高専の数学解説

第1章 1.1

(1)

y=x3+2x2−3x−1y = x^3 + 2x^2 - 3x - 1

y′=3x2+4x−3y' = 3x^2 + 4x - 3

y′′=6x+4y'' = 6x + 4

(2)

y=12x+1=(2x+1)−1y = \frac{1}{2x+1} = (2x+1)^{-1}

y′=−2(2x+1)−2y' = -2(2x+1)^{-2}

y′′=−2⋅(−2)⋅(2x+1)−3⋅2=8(2x+1)−3=8(2x+1)3y'' = -2 \cdot (-2) \cdot (2x+1)^{-3} \cdot 2 = 8(2x+1)^{-3} = \frac{8}{(2x+1)^3}

(3)

y=sin⁡3xy = \sin 3x

y′=3cos⁡3xy' = 3\cos 3x

y′′=−9sin⁡3xy'' = -9\sin 3x

(4)

y=cos⁡2xy = \cos^2 x

y′=2cos⁡x⋅(−sin⁡x)=−2cos⁡x⋅sin⁡xy' = 2\cos x \cdot (-\sin x) = -2\cos x \cdot \sin x

y′′=−2{−sin⁡x⋅sin⁡x+cos⁡x⋅cos⁡x}=2(sin⁡2x−cos⁡2x)y'' = -2\{-\sin x \cdot \sin x + \cos x \cdot \cos x\} = 2(\sin^2 x - \cos^2 x)

(5)

y=e−2xy = e^{-2x}

y′=−2e−2xy' = -2e^{-2x}

y′′=−2(−2e−2x)=4e−2xy'' = -2(-2e^{-2x}) = 4e^{-2x}

(6)

y=xexy = xe^x

y′=ex+xex=(1+x)exy' = e^x + xe^x = (1+x)e^x

y′′=ex+(1+x)ex=(x+2)exy'' = e^x + (1+x)e^x = (x+2)e^x

(7)

y=log⁡(x+1)y = \log(x+1)

y′=1x+1y' = \frac{1}{x+1}

y′′=−1(x+1)2y'' = \frac{-1}{(x+1)^2}

(8)

y=log⁡sin⁡xy = \log \sin x

y′=cos⁡xsin⁡xy' = \frac{\cos x}{\sin x}

y′′=−sin⁡x⋅sin⁡x−cos⁡x⋅cos⁡xsin⁡2x=−sin⁡2x−cos⁡2xsin⁡2x=−1sin⁡2xy'' = \frac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{\sin^2 x} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{-1}{\sin^2 x}

解説: r31bn1z