8.

已知

用数学归纳法证明，对于 $n \in \mathbb{N}$，

先验证 $n=1$。在归纳步骤中，把两个系数分别记作 $A_k$ 和 $B_k$；对假设式求导后，乘积法则会把它们变成 $3A_k+2B_k$ 和 $2A_k+3B_k$。化简这两个新系数，即可得到 $n=k+1$ 时的目标形式。

For $n=1$,

Therefore, the result is true for $n=1$.

Assume that the result is true for $n=k$, where $k\in\mathbb N$. Let

Then the induction hypothesis is

Differentiating both sides gives

Now

and

Hence

Thus, if the result is true for $n=k$, it is true for $n=k+1$. Since it is true for $n=1$, the result follows by induction for all $n\in\mathbb N$.

利用 $\cosh 2x=\frac12(e^{2x}+e^{-2x})$，先把 $y$ 化为两个指数函数之和。它的 $n$ 阶导数形式很容易用归纳法证明；最后再将结果重新组合成 $\cosh 2x$ 与 $\sinh 2x$，便得到题目所要求的式子。

Using the exponential form of $\cosh 2x$,

We first prove by induction that

For $n=1$,

so the result is true for $n=1$.

Assume that it is true for $n=k$. Then

Differentiating both sides,

Therefore, by induction,

for all $n\in\mathbb N$.

Finally,

as required.