06　種々の関数の微分演習

\(f'(x)\begin{array}[t]{l} = (x + 1)'\cdot (x^2 - x + 1) + (x + 1) \cdot (x^2 - x + 1)' \\ = 1 \cdot (x^2 - x + 1) + (x + 1) \cdot (2x - 1) \\ = x^2 - x + 1 + 2x^2 + x - 1 \\ = 3x^2 \end{array}\)

\(f'(x)\begin{array}[t]{l} = 3(x^2 + 2x - 3)^2\cdot(2x + 2) \\ = 6(x + 1)(x^2 + 2x - 3)^2 \end{array}\)

\(f'(x)\begin{array}[t]{l} = \{(x - 1)^2\}' (x + 1)^3 + (x - 1)^2 \{(x + 1)^3\}' \\ = 2(x - 1)(x + 1)^3 + (x - 1)^2 \cdot 3(x + 1)^2 \\ = (x - 1)(x + 1)^2\{2(x + 1) + 3(x - 1)\} \\ = (x - 1)(x + 1)^2(5x - 1) \end{array}\)

\(f'(x)\begin{array}[t]{l} = (x^3 - 3x^2 + 2x)' \\ = 3x^2 - 6x + 2 \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = 2\cdot\frac{x}{x - 1}\cdot\left(\frac{x}{x - 1}\right)' \\ \displaystyle = 2\cdot\frac{x}{x - 1}\cdot\frac{(x)'\cdot(x -1 ) - x\cdot(x - 1)'}{(x - 1)^2} \\ \displaystyle = 2\cdot\frac{x}{x - 1}\cdot\frac{x - 1 - x}{(x - 1)^2} \\ \displaystyle = -\frac{2x}{(x - 1)^3} \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = 3\left(2 - \frac{1}{x}\right)^2 \left(2 - \frac{1}{x}\right)' \\ \displaystyle = 3\left(2 - \frac{1}{x}\right)^2 \cdot \frac{1}{x^2} \\ \displaystyle = \frac{3(2x - 1)^2}{x^4} \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{(x - 1)'(x^2 + x + 1) - (x - 1)(x^2 + x + 1)'}{(x^2 + x + 1)^2} \\ \displaystyle = \frac{(x^2 + x + 1) - (x - 1)(2x + 1)}{(x^2 + x + 1)^2} \\ \displaystyle = \frac{(x^2 + x + 1) - (2x^2 - x - 1)}{(x^2 + x + 1)^2} \\ \displaystyle = -\frac{x^2 - 2x - 2}{(x^2 + x + 1)^2} \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{(x)'(x + 2)^2 - x\{(x + 2)^2\}'}{(x + 2)^4} \\ \displaystyle = \frac{(x + 2)^2 - x\{2(x + 2)\}}{(x + 2)^4} \\ \displaystyle = \frac{(x + 2)^2 - 2x(x + 2)}{(x + 2)^4} \\ \displaystyle = \frac{(x + 2) - 2x}{(x + 2)^3} \\ \displaystyle = -\frac{x - 2}{(x + 2)^3} \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{(x)'\sqrt{x^2 + 1} - x(\sqrt{x^2 + 1})'}{x^2 + 1} \\ \displaystyle = \frac{\sqrt{x^2 + 1} - \displaystyle \frac{\cancel{2}x^2}{\cancel{2}\sqrt{x^2 + 1}}}{x^2 + 1} \\ \displaystyle = \frac{x^2 + 1 - x^2}{(x^2 + 1)\sqrt{x^2 + 1}} \\ \displaystyle = \frac{1}{(x^2 + 1)\sqrt{x^2 + 1}} \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{1}{3}(2x^2 - 3)^{-\frac{2}{3}} \cdot (2x^2 - 3)' \\ \displaystyle = \frac{4x}{3\sqrt[3]{(2x^2 - 3)^2}} \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{3}{4}(3x^2 + 2x + 1)^{-\frac{1}{4}} \cdot (3x^2 + 2x + 1)' \\ \displaystyle = \frac{3}{4}(3x^2 + 2x + 1)^{-\frac{1}{4}} \cdot (6x + 2) \\ \displaystyle = \frac{3(3x + 1)}{2\sqrt[4]{3x^2 + 2x + 1}} \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{1}{2\sqrt{\displaystyle\frac{1 - x}{1 + x}}}\cdot\left(\frac{1 - x}{1 + x}\right)' \\ \displaystyle = \frac{1}{2}\sqrt{\frac{1 + x}{1 - x}} \cdot \frac{(1 - x)'(1 + x) - (1 - x)(1 + x)'}{(1 + x)^2} \\ \displaystyle = \frac{1}{2} \sqrt{\frac{1 + x}{1 - x}} \cdot \frac{-(1 + x) - (1 - x)}{(1 + x)^2} \\ \displaystyle = \frac{1}{\cancel{2}} \cdot \sqrt{\frac{1 + x}{1 - x}} \cdot \frac{-\cancel{2}}{(1 + x)^2} \\ \displaystyle = -\frac{\sqrt{1 + x}}{(1 + x)^2 \sqrt{1 - x}} \\ \displaystyle \left[ = - \frac{1}{(1 + x) \sqrt{(1 + x)(1 - x)}}\right] \end{array}\)

\(f'(x)\begin{array}[t]{l} = 2\sin x \cdot (\sin x)' \\ = 2\sin x \cos x \end{array}\)

\(f'(x)\begin{array}[t]{l} = 3\cos^2 2x \cdot (\cos 2x)' \\ = 3\cos^2 2x \cdot (-\sin 2x) \cdot (2x)' \\ = -6\cos^2 2x \, \sin 2x \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{1}{2\sqrt{\sin^2 x + 1}} \cdot (\sin^2 x + 1)' \\ \displaystyle = \frac{1}{2\sqrt{\sin^2 x + 1}} \cdot 2\sin x \cdot (\sin x)' \\ \displaystyle = \frac{1}{\cancel{2}\sqrt{\sin^2 x + 1}} \cdot \cancel{2}\sin x \cos x \\ \displaystyle = \frac{\sin x \cos x}{\sqrt{\sin^2 x + 1}} \end{array}\)

\(f(x)\begin{array}[t]{l} = \sin^2 x (1 - 2\sin^2 x) \\ = \sin^2 x - 2\sin^4 x \end{array}\)
\(f'(x)\begin{array}[t]{l} = 2\sin x \cdot (\sin x)' - 8\sin^3 x \cdot (\sin x)' \\ = 2\sin x \cos x (1 - 4\sin^2 x) \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{1}{x} \end{array}\)

\(f'(x)\begin{array}[t]{l} = e^x + e^{-x}(-x)' \\ = e^x - e^{-x} \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{1}{\sin x} \cdot (\sin x)' \\ \displaystyle = \frac{\cos x}{\sin x} \end{array}\)

\(f'(x)\begin{array}[t]{l} = (x^2 + x + 1)'e^{2x} + (x^2 + x + 1)(e^{2x})' \\ (2x + 1)e^{2x} + (x^2 + x + 1)e^{2x}(2x)' \\ = (2x + 1)e^{2x} + 2(x^2 + x + 1)e^{2x} \\ = (2x^2 + 4x + 3)e^{2x} \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{1}{\log x} \cdot (\log x)' \\ = \frac{1}{x\log x} \end{array}\)

\(f'(x)\begin{array}[t]{l} = (e^{-2x})'\cos 2x + e^{-2x}(\cos 2x)' \\ = e^{-2x}(-2x)'\cos 2x + e^{-2x} (-\sin 2x)(2x)' \\ = -2e^{-2x}(\cos 2x + \sin 2x) \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{1}{\sqrt{\displaystyle{\frac{x + 1}{x - 1}}}}\left(\sqrt{\frac{x + 1}{x - 1}}\right)' \\ \displaystyle = \sqrt{\frac{x - 1}{x + 1}} \cdot \frac{1}{2 \sqrt{\displaystyle \frac{x + 1}{x - 1}}} \cdot \left(\frac{x + 1}{x - 1}\right)' \\ \displaystyle = \sqrt{\frac{x - 1}{x + 1}} \cdot \frac{1}{2} \sqrt{\frac{x - 1}{x + 1}} \cdot \frac{(x + 1)'(x - 1) - (x + 1)(x - 1)'}{(x - 1)^2} \\ \displaystyle = \frac{1}{2} \cdot \frac{\cancel{x - 1}}{x + 1} \cdot \frac{(x - 1) - (x + 1)}{(x - 1)^{\cancel{2}}} \\ \displaystyle = \frac{1}{\cancel{2}} \cdot \frac{-\cancel{2}}{(x + 1)(x - 1)} \\ \displaystyle = -\frac{1}{(x + 1)(x - 1)} \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{(e^{x^2})'x - e^{x^2}(x)'}{x^2} \\ \displaystyle = \frac{e^{x^2} \cdot (x^2)' \cdot x - e^{x^2} \cdot 1}{x^2} \\ \displaystyle = \frac{2x^2 e^{x^2} - e^{x^2}}{x^2} \\ \displaystyle = \frac{(2x^2 - 1)\,e^{x^2}}{x^2} \end{array}\)

\(f'(x)\begin{array}[t]{l} \displaystyle = \frac{1}{x + \sqrt{x^2 + 1}}\left(x + \sqrt{x^2 + 1}\right)' \\ \displaystyle = \frac{1}{x + \sqrt{x^2 + 1}} \left(1 + \frac{(x^2 + 1)'}{2\sqrt{x^2 + 1}}\right) \\ \displaystyle = \frac{1}{x + \sqrt{x^2 + 1}} \left(1 + \frac{\cancel{2}x}{\cancel{2}\sqrt{x^2 + 1}}\right) \\ \displaystyle = \frac{1}{x + \sqrt{x^2 + 1}} \cdot \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}} \\ \displaystyle = \frac{1}{\cancel{x + \sqrt{x^2 + 1}}} \cdot \frac{\cancel{x + \sqrt{x^2 + 1}}}{\sqrt{x^2 + 1}} \\ \displaystyle = \frac{1}{\sqrt{x^2 + 1}} \end{array}\)