さまざまな微分法

微分の定義より、 \begin{gather} f^\prime \left(x\right)=\lim_{h\rightarrow0}{\frac{f \left(x+h\right)-f \left(x\right)}{h}}=\lim_{h\rightarrow0}{\frac{c-c}{h}}=0 \end{gather} すなわち、定数関数の導関数は定数関数 $0$ となる。 $\blacksquare$

微分の定義より、 \begin{align} \left\{kf \left(x\right)\right\}^\prime&=\lim_{h\rightarrow0}{\frac{kf \left(x+h\right)-kf \left(x\right)}{h}}\\ &=k \left[\lim_{h\rightarrow0}{\frac{f \left(x+h\right)-f \left(x\right)}{h}}\right]\\ &=kf^\prime \left(x\right) \end{align} $\blacksquare$

微分の定義より、 \begin{align} \left\{f \left(x\right)+g \left(x\right)\right\}^\prime&=\lim_{h\rightarrow0}{\frac{ \left\{f \left(x+h\right)+g \left(x+h\right)\right\}- \left\{f \left(x\right)+g \left(x\right)\right\}}{h}}\\ &=\lim_{h\rightarrow0}{\frac{f \left(x+h\right)-f \left(x\right)}{h}}+\lim_{h\rightarrow0}{\frac{g \left(x+h\right)-g \left(x\right)}{h}}\\ &=f^\prime \left(x\right)+g^\prime \left(x\right) \end{align} $\blacksquare$

微分の定義より、 \begin{align} \left\{f \left(x\right)-g \left(x\right)\right\}^\prime&=\lim_{h\rightarrow0}{\frac{ \left\{f \left(x+h\right)-g \left(x+h\right)\right\}- \left\{f \left(x\right)-g \left(x\right)\right\}}{h}}\\ &=\lim_{h\rightarrow0}{\frac{f \left(x+h\right)-f \left(x\right)}{h}}-\lim_{h\rightarrow0}{\frac{g \left(x+h\right)-g \left(x\right)}{h}}\\ &=f^\prime \left(x\right)-g^\prime \left(x\right) \end{align} $\blacksquare$

微分の定義より、 \begin{align} \left\{f \left(x\right) \cdot g \left(x\right)\right\}^\prime&=\lim_{h\rightarrow0}{\frac{f \left(x+h\right) \cdot g \left(x+h\right)-f \left(x\right) \cdot g \left(x\right)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{ \left\{f \left(x+h\right) \cdot g \left(x+h\right)-f \left(x+h\right) \cdot g \left(x\right)\right\}+ \left\{f \left(x+h\right) \cdot g \left(x\right)-f \left(x\right) \cdot g \left(x\right)\right\}}{h}}\\ &=\lim_{h\rightarrow0}{\frac{f \left(x+h\right) \cdot g \left(x+h\right)-f \left(x+h\right) \cdot g \left(x\right)}{h}}+\lim_{h\rightarrow0}{\frac{f \left(x+h\right) \cdot g \left(x\right)-f \left(x\right) \cdot g \left(x\right)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{f \left(x+h\right) \left\{g \left(x+h\right)-g \left(x\right)\right\}}{h}}+\lim_{h\rightarrow0}{\frac{g \left(x\right) \left\{f \left(x+h\right)-f \left(x\right)\right\}}{h}}\\ &= \left\{\lim_{h\rightarrow0}{f \left(x+h\right)}\right\} \left\{\lim_{h\rightarrow0}{\frac{g \left(x+h\right)-g \left(x\right)}{h}}\right\}+g \left(x\right) \left\{\lim_{h\rightarrow0}{\frac{f \left(x+h\right)-f \left(x\right)}{h}}\right\}\\ &=f \left(x\right) \cdot g^\prime \left(x\right)+f^\prime \left(x\right) \cdot g \left(x\right) \end{align} $\blacksquare$

微分の定義より、 \begin{align} \left\{\frac{f \left(x\right)}{g \left(x\right)}\right\}^\prime&=\lim_{h\rightarrow0}{\frac{\frac{f \left(x+h\right)}{g \left(x+h\right)}-\frac{f \left(x\right)}{g \left(x\right)}}{h}}\\ &=\lim_{h\rightarrow0}{\frac{f \left(x+h\right) \cdot g \left(x\right)-g \left(x+h\right) \cdot f \left(x\right)}{h \cdot g \left(x\right) \cdot g \left(x+h\right)}}\\ &=\lim_{h\rightarrow0}{\frac{ \left\{f \left(x+h\right) \cdot g \left(x\right)-f \left(x\right) \cdot g \left(x\right)\right\}- \left\{g \left(x+h\right) \cdot f \left(x\right)-f \left(x\right) \cdot g \left(x\right)\right\}}{h \cdot g \left(x\right) \cdot g \left(x+h\right)}}\\ &=\lim_{h\rightarrow0}{ \left[\frac{1}{g \left(x\right) \cdot g \left(x+h\right)} \left\{\frac{f \left(x+h\right)-f \left(x\right)}{h} \cdot g \left(x\right)-\frac{g \left(x+h\right)-g \left(x\right)}{h} \cdot f \left(x\right)\right\}\right]}\\ &= \left[\lim_{h\rightarrow0}{\frac{1}{g \left(x\right) \cdot g \left(x+h\right)}}\right] \left[g \left(x\right) \cdot \lim_{h\rightarrow0}{\frac{f \left(x+h\right)-f \left(x\right)}{h}}-f \left(x\right) \cdot \lim_{h\rightarrow0}{\frac{g \left(x+h\right)-g \left(x\right)}{h}}\right]\\ &=\frac{f^\prime \left(x\right) \cdot g \left(x\right)-f \left(x\right) \cdot g^\prime \left(x\right)}{ \left\{g \left(x\right)\right\}^2} \end{align} $\blacksquare$

合成関数 $y=f \left\{g \left(x\right)\right\}$ を $x$ で微分すると、微分の定義式より、 \begin{gather} \frac{dy}{dx}=\lim_{h\rightarrow0}{\frac{f \left\{g \left(x+h\right)\right\}-f \left\{g \left(x\right)\right\}}{h}}\tag{P1} \end{gather} ここで、$g \left(x+h\right)=g \left(x\right)+u$ とおくと、 \begin{gather} u=g \left(x+h\right)-g \left(x\right)\tag{P2}\\ \lim_{h\rightarrow0}{u}=\lim_{h\rightarrow0}{ \left[g \left(x+h\right)-g \left(x\right)\right]}=g \left(x\right)-g \left(x\right)=0 \end{gather} したがって、 $h\rightarrow0 \quad $のとき$ \quad u\rightarrow0$ さらに、$g \left(x\right)=t$ とおくと、式 $(\mathrm{P}1)$ は、 \begin{gather} \frac{dy}{dx}=\lim_{h\rightarrow0}{\frac{f \left\{g \left(x\right)+u\right\}-f \left\{g \left(x\right)\right\}}{u} \cdot \frac{u}{h}} \end{gather} 式 $(\mathrm{P}2)$ より、$u=g \left(x+h\right)-g \left(x\right)$ を戻して、 \begin{align} \frac{dy}{dx}&=\lim_{h\rightarrow0,u\rightarrow0}{ \left[\frac{f \left(t+u\right)-f \left(t\right)}{u} \cdot \frac{g \left(x+h\right)-g \left(x\right)}{h}\right]}\\ &=\lim_{u\rightarrow0}{\frac{f \left(t+u\right)-f \left(t\right)}{u}} \cdot \lim_{h\rightarrow0}{\frac{g \left(x+h\right)-g \left(x\right)}{h}}\\ &=\frac{dy}{dt} \cdot \frac{dt}{dx} \end{align} $\blacksquare$

微分の定義より、 \begin{gather} \frac{dx}{dy}=\lim_{\Delta y\rightarrow0}\frac{g \left(y+\Delta y\right)-g \left(y\right)}{\Delta y} \end{gather} ここで、 \begin{gather} \Delta y=f \left(x+\Delta x\right)-f \left(x\right)\\ g \left(y+\Delta y\right)-g \left(y\right)=\Delta x \end{gather} とすると、 $\Delta y\rightarrow0 \quad $のとき、$ \quad \Delta x\rightarrow0$ となるので、 \begin{align} \frac{dx}{dy}&=\lim_{\Delta x\rightarrow0}\frac{\Delta x}{f \left(x+\Delta x\right)-f \left(x\right)}\\ &=\lim_{\Delta x\rightarrow0}\frac{1}{\frac{f \left(x+\Delta x\right)-f \left(x\right)}{\Delta x}}\\ &=\frac{1}{\lim_{\Delta x\rightarrow0}{\frac{f \left(x+\Delta x\right)-f \left(x\right)}{\Delta x}}}\\ &=\frac{1}{f^\prime \left(x\right)} \end{align} $\blacksquare$