1の証明

　$\{kf(x)\}'$

$\displaystyle =\lim_{h\to 0}\dfrac{kf(x+h)-kf(x)}{h}$

$\displaystyle =\lim_{h\to 0}k \cdot \dfrac{f(x+h)-f(x)}{h}$

$\displaystyle =kf'(x)$

2の証明

　$\{f(x)+g(x)\}'$

$\displaystyle =\lim_{h\to 0}\dfrac{f(x+h)+g(x+h)-f(x)-g(x)}{h}$

$\displaystyle =\lim_{h\to 0}\left\{\dfrac{f(x+h)-f(x)}{h}+\dfrac{g(x+h)-g(x)}{h}\right\}$

$\displaystyle =f'(x)+g'(x)$

3の証明

1，2が証明できれば3も同様なので割愛します．

4，5の証明

当サイトの数学Ⅲのページを紹介します．

4(積の微分)の証明

5(合成関数の微分)の証明

☆の証明

合成関数の微分の証明が理解しづらい人向けです．こちらの証明は数学ⅡBまでで理解できます．

$f(x)=\left(ax+b\right)^{n}$ のとき

　$f'(x)$

$\displaystyle =\lim_{h\to 0}\dfrac{\left\{a(x+h)+b\right\}^{n}-\left(ax+b\right)^{n}}{h}$

$\displaystyle =\lim_{h\to 0}\dfrac{1}{h}\left\{\sum_{k=0}^{n}\hspace{0mm} _{n}{\rm C}_{k}a^{k}(x+h)^{k}b^{n-k}-\sum_{k=0}^{n}\hspace{0mm} _{n}{\rm C}_{k}a^{k}x^{k}b^{n-k}\right\}$

$\displaystyle =\lim_{h\to 0}\dfrac{1}{h}\left\{\sum_{k=0}^{n}\hspace{0mm} _{n}{\rm C}_{k}a^{k}(x^{k}+\hspace{0mm} _{k}{\rm C}_{1}x^{k-1}h+\cdots+h^{k})b^{n-k}-\sum_{k=0}^{n}\hspace{0mm} _{n}{\rm C}_{k}a^{k}x^{k}b^{n-k}\right\}$

$\displaystyle =\lim_{h\to 0}\dfrac{1}{h}\left\{\sum_{k=0}^{n}\hspace{0mm} _{n}{\rm C}_{k}a^{k}(\hspace{0mm} _{k}{\rm C}_{1}x^{k-1}h+\cdots+h^{k})b^{n-k}\right\}$

$\displaystyle =\lim_{h\to 0}\left\{\sum_{k=0}^{n}\hspace{0mm} _{n}{\rm C}_{k}a^{k}(\hspace{0mm} _{k}{\rm C}_{1}x^{k-1}+\cdots+h^{k-1})b^{n-k}\right\}$

$\displaystyle =\sum_{k=0}^{n}\hspace{0mm} _{n}{\rm C}_{k}a^{k}\hspace{0mm} _{k}{\rm C}_{1}x^{k-1}b^{n-k}$

$\displaystyle =a\sum_{k=0}^{n}\dfrac{n!}{k!(n-k)!}k(ax)^{k-1}b^{n-k}$

$\displaystyle =an\sum_{k=1}^{n}\dfrac{(n-1)!}{(k-1)!(n-k)!}(ax)^{k-1}b^{n-k}$

$\displaystyle =an\sum_{k=1}^{n}\hspace{0mm} _{n-1}{\rm C}_{k-1}(ax)^{k-1}b^{n-k}$

$=an\left(ax+b\right)^{n-1}$