次の導関数を微分せよ．
\( (1) \) \( y = \sqrt{x^2 + 1} \)
\( (2) \) \( y = \dfrac{1}{\tan x} \)
\( (3) \) \( y = \sin^2 2x \)
\( (4) \) \( y = e^{3x} \cos 4x \)
＊微分公式＊
\( 1^\circ \)[微分の線形性] \( \{ k f(x) + l g(x) \}’ = k f'(x) + l g'(x) \)
\( 2^\circ \)[積の微分公式]\( \{ f(x) g(x) \}’ = f'(x) g(x) + f(x) g'(x) \)
\( 3^\circ \)[商の微分公式] \( \left\{ \dfrac{f(x)}{g(x)} \right\}’ = \dfrac{f'(x) g(x) – f(x) g'(x)}{\{ g(x) \}^2} \)
\( 4^\circ \)[ \( x \) の累乗] \( (x^n)’ = nx^{n-1} \)
\( 5^\circ \)[三角関数] \( (\sin x)’ = \cos x \) ， \( (\cos x)’ = – \sin x \) ， \( (\tan x)’ = \dfrac{1}{\cos^2} \)
\( 6^\circ \)[指数関数] \( (e^x)’ = e^x \) ， \( (a^x)’ = a^x \log a \)
\( 7^\circ \)[対数関数] \( (\log x)’ = \dfrac{1}{x} \) ， \( (\log_a x)’ = \dfrac{1}{x \log a} \)
＊合成関数の微分法＊
\( \{ g(f(x)) \}’ = g'(f(x)) \cdot f'(x) \)
\( \{ h(g(f(x))) \}’ = h'(g(f(x))) \cdot g'(f(x)) \cdot f'(x) \)
\( (1) \)
\( y = (x^2 + 1)^{\frac{1}{2}} \) であるから
\( y’ = \dfrac{1}{2} (x^2+1)^{-\frac{1}{2}} \cdot (x^2+1)’ \)
\( = \dfrac{1}{2\sqrt{x^2 + 1}} \cdot 2x \)
\( = \dfrac{x}{\sqrt{x^2 + 1}} \cdots \) 答
\( (2) \)
\( y = (\tan x)^{-1} \) であるから
\( y’ = -(\tan x)^{-2} \cdot (\tan x)’ \)
\( = -\dfrac{1}{\tan^2 x} \cdot \dfrac{1}{\cos^2 x} \)
\( = -\dfrac{1}{\sin^2 x} \cdots \) 答
\( (3) \)
\( y = (\sin 2x)^2 \) であるから
\( y’ = 2 \sin 2x \cdot (\sin 2x)’ \)
\( = 2 \sin 2x \cdot \cos 2x \cdot (2x)’ \)
\( = 4 \sin 2x \cos 2x \cdots \) 答
\( (4) \)
積の微分公式により
\( y’ = (e^{3x})’ \cos 4x + e^{3x} (\cos 4x)’ \)
\( = \{ e^{3x} \cdot (3x)’ \} \cos 4x + e^{3x} \{ (-\sin 4x) \cdot (4x)’ \} \)
\( = 3 e^{3x} \cos 4x – 4 e^{3x} \sin 4x \cdots \) 答
