【数学ⅡB】導関数と微分【大阪工業大・岡山理科大・福島県立医科大】

$$\begin{align*} f'(x)=3x^2-2x+a \end{align*}$$

$$\begin{align*} &3a^2-a=2(a^3-a^2+a^2+3) \\[4pt] &2a^3-3a^2+a+6=0 \\[4pt] &(a+1)(a^2-4a+6)=0 \end{align*}$$

$$\begin{align*} a^2-4a+6=(a-2)^2+2>0 \end{align*}$$

$$\begin{align*} &a+1=0 \\[4pt] &a=-1 \end{align*}$$

$$\begin{align*} f'(x)&=\dlim{h\to0}\dfrac{f(x+h)-f(x)}{h} \\[4pt] &=\dlim{h\to0}\dfrac{(x+h)^5-x^5}{h} \\[4pt] &=\dlim{h\to0}\dfrac{\nCk{5}{1}x^4h+\nCk{5}{2}x^3h^2+\nCk{5}{3}x^2h^3+\nCk{5}{4}xh^4+h^5}{h} \\[4pt] &=\dlim{h\to0}(\nCk{5}{1}x^4+\nCk{5}{2}x^3h+\nCk{5}{3}x^2h^2+\nCk{5}{4}xh^3+h^4) \\[4pt] &=\nCk{5}{1}x^4 \\[4pt] &=5x^4 \end{align*}$$

$$\begin{align*} f'(x)&=\dlim{h\to0}\dfrac{f(x+h)-f(x)}{h} \\[4pt] &=\dlim{h\to0}\dfrac{(x+h)^n-x^n}{h} \\[4pt] &=\dlim{h\to0}\dfrac{\nCk{n}{1}x^{n-1}h+\nCk{n}{n-2}x^{n-2}h^2+\cdots+\nCk{n}{n-1}xh^{n-1}+h^n}{h} \\[4pt] &=\dlim{h\to0}(\nCk{n}{1}x^{n-1}+\nCk{n}{n-2}x^{n-2}h+\cdots+\nCk{n}{n-1}xh^{n-2}+h^{n-1}) \\[4pt] &=\nCk{n}{1}x^{n-1} \\[4pt] &=nx^{n-1} \end{align*}$$