2.1 解答

2.1

1.

(a) $${f'(x) := \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}}$$より，

$$\lim_{h \to 0}\frac{\cos(3(x+h)) - \cos(3x)}{h} = \lim_{h \to 0}(\frac{\cos{3x}(\cos{3h} - 1)}{h} - \frac{\sin{3x}\sin{3h}}{h}) = -3\sin{3x}$$

(b)

$$\frac{(x+2+h)^{n} - (x+2)^n}{h} = \frac{n(x+2)^{n-1}h}{h} + h(\cdots) \rightarrow n(x+2)^{n-1}$$

2.

(a) $df = f^{\prime}(x) dx$より $df = 4x^3 dx$

(b) $df = e^{x} dx$

3.

(a) $${f_{-}^{'}(0) = \lim_{h \to 0-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0-} \frac{\vert h^2 + h\vert}{h} = \lim_{h \to 0-} \frac{-(h^2 + h)}{h} = -1}$$

$${f_{+}^{'}(0) = \lim_{h \to 0+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0+} \frac{\vert h^2 + h\vert}{h} = \lim_{h \to 0+} \frac{h^2 + h}{h} = 1}$$

(b) $${f_{-}^{'}(0) = \lim_{h \to 0-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0-} \frac{h^2 \sin{\frac{1}{h}}}{h} = \lim_{h \to 0-} h \sin{\frac{1}{h}} = 0}$$

$${f_{+}^{'}(0) = \lim_{h \to 0+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0+} \frac{h^2 \sin{\frac{1}{h}}}{h} = \lim_{h \to 0+} h \sin{\frac{1}{h}} = 0}$$

(c) $${f_{-}^{'}(0) = \lim_{h \to 0-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0-} \frac{\sqrt{h^3 + h^2}}{h} = \lim_{h \to 0-} \sqrt{1 + h} = 1}$$

$${f_{+}^{'}(0) = \lim_{h \to 0+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0+} \frac{\sqrt{h^3 + h^2}}{h} = \lim_{h \to 0+} \sqrt{1 + h} = 1}$$

4.

(a) $${\left(\frac{3x - 1}{x^2 + 1}\right)^{\prime} = \frac{3(x^2 + 1) - (3x-1)(2x)}{(x^2 + 1)^2} = \frac{-3x^2 + 2x + 3}{(x^2 + 1)^2}}$$

(b) $${(\sec{x})^{\prime} = \left(\frac{1}{\cos{x}}\right)^{\prime} = \sec{x}\tan{x}}$$

(c) $${(\csc{x})^{\prime} = \left(\frac{1}{\sin{x}}\right)^{\prime} = - \csc{x}\cot{x}}$$

(d) $${(\cot{x})^{\prime} = \left(\frac{\cos{x}}{\sin{x}}\right)^{\prime} = -\frac{1}{\sin^{2}{x}}}$$

(e) $${(x^{2}e^{x})^{\prime} = 2xe^{x} + x^{2}e^{x} = x(x+2)e^{x}}$$

(f) $${(e^{x}\sin{x})^{\prime} = e^{x}\sin{x} + e^{x}\cos{x} = e^{x}(\sin{x} + \cos{x})}$$

(g) $${(\frac{e^{x}}{\sin{x}})^{\prime} = \frac{e^{x}\sin{x} - e^{x}\cos{x}}{\sin^{2}{x}} = \frac{e^{x}(\sin{x} - \cos{x})}{\sin^{2}{x}}}$$