T12 变上限积分求二阶偏导

设 $f (x, y) = \int_{0}^{x y} e^{x t^{2}} d t$ ，求 $\frac{\partial ^{2} f}{\partial x \partial y} ∣_{(1, 1)}$

:::{note} 计算错误

$e^{x t^{2}}$ 中 $x$ 不能把 $e^{x}$ 提出来。

如果要对 $x$ 求偏导，需要先对变限积分换元

$x t = u \Rightarrow t = \frac{u}{x}$

:::

考虑先 $y$ 后 $x$ ：

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \int_{0}^{x y} e^{x t^{2}} d t = e^{x (x y)^{2}} \cdot x = x e^{x^{3} y^{2}}$
$\frac{\partial ^{2} f}{\partial y \partial x} = \frac{\partial}{\partial x} (x e^{x^{3} y^{2}}) = e^{x^{3} y^{2}} + x \cdot e^{x^{3} y^{2}} \cdot 3 x^{2} y^{2} = e^{x^{3} y^{2}} (1 + 3 x^{3} y^{2})$
故 $\frac{\partial ^{2} f}{\partial x \partial y} ∣_{(1, 1)} = \frac{\partial ^{2} f}{\partial y \partial x} ∣_{(1, 1)} = 4 e$

Okay, three, two, one, let's jam