高専の数学解説

第6章 6.2

(1)

y=e2xy = e^{2x}

y′=(2x)′e2x=2e2xy' = (2x)'e^{2x} = 2e^{2x}

(2)

y=e2−3xy = e^{2-3x}

y′=(2−3x)′e2−3x=−3e2−3xy' = (2-3x)'e^{2-3x} = -3e^{2-3x}

(3)

y=xe−2xy = xe^{-2x}

y′=e−2x+x(−2x)′e−2x=e−2x−2xe−2x=(1−2x)e−2xy' = e^{-2x} + x(-2x)'e^{-2x} = e^{-2x} - 2xe^{-2x} = (1-2x)e^{-2x}

(4)

y=xe−x2=e−x2+x(−x2)′e−x2=e−x2−2x2e−x2=(1−2x2)e−x2y = xe^{-x^2} = e^{-x^2} + x(-x^2)'e^{-x^2} = e^{-x^2} - 2x^2e^{-x^2} = (1-2x^2)e^{-x^2}

(5)

y=(x+1)a2xy = (x+1)a^{2x} 両辺の log⁡\log をとって

log⁡y=log⁡(x+1)a2xlog⁡y=log⁡(x+1)+2x⋅log⁡a\log y = \log(x+1)a^{2x} \qquad \log y = \log(x+1) + 2x \cdot \log a

y′y=1x+1+2log⁡ay′=y(1x+1+2log⁡a)y′=(x+1)a2x(1x+1+2log⁡a)\frac{y'}{y} = \frac{1}{x+1} + 2\log a \qquad y' = y\left(\frac{1}{x+1} + 2\log a\right) \qquad y' = (x+1)a^{2x}\left(\frac{1}{x+1} + 2\log a\right)

y′={1+2(x+1)log⁡a}a2xy' = \{1 + 2(x+1)\log a\}a^{2x}

(6)

y=11−e2xy = \frac{1}{1-e^{2x}}

y′=(1−e2x)−1′=−(1−e2x)′(1−e2x)−2=−(−2e2x)(1−e2x)−2=2e2x(1−e2x)2y' = (1-e^{2x})^{-1\prime} = -(1-e^{2x})'(1-e^{2x})^{-2} = -(-2e^{2x})(1-e^{2x})^{-2} = \frac{2e^{2x}}{(1-e^{2x})^2}

解説: r31bn1z