第6章 · 第6-3-5题

exercise

📝 题目

5.设 $u=\sin x+F(\sin y-\sin x)$ ,其中 $F$ 是可微函数,证明:

$$ \frac{\partial u}{\partial x} \cos y+\frac{\partial u}{\partial y} \cos x=\cos x \cdot \cos y . $$

💡 答案与解析

[AI解答]

[AI解答]

我们已知 $$ u = \sin x + F(\sin y - \sin x), $$ 其中 $F$ 是可微函数。令中间变量 $$ t = \sin y - \sin x, $$ 则 $$ u = \sin x + F(t). $$

首先计算偏导数 $\displaystyle\frac{\partial u}{\partial x}$ 和 $\displaystyle\frac{\partial u}{\partial y}$。

对 $x$ 求偏导时,将 $y$ 视为常数: $$ \frac{\partial u}{\partial x} = \cos x + F'(t) \cdot \frac{\partial t}{\partial x}, $$ 而 $$ \frac{\partial t}{\partial x} = -\cos x, $$ 所以 $$ \frac{\partial u}{\partial x} = \cos x - F'(t) \cos x = \cos x \bigl(1 - F'(t)\bigr). $$

对 $y$ 求偏导时,将 $x$ 视为常数: $$ \frac{\partial u}{\partial y} = 0 + F'(t) \cdot \frac{\partial t}{\partial y}, $$ 而 $$ \frac{\partial t}{\partial y} = \cos y, $$ 所以 $$ \frac{\partial u}{\partial y} = F'(t) \cos y. $$

现在计算表达式 $$ \frac{\partial u}{\partial x} \cos y + \frac{\partial u}{\partial y} \cos x. $$

代入上面结果: $$ \begin{aligned} &\frac{\partial u}{\partial x} \cos y + \frac{\partial u}{\partial y} \cos x \\ &= \bigl[\cos x (1 - F'(t))\bigr] \cos y + \bigl[F'(t) \cos y\bigr] \cos x \\ &= \cos x \cos y (1 - F'(t)) + \cos x \cos y \, F'(t) \\ &= \cos x \cos y \bigl(1 - F'(t) + F'(t)\bigr) \\ &= \cos x \cos y. \end{aligned} $$

因此等式成立,证毕。

难度:★★☆☆☆