【微分積分】3-4-3 n階導関数の計算｜問題集

\(\displaystyle f(x)=-x^2-x-1+\frac{1}{1-x}\)
\(\displaystyle f^{(1)}(x)=-2x-1+(1-x)^{-2}\)
\(\displaystyle f^{(2)}(x)=-2+2(1-x)^{-3}\)
\(\displaystyle f^{(n)}(x)=n!(1-x)^{-(n+1)}\)

\(\displaystyle f^{(n)}(x)=\sum_{k=0}^{n}{}_{n}\mathrm{C}_k(e^x)^{(k)}(x^{-1})^{(n-k)}\)
\(\displaystyle =\sum_{k=0}^{n}\frac{n!}{k!(n-k)!}e^x\frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}\)
\(\displaystyle =\frac{(-1)^nn!e^x}{x^{n+1}}\sum_{k=0}^{n}\frac{(-x)^k}{k!}\)

\(f^{(n)}(x)\)
\(\displaystyle =\frac{1}{2}・-\frac{1}{2}・-\frac{3}{2}・\cdots・-\frac{2n-3}{2}(x-1)^{\frac{1}{2}-n}\)
\(\displaystyle =\frac{(-1)^{n-1}(2n-3)!!}{2^n(x-1)^{\frac{2n-1}{2}}}\)

\(f^{(1)}(x)=1+\log x\)
\(\displaystyle f^{(n)}(x)=(x^{-1})^{(n-2)}=\frac{(-1)^n(n-2)!}{x^{n-1}}\)

\(f'(x)=e^x\cos x-e^x\sin x\)
\(=e^x(\cos x-\sin x)\)
\(\displaystyle =\sqrt{2}e^x\cos\left(x+\frac{\pi}{4}\right)\)
\(\displaystyle f^{(n)}(x)=(\sqrt{2})^ne^x\cos\left(x+\frac{n\pi}{4}\right)\)