无穷阶求导

泰勒展开式

ex=∑n=0∞xnn!=1+x+x22!+⋯e^x = \sum^\infty_{n=0}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \cdots 1x+1=∑n=0∞(−1)nxn=1−x+x2+⋯\frac{1}{x+1} = \sum^\infty_{n=0}(-1)^nx^n = 1-x+x^2+ \cdots 11−x=∑n=0∞xn=1+x+x2+⋯\frac{1}{1-x} = \sum^\infty_{n=0}x^n = 1 + x + x^2 + \cdots ln⁡(1+x)=∑n=1∞(−1)n−1xnn=x−x22+x33−⋯\ln(1+x) = \sum^\infty_{n=1}(-1)^{n-1}\frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots sin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!=x−x33!+x55!+⋯\sin x = \sum^\infty_{n=0}(-1)^n\frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots cos⁡x=∑n=0∞(−1)nx2n(2n)!=1−x22!+x44!−⋯\cos x = \sum^\infty_{n=0}(-1)^n\frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots (1+x)α=1+αx+α(α−1)2!x2+⋯(1+x)^{\alpha} = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}x^2 + \cdots

无穷阶求导公式

(αx)(n)=αx(ln⁡α)n(\alpha^x)^{(n)} = \alpha^x(\ln \alpha)^n (ex)(n)=ex(e^x)^{(n)} = e^x (sin⁡kx)(n)=knsin⁡(kx+nπ2)(\sin kx)^{(n)} = k^n\sin(kx + \frac{n\pi}{2}) (cos⁡kx)(n)=kncos⁡(kx+nπ2)(\cos kx)^{(n)} = k^n\cos(kx + \frac{n\pi}{2}) (ln⁡x)(n)=(−1)n−1(n−1)!xn(\ln x)^{(n)} = (-1)^{n-1}\frac{(n-1)!}{x^n} [ln⁡(1+x)](n)=(−1)n−1(n−1)!(1+x)n[\ln(1+x)]^{(n)} = (-1)^{n-1}\frac{(n-1)!}{(1+x)^n} [(x+x0)m](n)=m(m−1)(m−2)⋯(m−n+1)(x+x0)m−n[(x+x_0)^m]^{(n)} = m(m-1)(m-2)\cdots(m-n+1)(x+x_0)^{m-n} (1x+a)(n)=(−1)nn!(x+a)n+1(\frac{1}{x+a})^{(n)} = \frac{(-1)^nn!}{(x+a)^{n+1}}