高数解题，要解题步骤

如图先答个第一问

提示：为什么分母的arctanx 可以先化为 x+o(x)

当极限式最外层是一个0/0型分式时，将其分子分母都完全泰勒展开，则其结果仅由最低阶无穷小决定！因为低阶无穷小±高阶无穷小=低阶无穷小，高阶无穷小在比值中完全不影响结果。

所以分母的arctanx麦克劳林展开时只需展开最低阶无穷小。而分子因为有对c*x 和 d*x²的加减运算，所以要展开到x²及以上的无穷小，才能确定精度！

未完待续。如图，如有疑问或不明白请追问哦！

（3）原式=∫(1,2)dy∫(y,y^2)sin(πx/2y)dx =∫(1,2)dy*[-(2y/π)*cos(πx/2y)|(y,y^2)] =∫(1,2) (-2y/π)*cos(πy/2)dy =∫(1,2) (-4y/π^2)*d[sin(πy/2)] =(-4y/π^2)*sin(πy/2)|(1,2)+∫(1,2) (4/π^2)*sin(πy/2)dy =4/π^2-(8/π^3)*cos(πy/2)|(1,2) =4/π^2+8/π^3

1. 原式 = lim(arctanx-x+πx^2)/x^2 (0/0)
= lim[1/(1+x^2)-1+2πx]/(2x)
= lim(1-1-x^2+2πx+2πx^3)/[2x(1+x^2)]
= lim(-x+2π+2πx^2)/2 = π;
2. 原式 = lim<(x, y)→(0, 0)> -xy{√[2-e^(xy)]+1}/[e^(xy)-1]
= lim<(x, y)→(0, 0)> -2xy/(xy) = -2.
3. z = x^3 y^2 - 2xy^3 + x^2 + xy
∂z/∂x = 3x^2 y^2 - 2y^3 + 2x
∂z/∂y = 2yx^3 - 6xy^2 + x
∂^2z/∂x^2 = 6xy^2 + 2
∂^2z/∂x∂y = ∂^2z/∂y∂x = 6yx^2 - 6y^2
∂^2z/∂y^2 = 2x^3 - 12xy
4. 记 u = yx^2, v = y/x
∂z/∂x = (∂f/∂u)(∂u/∂x) + (∂f/∂v)(∂v/∂x) = 2xy(∂f/∂u) - (y/x^2)(∂f/∂v)
∂^2z/∂x∂y = 2x(∂f/∂u) + 2xy[(∂^2f/∂u^2)(∂u/∂y) + (∂^2f/∂u∂v)(∂v/∂y)]
- (1/x^2)(∂f/∂v) - (y/x^2)[(∂^2f/∂v∂u)(∂u/∂y) + (∂^2f/∂v^2)(∂v/∂y)]
= 2x(∂f/∂u) - (1/x^2)(∂f/∂v) + 2yx^3(∂^2f/∂u^2) + y(∂^2f/∂u∂v) - (y/x^3)(∂^2f/∂v^2).
x = y = 1 时，∂^2z/∂x∂y = 2∂f/∂u - ∂f/∂v + 2∂^2f/∂u^2 + ∂^2f/∂u∂v - ∂^2f/∂v^2.