dfrac{-2x}{(1+x^2)^2}$

$y''' = \dfrac{-2(1+x^2)^2+2x\cdot 2(1+x^2)\cdot 2x}{(1+x^2)^4} = \dfrac{-2(1+x^2)^2+8x^2(1+x^2)}{(1+x^2)^4}$

$= \dfrac{-2(1+x^2)+8x^2}{(1+x^2)^3} = \dfrac{6x^2-2}{(1+x^2)^3} = \dfrac{2(3x^2-1)}{(x^2+1)^3}$