0. 基础 n 阶四公式形式
高阶导数
高阶导数四大解法：
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变形成 n 阶四公式形式
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莱布尼茨公式（常需利用 n 阶四公式）
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泰勒公式化得多项式
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观察规律法
首先，要想解高阶导数又快又准，n 阶四公式绝对是基础中的基础，所以，请务必记住 n 阶四公式：
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\({(x^m)}^{(n)}=m(m−1)(m−2)\cdots(m−n+1)x^{m−n}\ (m \ge n)\)
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\({(a^x)}^{(n)}={(\ln_{}{a})}^{n}a^{x}\)
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\({(lnx)}^{(n)}=\frac{{(−1)}^{n−1}(n−1)!}{x^n}\) （由 \({(ln_{}{x})}^{\prime}=\frac{1}{x}\) ，有 \({(\frac{1}{x})}^{(n)}=\frac{{(−1)}^{n}n!}{x^{n+1}}\) ）
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所谓 n 阶四公式，即幂函数、指数函数、对数函数、三角函数最简单形式的 n 阶导数的值。
但是通常，题目不会直接让我们求这四个函数，一般我们要求的，都是 n 阶四公式形式的函数，比如说，求的是 \({(ax+b)}^{(n)}\) ，\({[\ln_{}{(ax+b)}]}^{(n)}\)，\({[\sin (kx+b)]}^{(n)}\)。
我们只要记住了形式简单的 n 阶四公式，就可以很快地推出 n 阶四公式形式的函数。
所以，现在，请立刻开始把 n 阶四公式记住，不要说留到后面再背，告诉自己，我现在就要记住 n 阶四公式，并且我不会忘了。
只有我们有坚定说要去记住，才真的更容易记牢，这是我自己的感受。
好，现在我们记住了 n 阶四公式。因为是最简单的形式，所以记起来也还行。
ok，前面说了这么多，其实就讲了一样东西，叫 n 阶四公式。为了检验你是否掌握，请你拿出纸笔，求：
\(𝑓(𝑥)=\ln_{}{(1−x)}\) 的 n 阶导数。（答案在文末，题号为 ① ）
上面的问题你答对了吗？答对了就点个赞吧！
好，现在，停下来，休息一下，放松地思考一下，下面两个问题：
\({[{(ax+b)}^𝑚]}^{(n)}\) ，\({[\ln_{}{(ax+b)}]}^{(n)}\) ， \({[\sin (kx+b)]}^{(n)}\)
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上面的三个 n 阶导数求出来是什么？
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它们比对应的原来的 n 阶四公式多出了什么？或者说有什么联系和区别。
问题难度适中，相信你能思考出来。
如果你已经掌握了 n 阶四公式，我们就进入本文的正题吧——高阶导数题的四大解法。
（注：本文的例题和习题都是经过挑选，觉得很经典和不错的题目，绝对值得一做）
1. 变形成 n 阶四公式形式
例题：已知函数 \(f(x)=ln_{}{\frac{1−3𝑥}{1+2𝑥}}\) ，则 \(f^{(n)}(0)=()\) .
解：如果按原来的 𝑓(𝑥) 的形式一次次求导，感觉会有点复杂，尝试将其化成 n 阶四公式形式，即
\(f(x)=\ln{}{\frac{1−3x}{1+2x}}=\ln_{}{(1−3x)}−\ln_{}{(1+2x)}\)
这样一来，对 \(f(x)\) 求 n 阶导就相当于分别对 \(\ln_{}{(1−3x)}\) 和 \(\ln_{}{(1+2x)}\) 求 n 阶导然后相减。
根据 n 阶四公式中对数函数的 n 阶导数的值，我们可以快速推导出
\({[\ln_{}{(1+𝑎𝑥)}]}^{(n)}=\frac{{(−1)}^{n−1}a^n(n−1)!}{{(1+ax)}^n}\) ，所以
\(f^{(n)}(x)={(−1)}^{n−1}(n−1)!\left [\frac{{(−3)}^n}{{(1−3x)}^n}−\frac{2^n}{{(1+2x)}^n}\right ]\) ，则有
\(f^{(n)}(0)={(−1)}^{n−1}(n−1)![{(−3)}^n−2^n]\)
小结：问题解答完成，由此可见，对于一个复杂的函数，可以化成 多个n 阶四公式形式的函数，再利用 n 阶四公式，快速推导其出 n 阶导数的值，进而求得复杂函数的 n 阶导数的值。
拿出纸笔，快速做一下下面这道题练练手吧。
习题：已知函数 \(𝑓(𝑥)=\frac{1}{x^2−1}\) ，求 \(f^{(5)}(x)\) .（答案在文末，题号为②）
耐心一点，请确保能解开上一道习题再继续阅读。
开始第二种方法之前，请问自己，是否了解过莱布尼茨公式，公式是怎么样的，是用来干什么的。如果答案是否定的，请停止阅读本文，通过书本或上网对莱布尼茨公式进行初步了解，能够回答这两个问题后，再继续阅读。
（学习不总是“线性”的。通过一个问题，可以牵带出相关的问题。然后带着对相关问题的好奇，去了解它们，逐步深入，这是我自己很喜欢的一种学习的方式，好奇和求知欲也应该并且可以成为我们学习的动力。）
2. 莱布尼茨公式
例题：已知函数 \(f(x)=x^2\ln{}{(1-x)}\) ，当 \(n \ge 3\) 时， \(f^{(n)}(0)=()\) .
解：看着像两个函数相乘的形式，并且 \(x^2\) 在导数大于2后为0，所以可以考虑一下使用莱布尼茨公式。则有，
\(f^{(𝑛)}(x)=C_{n}^{0}x^2{[\ln_{}{(1−x)}]}^{(n)}+C_{n}^{1}(x^2)^{\prime}{[\ln_{}{(1−x)}]}^{(n-1)}+C_{n}^{2}(x^2)^{\prime\prime}{[\ln_{}{(1−x)}]}^{(n-2)}\)
因为 \({[\ln_{}{(1−x)}]}^{(n)}=\frac{(n-1)!(-1)}{{(1-x)}^n}\) ，
\({[\ln_{}{(1−x)}]}^{(n-1)}=\frac{(n−2)!(−1)}{{(1−x)}^{𝑛−1}}\) ，
\({[\ln_{}{(1−x)}]}^{(n-2)}=\frac{(n−3)!(−1)}{{(1−x)}^{𝑛−2}}\) ，
\({(x^2)}^{\prime}=2x\) ， \({(x^2)}^{\prime\prime}=2\) .
所以 \(f^{(n)}(x)=x^2\frac{(n-1)!(-1)}{{(1-x)}^n}+2xn\frac{(n−2)!(−1)}{{(1−x)}^{n−1}}+2\frac{n(n−1)}{2}\frac{(n−3)!(−1)}{{(1−x)}^{n−2}}\)
故 \(f^{(n)}(0)=−\frac{n!}{n−2}\) .
小结：例题解答完成，通过观察我们可以发现，莱布尼茨公式解 n 阶导数适用于：
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两个函数相乘
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其中一个函数在求高阶导数后会变为 0
这样的函数。
当遇到这样的函数时，使用莱布尼茨公式。对于求其中需要求 n 阶导数的（本题中是 \(\ln_{}{(1−x)}\) ），需要变形成 n 阶四公式形式（即方法一），然后利用 n 阶四公式推导进行解决。
此外，当要求的是 \(f^{(𝑛)}(0)\) 时，有的项其实并不需要计算。比如在本题中， \(f^{(𝑛)}(𝑥)\) 的前面两项是不需要计算的。
例题讲完了，你现在能解决下面这一道习题吗？
习题：函数 \(f(x)=x^22^𝑥\) 在 \(x=0\) 处的 𝑛 阶导数 \(f^{(n)}(0)=()\) .（答案在文末，题号为③）
能够看到这里，高阶导数的类型题基本也能解决一大半了，在坚持一下吧，后面的内容不会很难。
下面的方法会涉及泰勒公式和麦克劳林公式，统称为泰勒公式化多项式法。这里还是默认大家已经掌握了泰勒公式。
3. 泰勒公式化多项式
例题：已知函数 \(f(x)=x^{100}e^{x^2}\) ，则 \(f^{(200)}(0)=()\) .
解：使用前面两种方法似乎解决不了本题，尝试使用泰勒公式。
由麦克劳林公式，得
\(f(x)=x^{100}(\sum_{k=0}^{n}\frac{x^{2k}}{k!}+o(x^{2n}))=\sum_{k=0}^{n}\frac{x^{2𝑘}+100}{k!}+o(x^{2n+100})\)
取 \(n=50\) ，根据多项式求导的性质，由
展开式中 \(x^{200}\) 项的系数是 \(\frac{1}{50!}\) \(\Rightarrow\) \(f^{200}(0)=200!\frac{1}{50!}\)
小结：根据我个人的经验，利用莱布尼茨公式法和泰勒公式化多项式是考察比较多的，而且常常这两种方法是同一道题的两种解法。
之所以把变形成 n 阶四公式法放在第一个，是因为这种方法最基础的方法，并且在一些时候，这种方法也是后面这两种方法的基础，起到辅助解题的作用。
这次，请你尝试用泰勒公式化多项式法，解决一下，上面莱布尼茨公式法的习题吧。
习题：函数 \(f(x)=x^22^x\) 在 \(x=0\) 处的 𝑛 阶导数 \(f^{(n)}(0)=()\) .（答案在文末，题号为④）
当上面三种方法，三种通用套路，看起来都没办法解决的时候，我们只能祭出观察规律法。
数学难题常常就出现在需要观察的题目上。有时候，一个函数，你观察到它潜在的某种特点、或者可以进行某种不常见的变形、或者有某种几何性质，便能够很快速地解决掉题目，但如果观察不出来，就真的绞经脑汁也解不出来。观察题和套路题一定程度上是对立的（当然对于身经百战的解题者，观察题也是套路题）。观察题是让人又爱又恨的，它常常是让人困惑的，但也常常是美妙的。
废话有点多了，下面进入第四种方法——观察规律法。
4. 观察规律法
例题：已知函数 \(f(x)\) 在 \((-\infty,+\infty)\) 上连续，且 \(f(x)={(x+1)}^2+2\int_{0}^{x}f(t)dt\) ，当 \(n\ge2\) 时， \(f^{(n)}(0)=()\).
解：尝试按上面三种套路思考，发现似乎都行不通，于是尝试使用观察规律法。
观察、思考，不难想到 \({(x+1)}^2\) 在导数等于 3 后会变成 0。于是尝试对 \(f(x)\) 求前 3 次导数。
\(f^{\prime}(x)=2(x+1)+2f(x)\) ，
\(f^{\prime\prime}(x)=2+2f^{\prime}(x)\) ，
\(f^{\prime\prime\prime}(x)=2f^{\prime\prime}(x)\) ，则
\(f^{(n)}(x)=2f^{(n-1)}(x)=2^2f^{(n-2)}(x)=\cdots=2^{n-2}f^{\prime\prime}(x)\ (n\ge2)\)
而 \(f(0)=1\) ， \(f^{\prime}(0)=2+2=4\) ， \(f^{\prime\prime}(0)=10\)
因此 \(f^{(n)}(0)=5*2^{n-1}\)
小结：观察规律法，通常是尝试求前几阶导数，然后进行观察，总结导数可能存在的规律。而且有时后，在求解前一二阶导数后，便可以使用前面的三种方法解决了。
习题：已知函数 \(f(x)=\arctan x\) ，求 \(f^{(n)}(0)\) .
5. 下一步做什么
找书本或者网上的高阶导数的题来做吧。认真做个15道高阶导数的题感觉就没什么太大问题了。
最后的提醒，解决高阶导数的问题，几乎就这四种方法不会错。不过有的时候，稍微复杂的题目并不一定是一定用单一的方法解决的，对这四种方法融会贯通，学会综合运用，方能立于不败之地。
答案
① \(\frac{(n-1)!(-1)}{{(1-x)}^n}\)
② \(60\left [ \frac{1}{{(x+1)}^6} - \frac{1}{{(x-1)}^6} \right ]\)
③ \(n(n-1){(\ln_{}{2})}^{n-2}\ (n=1,2,3,\cdots)\)
④ \(n(n-1){(\ln_{}{2})}^{n-2}\ (n=1,2,3,\cdots)\) 。注意化 \(f(x)=x^22^x=x^2e^{x\ln_{}{2}}\) ，然后再泰勒展开。
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