📝 题目
例 10 求 $f\left( x\right) = \frac{1}{{a}^{2} - {b}^{2}{x}^{2}}\left( {a \neq 0}\right)$ 的 $n$ 阶导数.
💡 答案与解析
解 $f\left( x\right) = \frac{1}{{a}^{2} - {b}^{2}{x}^{2}} = \frac{1}{2a}\left( {\frac{1}{a + {bx}} + \frac{1}{a - {bx}}}\right)$ ,
$$ {f}^{\prime }\left( x\right) = \frac{b}{2a}\left( {\frac{1}{{\left( a - bx\right) }^{2}} + \frac{-1}{{\left( a + bx\right) }^{2}}}\right) , $$
$$ {f}^{\prime \prime }\left( x\right) = \frac{2!{b}^{2}}{2a}\left( {\frac{1}{{\left( a - bx\right) }^{3}} + \frac{{\left( -1\right) }^{2}}{{\left( a + bx\right) }^{3}}}\right) , $$
$$ {f}^{\left( n\right) }\left( x\right) = \frac{n!{b}^{n}}{2a}\left( {\frac{1}{{\left( a - bx\right) }^{n + 1}} + \frac{{\left( -1\right) }^{n}}{{\left( a + bx\right) }^{n + 1}}}\right) . $$