复合函数求偏导：循环复用，逐渐化简

一、题目

已知函数 $y = f(x)$ 由 $y=\sin (x+y)$ 确定，则 $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2} = ?$

难度评级：

二、解析

$$
\frac{\mathrm
{d} y}{ \mathrm{d} x } = \cos (x + y) \cdot [ 1 + \frac{\mathrm
{d} y}{ \mathrm{d} x } ] \Rightarrow
$$

$$
\frac{\mathrm
{d} y}{ \mathrm{d} x } = \cos (x + y) + \cos (x + y) \cdot \frac{\mathrm
{d} y}{ \mathrm{d} x } \Rightarrow
$$

$$
\frac{\mathrm
{d} y}{ \mathrm{d} x } \cdot [1 – \cos (x + y)] = \cos (x + y) \Rightarrow
$$

$$
\frac{\mathrm
{d} y}{ \mathrm{d} x } = \frac{\cos (x + y)}{1 – \cos (x + y)}.
$$

Next

进而：

$$
\frac{\mathrm
{d} y}{ \mathrm{d} x } = \frac{\cos (x + y)}{1 – \cos (x + y)} \Rightarrow
$$

$$
\frac{\mathrm
{d} ^{2} y}{ \mathrm{d} x ^{2} } =
$$

$$
\frac{-\sin (x + y) [1 + \frac{\mathrm{d} y}{\mathrm{d} x}] \cdot [1 – \cos (x + y)] – \cos (x + y) \cdot \sin (x + y) [1 + \frac{\mathrm{d} y}{\mathrm{d} x}] }{ [1 – \cos (x + y)] ^{2} } \Rightarrow
$$

$$
\frac{[1 + \frac{\mathrm{d} y}{\mathrm{d} x}] \cdot \big\{ [-\sin (x + y)] \cdot [1 – \cos (x + y)] – \cos (x + y) \cdot \sin (x + y) \big\} }{ [1 – \cos (x + y)] ^{2} } \Rightarrow
$$

Next

继续化简 $\Rightarrow$

$$
\frac{[1 + \frac{\mathrm{d} y}{\mathrm{d} x}] \cdot \sin (x + y) \cdot \big\{ \cos (x + y) – 1 – \cos (x + y) \big\} }{ [1 – \cos (x + y)] ^{2} } \Rightarrow
$$

$$
\frac{ – [1 + \frac{\mathrm{d} y}{\mathrm{d} x}] \cdot \sin (x + y) }{ [1 – \cos (x + y)] ^{2} } \Rightarrow
$$

Next

代入前面得到的 $\frac{\mathrm{d} y}{\mathrm{d} x}$ 的值 $\Rightarrow$

$$
\frac{ – [1 + \frac{\cos (x + y)}{1 – \cos (x + y)} ] \cdot \sin (x + y) }{ [1 – \cos (x + y)] ^{2} } \Rightarrow
$$

$$
\frac{ – [ \frac{1 – \cos (x + y) + \cos (x + y)}{1 – \cos (x + y)} ] \cdot \sin (x + y) }{ [1 – \cos (x + y)] ^{2} } \Rightarrow
$$

$$
\frac{ – [ \frac{1}{1 – \cos (x + y)} ] \cdot \sin (x + y) }{ [1 – \cos (x + y)] ^{2} } \Rightarrow
$$

$$
\frac{ \frac{- \sin (x + y)}{1 – \cos (x + y)} }{ [1 – \cos (x + y)] ^{2} } \Rightarrow
$$

$$
\frac{ – \sin (x + y) }{ [1 – \cos (x + y)] ^{3} }.
$$

Next

综上可得：

$$
\frac{\mathrm
{d} ^{2} y}{ \mathrm{d} x ^{2} } = \frac{ – \sin (x + y) }{ [1 – \cos (x + y)] ^{3} } \Rightarrow
$$

$$
\frac{\mathrm
{d} ^{2} y}{ \mathrm{d} x ^{2} } = \frac{ – y }{ [1 – \cos (x + y)] ^{3} }.
$$

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