高専の数学解説

第1章 1.21

(1)

y=sin⁡−11−x2y = \sin^{-1}\sqrt{1-x^2}

y′=−2x21−x21−(1−x2)=−x1−x2x2=−x∣x∣1−x2x∣x∣=±1 からy' = \dfrac{\dfrac{-2x}{2\sqrt{1-x^2}}}{\sqrt{1-(1-x^2)}} = \dfrac{\dfrac{-x}{\sqrt{1-x^2}}}{\sqrt{x^2}} = \dfrac{-x}{|x|\sqrt{1-x^2}} \qquad \dfrac{x}{|x|} = \pm 1 \text{ から}

=±11−x2= \pm\dfrac{1}{\sqrt{1-x^2}}

(2)

y=tan⁡−1x−1x+1y = \tan^{-1}\dfrac{x-1}{x+1}

y′=x+1−(x−1)(x+1)2(x−1x+1)2+1=2(x−1)2+(x+1)2=22x2+2=1x2+1y' = \dfrac{\dfrac{x+1-(x-1)}{(x+1)^2}}{\left(\dfrac{x-1}{x+1}\right)^2+1} = \dfrac{2}{(x-1)^2+(x+1)^2} = \dfrac{2}{2x^2+2} = \dfrac{1}{x^2+1}

(3)

y=x⋅sin⁡−1xy = x \cdot \sin^{-1}x

y′=sin⁡−1x+x1−x2y' = \sin^{-1}x + \dfrac{x}{\sqrt{1-x^2}}

(4)

y=(1+x2)tan⁡−1xy = (1+x^2)\tan^{-1}x

y′=2x⋅tan⁡−1x+(1+x2)⋅11+x2=2x⋅tan⁡−1x+1y' = 2x \cdot \tan^{-1}x + (1+x^2) \cdot \dfrac{1}{1+x^2} = 2x \cdot \tan^{-1}x + 1

(5)

y=x⋅sin⁡−1x+1−x2y = x \cdot \sin^{-1}x + \sqrt{1-x^2}

y′=sin⁡−1x+x1−x2+−2x21−x2=sin⁡−1xy' = \sin^{-1}x + \dfrac{x}{\sqrt{1-x^2}} + \dfrac{-2x}{2\sqrt{1-x^2}} = \sin^{-1}x

(6)

y=tan⁡−1x+log⁡(1+x2)y = \tan^{-1} x + \log(1+x^2)

y′=1x2+1+2x1+x2=2x+1x2+1y' = \frac{1}{x^2+1} + \frac{2x}{1+x^2} = \frac{2x+1}{x^2+1}

解説: r31bn1z