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微分(2)
\( y = f(u) \quad u = g(v) \quad v = h(x) \) の場合の
\( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} = f'(u) \cdot g'(v) \cdot h'(x)\)
を計算すればよい.
\( y = \log | \cos( 2x+3 ) | \)
\( y = \log |u| \quad u = \cos(v) \quad v = 2x+3 \) に対応させて,
\( y = \log |u| \quad f'(u) = \frac{dy}{du} = \frac{1}{u} \)
\( u = \cos(v) \quad u'(v) = \frac{du}{dv} = -\sin(v) \)
\( v = 2x+3 \quad v'(x) = \frac{dv}{dx} = 2 \)
\( \frac{dy}{dx} = \frac{1}{u} \times -\sin(v) \times 2 = \frac{1}{\cos(2x+3)} \times -\sin(2x+3) \times 2\)
\( = -2\frac{\sin(2x+3)}{\cos(2x+3)}\)
\( = -2\tan(2x+3)\)
選択肢
(1) | \(\frac{3 \cos{\left(3 x \right)}}{x} - \frac{\sin{\left(3 x \right)}}{x^{2}}\) | (2) | \(2 e^{2 x}\) | (3) | \(- 2 \sin{\left(2 x + 3 \right)}\) |
(4) | \(2 \left(x + 1\right) e^{x^{2} + 2 x + 1}\) | (5) | \(\frac{2}{\tan{\left(2 x + 3 \right)}}\) | (6) | \(3 \cos{\left(3 x \right)}\) |
(7) | \(- 6 x \left(2 - x\right)^{5} + \left(2 - x\right)^{6}\) | (8) | \(6 \left(x - 2\right)^{5}\) | (9) | \(5 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\) |