第1章 1.25

(1)

$y = x\sin^{-1}\sqrt{1-x^2}$

$y' = \sin^{-1}\sqrt{1-x^2} + x\cdot\frac{\dfrac{-2x}{2\sqrt{1-x^2}}}{\sqrt{1-(1-x^2)}} = \sin^{-1}\sqrt{1-x^2} + \frac{-x^2}{\sqrt{x^2}\sqrt{1-x^2}}$

$= \sin^{-1}\sqrt{1-x^2} + \frac{-x^2}{|x|}\cdot\frac{1}{\sqrt{1-x^2}} = \sin^{-1}\sqrt{1-x^2} - \frac{|x|}{\sqrt{1-x^2}}$

(2)

$y = \sin^{-1}e^x$

$y' = \frac{e^x}{\sqrt{1-e^{2x}}}$

(3)

$y = \tan^{-1}\sqrt{\frac{x-1}{x+1}}$

$y' = \frac{1}{1+\dfrac{x-1}{x+1}}\left(\sqrt{\frac{x-1}{x+1}}\right)' = \frac{1}{\dfrac{x+1+x-1}{x+1}}\cdot\frac{\dfrac{\sqrt{x+1}}{2\sqrt{x-1}}-\dfrac{\sqrt{x-1}}{2\sqrt{x+1}}}{x+1}$

$= \frac{x+1}{2x}\cdot\frac{\dfrac{x+1-x+1}{2\sqrt{x-1}\sqrt{x+1}}}{x+1} = \frac{1}{2x}\cdot\frac{2}{2\sqrt{(x-1)(x+1)}} = \frac{1}{2x\sqrt{x^2-1}}$

(4)

$y = \tan^{-1}\left(2\tan\frac{x}{2}\right)$

$y' = \frac{\left(2\tan\dfrac{x}{2}\right)'}{1+\left(2\tan\dfrac{x}{2}\right)^2} = \frac{1}{1+4\tan^2\dfrac{x}{2}}\cdot 2\cdot\frac{\dfrac{1}{2}}{\cos^2\dfrac{x}{2}} = \frac{1}{1+4\dfrac{\sin^2(x/2)}{\cos^2(x/2)}}\cdot\frac{1}{\cos^2(x/2)}$

$= \frac{1}{\cos^2\dfrac{x}{2}+4\sin^2\dfrac{x}{2}} = \frac{1}{\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}+3\sin^2\dfrac{x}{2}} = \frac{1}{1+3\sin^2\dfrac{x}{2}}$