📝 题目
例 26 设 $y = \sin x$ ,求 ${y}^{\left( n\right) }$ .
💡 答案与解析
解
$$ {y}^{\prime } = \cos x,\;{y}^{\prime \prime } = - \sin x, $$
$$ {y}^{\prime \prime \prime } = - \cos x,\;{y}^{\left( 4\right) } = \sin x, $$
................................
$$ {y}^{\left( 2k - 1\right) } = {\left( -1\right) }^{k - 1}\cos x, $$
$$ {y}^{\left( 2k\right) } = {\left( -1\right) }^{k}\sin x. $$
为了用统一的公式写出求导结果, 可采用以下办法:
$$ {y}^{\prime } = \cos x = \sin \left( {x + \pi /2}\right) , $$
$$ {y}^{\prime \prime } = \cos \left( {x + \pi /2}\right) = \sin \left( {x + 2 \cdot \pi /2}\right) , $$
............
$$ {y}^{\left( n\right) } = \sin \left( {x + \frac{n\pi }{2}}\right) . $$
对于 $z = \cos x$ ,同样可得
$$ {z}^{\left( n\right) } = \cos \left( {x + \frac{n\pi }{2}}\right) . $$
作为乘积求导公式
$$ {\left( uv\right) }^{\prime } = {u}^{\prime }v + u{v}^{\prime } $$
的推广, 我们有以下的莱布尼茨公式.
定理 4 设函数 $u$ 和 $v$ 都在 ${x}_{0}$ 点 $n$ 次可导,则这两函数的乘积 ${uv}$ 也在 ${x}_{0}$ 点 $n$ 次可导,并且在该点有
$$ {\left( uv\right) }^{\left( n\right) } = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) {u}^{\left( n - k\right) }{v}^{\left( k\right) }, $$
这里 $\left( \begin{array}{l} n \\ k \end{array}\right) = {\mathrm{C}}_{n}^{k}$ 是二项式系数,即
$$ \left( \begin{array}{l} n \\ 0 \end{array}\right) = 1 $$
$$ \left( \begin{array}{l} n \\ k \end{array}\right) = \frac{n\left( {n - 1}\right) \cdots \left( {n - k + 1}\right) }{k!} $$
$$ \left( {k = 1,2,\cdots ,n}\right) \text{ . } $$
证明 我们用归纳法证明莱布尼茨公式. 证明中关键的一步将用到以下恒等式:
$$ \left( \begin{array}{l} n \\ k \end{array}\right) + \left( \begin{matrix} n \\ k - 1 \end{matrix}\right) = \left( \begin{matrix} n + 1 \\ k \end{matrix}\right) $$
这关系可以直接用定义加以验证.
对于 $n = 1$ 的情形,莱布尼茨公式即熟知的乘积求导公式. 假设对于 $n \in \mathbb{N}$ 已经证明了莱布尼茨公式. 我们来考察 $n + 1$ 的情形.
$$ {\left( uv\right) }^{\left( n + 1\right) } = {\left( {\left( uv\right) }^{\left( n\right) }\right) }^{\prime } $$
$$ = {\left( \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) {u}^{\left( n - k\right) }{v}^{\left( k\right) }\right) }^{\prime } $$
$$ = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) \left( {{u}^{\left( n - k + 1\right) }{v}^{\left( k\right) } + {u}^{\left( n - k\right) }{v}^{\left( k + 1\right) }}\right) $$
$$ = {\sigma }_{1} + {\sigma }_{2}, $$
这里
$$ {\sigma }_{1} = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) {u}^{\left( n + 1 - k\right) }{v}^{\left( k\right) }, $$
$$ {\sigma }_{2} = \mathop{\sum }\limits_{{j = 0}}^{n}\left( \begin{array}{l} n \\ j \end{array}\right) {u}^{\left( n - j\right) }{v}^{\left( j + 1\right) }. $$
在 ${\sigma }_{2}$ 的表示式中,令 $j = k - 1$ 可得
$$ {\sigma }_{2} = \mathop{\sum }\limits_{{k = 1}}^{{n + 1}}\left( \begin{matrix} n \\ k - 1 \end{matrix}\right) {u}^{\left( n + 1 - k\right) }{v}^{\left( k\right) }. $$
于是, 我们得到
$$ {\left( uv\right) }^{\left( n + 1\right) } = {\sigma }_{1} + {\sigma }_{2} $$
$$ = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) {u}^{\left( n + 1 - k\right) }{v}^{\left( k\right) } + \mathop{\sum }\limits_{{k = 1}}^{{n + 1}}\left( \begin{matrix} n \\ k - 1 \end{matrix}\right) {u}^{\left( n + 1 - k\right) }{v}^{\left( k\right) } $$
$$ = {u}^{\left( n + 1\right) }v + \mathop{\sum }\limits_{{k = 1}}^{n}\left( {\left( \begin{array}{l} n \\ k \end{array}\right) + \left( \begin{matrix} n \\ k - 1 \end{matrix}\right) }\right) {u}^{\left( n + 1 - k\right) }{v}^{\left( k\right) } $$
$$ + u{v}^{\left( n + 1\right) } $$
$$ = \mathop{\sum }\limits_{{k = 0}}^{{n + 1}}\left( \begin{matrix} n + 1 \\ k \end{matrix}\right) {u}^{\left( n + 1 - k\right) }{v}^{\left( k\right) }. $$
在结束本节之前,我们对复合函数、反函数以及参数式表示的函数的高阶导数的求法, 做简单的说明. 本来, 这些情形下高阶导数的计算,都只是相应情形下求一阶导数手续的重复使用,但对高阶导数的计算, 初学者容易犯错误, 所以仍有必要特别提请注意.
设函数 $f$ 和 $g$ 都至少是二阶可导的,并且 $g$ 与 $f$ 可复合,这时复合函数 $h = g \circ f$ 也至少是二阶可导的,其二阶导数可按以下办法计算:
$$ {h}^{\prime }\left( x\right) = {g}^{\prime }\left( {f\left( x\right) }\right) {f}^{\prime }\left( x\right) , $$
$$ {h}^{\prime \prime }\left( x\right) = {\left( {h}^{\prime }\left( x\right) \right) }^{\prime } $$
$$ = {\left( {g}^{\prime }\left( f\left( x\right) \right) {f}^{\prime }\left( x\right) \right) }^{\prime } $$
$$ = {\left( {g}^{\prime }\left( f\left( x\right) \right) \right) }^{\prime }{f}^{\prime }\left( x\right) + {g}^{\prime }\left( {f\left( x\right) }\right) {\left( {f}^{\prime }\left( x\right) \right) }^{\prime } $$
$$ = {g}^{\prime \prime }\left( {f\left( x\right) }\right) {\left( {f}^{\prime }\left( x\right) \right) }^{2} + {g}^{\prime }\left( {f\left( x\right) }\right) {f}^{\prime \prime }\left( x\right) . $$
更高阶的导数也可用类似的办法计算.
设在开区间 $I$ 上,函数 $y = F\left( x\right)$ 严格单调,至少二阶可导,并且满足条件 ${F}^{\prime }\left( x\right) \neq 0$ . 则 $F$ 的反函数 $x = G\left( y\right)$ 在 $I$ 也至少是二阶可导的. 我们已经知道
$$ {G}^{\prime }\left( y\right) = \frac{1}{{F}^{\prime }\left( {G\left( y\right) }\right) }. $$
对这式再求导就得到
$$ {G}^{\prime \prime }\left( y\right) = - \frac{{\left( {F}^{\prime }\left( G\left( y\right) \right) \right) }^{\prime }}{{\left( {F}^{\prime }\left( G\left( y\right) \right) \right) }^{2}} $$
$$ = - \frac{{F}^{\prime \prime }\left( {G\left( y\right) }\right) {G}^{\prime }\left( y\right) }{{\left( {F}^{\prime }\left( G\left( y\right) \right) \right) }^{2}} $$
$$ = - \frac{{F}^{\prime \prime }\left( {G\left( y\right) }\right) }{{\left( {F}^{\prime }\left( G\left( y\right) \right) \right) }^{3}}. $$
更高阶的导数也可用类似的办法求得.
设函数 $\varphi \left( t\right)$ 和 $\psi \left( t\right)$ 在开区间 $J$ 至少二阶可导; 函数 $\varphi \left( t\right)$ 在 $J$ 严格单调并且满足条件 ${\varphi }^{\prime }\left( t\right) \neq 0$ . 我们来考察由参数式
$$ x = \varphi \left( t\right) ,\;y = \psi \left( t\right) ,\;t \in J $$
所定义的函数
$$ y = f\left( x\right) = \psi \left( {{\varphi }^{-1}\left( x\right) }\right) . $$
已经知道, 该函数的一阶导数可以表示为
$$ \frac{\mathrm{d}y}{\mathrm{\;d}x} = {f}^{\prime }\left( x\right) = \frac{{\psi }^{\prime }\left( t\right) }{{\varphi }^{\prime }\left( t\right) }. $$
为了求二阶导数,我们把一阶导数 $\frac{\mathrm{d}y}{\mathrm{\;d}x}$ 看成参数式表示的函数
$$ x = \varphi \left( t\right) ,\;\frac{\mathrm{d}y}{\mathrm{\;d}x} = \frac{{\psi }^{\prime }\left( t\right) }{{\varphi }^{\prime }\left( t\right) },\;t \in J. $$
对该函数又可应用参数表示函数的求导公式,
$$ \frac{{\mathrm{d}}^{2}y}{\mathrm{\;d}{x}^{2}} = \frac{{\left( \frac{{\psi }^{\prime }\left( t\right) }{{\varphi }^{\prime }\left( t\right) }\right) }^{\prime }}{{\varphi }^{\prime }\left( t\right) }. $$
这样, 我们求得
$$ \frac{{\mathrm{d}}^{2}y}{\mathrm{\;d}{x}^{2}} = {f}^{\prime \prime }\left( x\right) = \frac{{\psi }^{\prime \prime }\left( t\right) {\varphi }^{\prime }\left( t\right) - {\psi }^{\prime }\left( t\right) {\varphi }^{\prime \prime }\left( t\right) }{{\left( {\varphi }^{\prime }\left( t\right) \right) }^{3}}. $$
更高阶的导数也可用类似的办法求出.
对于参数式表示的函数的二阶导数, 有的初学者误以为
$$ \frac{{\mathrm{d}}^{2}y}{\mathrm{\;d}{x}^{2}} = \frac{{\psi }^{\prime \prime }\left( t\right) }{{\varphi }^{\prime \prime }\left( t\right) }. $$
我们特别指出这一错误,希望读者引以为戒.