微积分
高阶导数
(a^x)^{(n)} = (\ln a)^n\cdot a^x
证明: 用数学归纳法
当 n = 1 时:
(a^x)' = (e^{x\ln a})' = \ln a \cdot e^{x\ln a} = \ln a \cdot a^x,成立
设 n = k 时,(a^x)^{(k)} = (\ln a)^k\cdot a^x 成立,则当 n = k + 1 时:
(a^x)^{(k+1)} = [(\ln a)^k\cdot a^x] = (\ln a)^{k+1}\cdot a^x
由第一数学归纳法,命题成立,证毕。
(e^x)^{(n)} = e^x
证明:
显然此命题为上一命题的简单推论
取 a = e,则 \ln e = 1,故
(e^x)^{(n)} = (\ln e)^n\cdot e^x = e^x
\displaystyle (\sin kx)^{(n)} = k^n\sin \left(\frac{n\pi}{2} + kx\right)
证明: 用数学归纳法
当 n = 1 时
$$
(\sin kx)’ =k \cos kx = k\sin \left(\frac{\pi}{2} + kx\right)
$$
当 n = 2 时(该步骤可以省略)
\begin{aligned}
&(\sin kx)'' \\
= & \left[k\sin \left(\frac{\pi}{2} + kx\right)\right]' \\
=& k^2\cos \left(\frac{\pi}{2} + kx\right) \\
=& k^2\sin \left(\frac{2\pi}{2} + kx\right) \\
\end{aligned}
设 n = t 时,
\displaystyle (\sin kx)^{(t)} =k^t\sin \left(\frac{t\pi}{2} + kx\right) 成立
当 n = t + 1 时
\begin{aligned}
&(\sin kx)^{(t + 1)} \\
= & \left[k^t\sin \left(\frac{t\pi}{2} + kx\right)\right]' \\
=& k^{t+1}\cos \left(\frac{t\pi}{2} + kx\right) \\
=& k^{t+1}\sin \left[\frac{(t + 1)\pi}{2} + kx\right] \\
\end{aligned}
故,由第一数学归纳法,命题成立,证毕。
\displaystyle (\cos kx)^{(n)} = k^n\cos \left(\frac{n\pi}{2} + kx\right)
证明:证明方法同上 (\sin kx)^{(n)},不再赘述。
\displaystyle (\ln x)^{(n)} = (-1)^{n-1}\frac{(n-1)!}{x^n}
证明: 用数学归纳法
当 n = 1 时
\displaystyle(\ln x)' = \frac{1}{x} = (-1)^{0}(1!)\frac{1}{x^1}
设 n = k 时
\displaystyle(\ln x)^{(k)} = (-1)^{k-1}(k-1)!\frac{1}{x^k} 成立
则当 n = k + 1 时
\begin{aligned}
&(\ln x)^{(k+1)} \\
= &\left[(-1)^{k-1}(k-1)!\frac{1}{x^k}\right]' \\
= &(-1)^{k}k!\frac{1}{x^{(k+1)}}
\end{aligned}
故,由第一数学归纳法,命题成立,证毕。
\displaystyle [\ln(1 + x)]^{(n)} = (-1)^{(n-1)}\frac{(n-1)!}{(1 + x)^n}
证明: 此命题为上一命题的简单推论
做变量替换 u = 1 + x 可得:
\displaystyle [\ln(1 + x)]^{(n)} = (-1)^{(n-1)}\frac{(n-1)!}{(1 + x)^n}
\displaystyle \left(\frac{1}{x}\right)^{(n)} = (-1)^{n}\frac{n!}{x^{n + 1}}
证明:
由于 \displaystyle(\ln x)' = \frac{1}{x},根据 \displaystyle (\ln x)^{(n)} = (-1)^{n-1}\frac{(n-1)!}{x^n} 可得:
\displaystyle \left(\frac{1}{x}\right)^{(n)} = (-1)^{n}\frac{n!}{x^{n + 1}}
\displaystyle \left(\frac{1}{x + a}\right)^{(n)} = (-1)^{n}\frac{n!}{(x + a)^{n + 1}}
证明: 由 x' = (x + a)' 易知,此命题为上一命题的简单推论。
\displaystyle (x^m)^{(n)} = m(m-1)\cdots(m - n + 1) x^{m - n}
证明: 用数学归纳法
当 n = 1 时
(x^m)' = m x^{m - 1},m - 1 + 1 = m 成立
设 n = k 时
(x^m)^{(k)} = m(m-1)\cdots(m - k + 1) x^{m - k} 成立
则当 n = k + 1 时
\begin{aligned}
& (x^m)^{(k + 1)} \\
= &[m(m-1)\cdots(m - k + 1) x^{m - k}]' \\
= &m(m-1)\cdots(m - k + 1)(m - k) x^{m - k - 1} \\
= &m(m-1)\cdots[m - (k + 1) - 1] x^{m - (k + 1)} \\
\end{aligned}
故,由第一数学归纳法,命题成立,证毕。
\displaystyle (u \pm v)^{(n)} = u^{(n)} \pm v^{(n)}
证明: 用数学归纳法
当 n = 1 时:
\begin{aligned}
& [u(x) \pm v(x)]' \\
= & \lim_{\Delta x\to 0} \frac{[u(x + \Delta x) \pm v(x + \Delta x)] - [u(x) \pm v(x)]}{\Delta x} \\
= & \lim_{\Delta x\to 0} \frac{u(x + \Delta x) - u(x)}{\Delta x} \pm \lim_{\Delta x\to 0} \frac{v(x + \Delta x) - v(x)}{\Delta x} \\
= & u(x)' \pm v(x)'
\end{aligned}
当 n = k 时:
\displaystyle (u \pm v)^{(k)} = u^{(k)} \pm v^{(k)}
则 n = k + 1 时:
\displaystyle (u \pm v)^{(k + 1)} = [u^{(k)} \pm v^{(k)}]' = u^{(k+1)} \pm v^{(k+1)}
故,由第一数学归纳法,命题成立,证毕。
莱布尼茨公式: \displaystyle (uv)^{(n)} = \sum_{k=0}^{n} \mathrm{C}_n^ku^{(n-k)}v^{(k)}
当 n = 1 时
\begin{aligned}
& [u(x)v(x)]' \\
= & \lim_{\Delta x\to 0} \frac{u(x + \Delta x)v(x + \Delta x) - u(x)v(x)}{\Delta x} \\
= & \lim_{\Delta x\to 0} \frac{u(x + \Delta x)v(x + \Delta x) - u(x)v(x + \Delta x) + u(x)v(x + \Delta x) - u(x)v(x)}{\Delta x} \\
= & \lim_{\Delta x\to 0} \frac{[u(x + \Delta x) - u(x)]v(x + \Delta x)}{\Delta x} + \lim_{\Delta x\to 0} \frac{u(x)[v(x + \Delta x) - v(x)]}{\Delta x} \\
= & \lim_{\Delta x\to 0} \frac{u(x + \Delta x) - u(x)}{\Delta x} \pm \lim_{\Delta x\to 0} \frac{v(x + \Delta x) - v(x)}{\Delta x} \\
= & u(x)'v(x) + u(x)v(x)'
\end{aligned}
设 n = u 时
\displaystyle (uv)^{(u)} = \sum_{k=0}^{u} \mathrm{C}_u^ku^{(u-k)}v^{(k)} 成立
则 n = u + 1 时:
\begin{aligned}
& (uv)^{(u + 1)} \\
= & \left[\sum_{k=0}^{u} \mathrm{C}_u^ku^{(u-k)}v^{(k)}\right]' \\
= & \left[
\mathrm{C}_u^0 u^{(u)}v^{(0)}
+ \mathrm{C}_u^1 u^{(u-1)}v^{(1)} + \cdots +
\mathrm{C}_u^u u^{(0)}v^{(u)}\right]' \\
= & \mathrm{C}_u^0 u^{(u + 1)}v^{(0)} + \mathrm{C}_u^0 u^{(u)}v^{(1)} + \\
& \mathrm{C}_u^1 u^{(u)}v^{(1)} + \mathrm{C}_u^1 u^{(u-1)}v^{(2)} + \cdots + \\
&\mathrm{C}_u^u u^{(1)}v^{(u)} + \mathrm{C}_u^u u^{(0)}v^{(u + 1)} \\
= & \mathrm{C}_{u+1}^0 u^{(u + 1)}v^{(0)} + \\
& [\mathrm{C}_u^0 u^{(u)}v^{(1)} + \mathrm{C}_u^1 u^{(u)}v^{(1)} ]+ \\
& [\mathrm{C}_u^1 u^{(u - 1)}v^{(2)} + \mathrm{C}_u^2 u^{(u-1)}v^{(2)}] + \cdots + \\
& [\mathrm{C}_u^1 u^{(1)}v^{(u)} + \mathrm{C}_u^u u^{(1)}v^{(u)}] + \\
= & \mathrm{C}_{u+1}^0 u^{(u + 1)}v^{(0)} + [\mathrm{C}_{u + 1}^1 u^{(u)}v^{(1)}]+ \\
& \mathrm{C}_{u + 1}^2 u^{(u - 1)}v^{(2)} + \cdots + \\
& \mathrm{C}_{u + 1}^u u^{(1)}v^{(u)} + \mathrm{C}_{u+1}^{u+1} u^{(0)}v^{(u + 1)} \\
= & \sum_{k=0}^{u + 1} \mathrm{C}_{u+1}^ku^{(u + 1 -k)}v^{(k)} \\
\end{aligned}
故,由第一数学归纳法,命题成立,证毕。
中值定理辅助函数
对于中值定理辅助函数的证明,只需要找到一个原函数即可,所以可以忽略不定积分的常数,题目中可能需要一个常数,这个需酌情添加。
f(x)f'(x) = 0 \to F(x) = f^2(x)
证明:
\begin{aligned}
&f(x)f'(x) = 0 \\
&\int f(x)f'(x) dx \\
=& \int f(x)d f(x) \\
=& \frac{1}{2} f^2(x) \\
\end{aligned}
于是可以取 F(x) = f^2(x),证毕。
[f'(x)]^2 + f(x)f''(x)= 0 \to F(x) = f'(x)f(x)
证明:
\begin{aligned}
&[f'(x)]^2 + f(x)f''(x)= 0 \\
& \frac{f'(x)}{f(x)} + \frac{f''(x)}{f'(x)}= 0 \\
& [\ln f(x) + \ln f'(x)]'= 0 \\
& \ln[f(x)f'(x)]'= 0 \\
\end{aligned}
于是可以取 F(x) = f(x)f'(x),证毕。
类似地,f(x)f''(x) - [f'(x)]^2= 0 \to \displaystyle F(x) = \frac{f'(x)}{f(x)}
f'(x) + f(x)\varphi'(x)= 0 \to F(x) = f(x)e^{\varphi(x)}
证明:
\begin{aligned}
&f'(x) + f(x)\varphi'(x)= 0 \\
& \frac{f'(x)}{f(x)} + \varphi'(x)= 0 \\
& [\ln f(x) + \ln e^{\varphi(x)}]'= 0 \\
& [\ln f(x)e^{\varphi(x)}]'= 0 \\
\end{aligned}
于是可以取 F(x) = f(x)e^{\varphi(x)},证毕。
类似地,可知:
f'(x) - f(x)\varphi'(x)= 0 \to \displaystyle F(x) = \frac{f(x)}{e^{\varphi(x)}}
xf'(x) + f(x)= 0 \to F(x) = xf(x)
证明:
\begin{aligned}
& xf'(x) + f(x)= 0 \\
& \frac{f'(x)}{f(x)} + \frac{1}{x}= 0 \\
& [\ln f(x) + \ln x]'= 0 \\
&[\ln xf(x)]'= 0
\end{aligned}
于是可以取 \displaystyle F(x) = xf(x),证毕。
类似地,可知:
xf'(x) - f(x)= 0 \to \displaystyle F(x) = \frac{f(x)}{x}
\displaystyle \int_a^b f(x) dx \to \displaystyle F(x) = \int_a^x f(t) dt
定积分常用公式
区间再线公式:\displaystyle \int_a^b f(x) dx = \int_a^b f(a + b - x) dx
证明:
\begin{aligned}
& \int_a^b f(x) dx \\
\xlongequal{x = a+b-t}& \int_b^a f(a+b-t) d(-t) \\
=& \int_a^b f(a+b-t) dt \\
=& \int_a^b f(a+b-x) dx
\end{aligned}
证毕。
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