## 2019年9月23日月曜日

### 数学 - Python - 微分積分学 - 微分法の公式 - 数学的帰納法 - 関数の積の導関数、一般化

1. $\begin{array}{l}\frac{d}{\mathrm{dx}}F\left(x\right)\\ =\frac{d}{\mathrm{dx}}\left({f}_{1}\left(x\right)\dots {f}_{n-1}\left(x\right){f}_{n}\left(x\right)\right)\\ =\frac{d}{\mathrm{dx}}\left({f}_{1}\left(x\right)·\dots ·{f}_{n-1}\left(x\right){f}_{n}\left(x\right)+{f}_{1}\left(x\right)·\dots ·{f}_{n-1}\left(x\right)\frac{d}{\mathrm{dx}}{f}_{n}\left(x\right)\\ ={f}_{1}\text{'}\left(x\right){f}_{2}\left(x\right)·\dots ·{f}_{n-1}\left(x\right){f}_{n}\left(x\right)\\ +{f}_{1}\left(x\right){f}_{2}\text{'}\left(x\right){f}_{3}\left(x\right)·\dots ·{f}_{n-1}\left(x\right){f}_{n}\left(x\right)\\ +\dots \\ +{f}_{1}\left(x\right)·\dots ·{f}_{n-1}\text{'}\left(x\right){f}_{n}\left(x\right)\\ +{f}_{1}\left(x\right)·\dots ·{f}_{n-1}\left(x\right){f}_{n}\text{'}\left(x\right)\end{array}$

よって、 帰納法によりすべての自然.'数について成り立つ。

（証明終）

コード

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, Function, product, Derivative

print('1.')

x = symbols('x')
f = 1

for i in range(1, 6):
print(f'n = {i}')
f *= Function(f'f_{i}')(x)
d = Derivative(f, x, 1)
for o in [d, d.doit()]:
pprint(o)
print()


$./sample1.py 1. n = 1 d ──(f₁(x)) dx d ──(f₁(x)) dx n = 2 d ──(f₁(x)⋅f₂(x)) dx d d f₁(x)⋅──(f₂(x)) + f₂(x)⋅──(f₁(x)) dx dx n = 3 d ──(f₁(x)⋅f₂(x)⋅f₃(x)) dx d d d f₁(x)⋅f₂(x)⋅──(f₃(x)) + f₁(x)⋅f₃(x)⋅──(f₂(x)) + f₂(x)⋅f₃(x)⋅──(f₁(x)) dx dx dx n = 4 d ──(f₁(x)⋅f₂(x)⋅f₃(x)⋅f₄(x)) dx d d f₁(x)⋅f₂(x)⋅f₃(x)⋅──(f₄(x)) + f₁(x)⋅f₂(x)⋅f₄(x)⋅──(f₃(x)) + f₁(x)⋅f₃(x)⋅f₄(x)⋅ dx dx d d ──(f₂(x)) + f₂(x)⋅f₃(x)⋅f₄(x)⋅──(f₁(x)) dx dx n = 5 d ──(f₁(x)⋅f₂(x)⋅f₃(x)⋅f₄(x)⋅f₅(x)) dx d d f₁(x)⋅f₂(x)⋅f₃(x)⋅f₄(x)⋅──(f₅(x)) + f₁(x)⋅f₂(x)⋅f₃(x)⋅f₅(x)⋅──(f₄(x)) + f₁(x)⋅ dx dx d d f₂(x)⋅f₄(x)⋅f₅(x)⋅──(f₃(x)) + f₁(x)⋅f₃(x)⋅f₄(x)⋅f₅(x)⋅──(f₂(x)) + f₂(x)⋅f₃(x)⋅ dx dx d f₄(x)⋅f₅(x)⋅──(f₁(x)) dx$