微分公式

$${ \frac{d}{dx}\{ A f(x)+B g(x) \} }$$

$$\frac{1}{\Delta x} \Delta[Af(x)+Bg(x)]$$

$$= A\frac{\Delta f(x)}{\Delta x} +B\frac{\Delta g(x)}{\Delta x}$$

$$\to A f'(x)+B g'(x)$$ (2)

$${ \frac{d}{dx} \left[ f(x)g(x) \right] }$$

$$\frac{1}{\Delta x} \left[\{f+\Delta f\}\{g+\Delta g\}-f\cdot g \right]$$

$$= \frac{1}{\Delta x} \left[f\cdot\Delta g +g\cdot\Delta f + \Delta f\cdot\Delta g \right]$$

$$= f\cdot\frac{\Delta g}{\Delta x}+\frac{\Delta f}{\Delta x}\cdot g +\frac{\Delta f}{\Delta x}\cdot\frac{\Delta g}{\Delta x}\cdot \Delta x$$

$$\to f(x) g'(x)+f'(x)g(x)$$ (3)

$${ \frac{d}{dx} \left\{ \frac{1}{f(x)} \right\} }$$

$$\frac{1}{\Delta x} \left[ \frac{1}{f+\Delta f}-\frac{1}{f} \right]$$

$$= \frac{1}{\Delta x} \cdot \frac{-\Delta f}{(f+\Delta f)f}$$

$$= -\frac{1}{(f+\Delta f)f}\frac{\Delta f}{\Delta x}$$

$$\to -\frac{f'(x)}{\{f(x)\}^2}$$ (4)

$$%% { \frac{d}{dx}\left\{\frac{f(x)}{g(x)}\right\} }$$

$$\frac{1}{\Delta x} \left[ \frac{f+\Delta f}{g+\Delta g}-\frac{f}{g} \right]$$

$$= \frac{1}{\Delta x} \frac{g\Delta f- f\Delta g}{(g+\Delta g)g}$$

$$= \frac{1}{(g+\Delta g)g} \cdot \left\{ \frac{\Delta f}{\Delta x}g - f\frac{\Delta g}{\Delta x} \right\}$$

$$\to \frac{f'(x)g(x)-f(x)g'(x)}{\{g(x)\}^2}$$ (5)

合成関数の微分

$$z=z(y),\qquad y=y(x)$$

のとき、

$$\frac{\Delta z}{\Delta x}=\frac{\Delta z}{\Delta y}\cdot \frac{\Delta y}{\Delta x}$$

$\Delta x\to 0$として、

$$\frac{dz}{dx} = \frac{dz}{dy}\times\frac{dy}{dx}$$

$$= z'(y)\times y'(x)$$ (6)

微分の定義にしたがって次の関数を$x$について微分せよ。

(1) $$y=x^2+3x+1$$

$$\begin{eqnarray*} \frac{dy}{dx}&=&\lim_{h\to0} \frac{\{(x+h)^2+3(x+h)+1\}-\{x^... ...to0}\frac{2xh+h^2+3h}{h}\\ &=&\lim_{h\to0}2x+h+3\\ &=& 2x+3 \end{eqnarray*}$$

(2) $$y=(x+3)(x-1)$$

$$\begin{eqnarray*} \frac{dy}{dx} &=&\lim_{h\to0}\frac{(x+h+3)(x+h-1)-(x+3)(x-1)}{h}\\ &=&\lim_{h\to0}\frac{h^2+(2x+2)h}{h}\\ &=& 2x+2 \end{eqnarray*}$$

(3) $$y=\frac{1}{x+1}$$

$$\begin{eqnarray*} \frac{dy}{dx} &=&\lim_{h\to0}\frac{\frac{1}{x+h+1}-\frac{1}{x}}{h} \end{eqnarray*}$$

通分して、整理する:

$$\begin{eqnarray*} \frac{dy}{dx} &=&\lim_{h\to0}\frac{(x+1)-(x+h+1)}{h(x+h+1)(x... ...=&\lim_{h\to0}\frac{-1}{(x+h+1)(x+1)}\\ &=&-\frac{1}{(x+1)^2} \end{eqnarray*}$$

(4) $$y=\frac{1+x}{x}$$

$$\begin{eqnarray*} y &=& \frac{1}{x}+1\\ \frac{dy}{dx} &=&\lim_{h\to0}\frac{\... ...h}\\ &=&\lim_{h\to0}\frac{-h}{h(x+h)x}\\ &=& -\frac{1}{x^2} \end{eqnarray*}$$

(5) $$y=x+2+\frac{1}{x}$$

$$\begin{eqnarray*} \frac{dy}{dx} &=&\lim_{h\to0}\frac{\{x+h+2+\frac{1}{x+h}\}-\... ...h\to0}\frac{h(x+h)x+x-(x+h)}{(x+h)xh}\\ &=& \frac{x^2-1}{x^2} \end{eqnarray*}$$

(6) $$y=(x+1)^3$$

$$\begin{eqnarray*} \lefteqn{ \frac{dy}{dx} }\\ &=&\lim_{h\to0}\frac{(x+h+1)^... ...0}\left\{(x+h+1)^2+(x+h+1)(x+1)+(x+1)^2\right\}\\ &=&3(x+1)^2 \end{eqnarray*}$$

[公式]

$$\frac{d}{dx}\left\{a f(x)+b g(x)\right\} =a f'(x)+b g'(x)$$

$$\frac{d}{dx}\left\{f(x)\cdot g(x)\right\} =f'(x)\cdot g(x)+f(x)\cdot g'(x)$$

$$\frac{d}{dx}\left\{\frac{1}{f(x)}\right\} = -\frac{f'(x)}{\left\{f(x)\right\}^2}$$

$$\frac{d}{dx}\left\{\frac{f(x)}{g(x)}\right\} = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left\{f(x)\right\}^2}$$

$$\frac{dz}{dx}=\frac{dz}{dy}\cdot \frac{dy}{dx}$$

逆関数の微分

$$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$$