同济高数 第2章 第2-2-8题

[AI解答]

[AI解答]

以下为各小题的求导过程：

(1) $y=\left(\arcsin \frac{x}{2}\right)^{2}$
令 $u=\arcsin\frac{x}{2}$，则 $y=u^{2}$。
$$
\frac{dy}{dx}=2u\cdot\frac{du}{dx}=2\arcsin\frac{x}{2}\cdot\frac{1}{\sqrt{1-\left(\frac{x}{2}\right)^{2}}}\cdot\frac{1}{2}
=\frac{\arcsin\frac{x}{2}}{\sqrt{1-\frac{x^{2}}{4}}}
=\frac{2\arcsin\frac{x}{2}}{\sqrt{4-x^{2}}}.
$$

(2) $y=\ln \tan \frac{x}{2}$
$$
y'=\frac{1}{\tan\frac{x}{2}}\cdot\sec^{2}\frac{x}{2}\cdot\frac{1}{2}
=\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}\cdot\frac{1}{\cos^{2}\frac{x}{2}}\cdot\frac{1}{2}
=\frac{1}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{1}{\sin x}=\csc x.
$$

(3) $y=\sqrt{1+\ln^{2}x}$
$$
y'=\frac{1}{2\sqrt{1+\ln^{2}x}}\cdot 2\ln x\cdot\frac{1}{x}
=\frac{\ln x}{x\sqrt{1+\ln^{2}x}}.
$$

(4) $y=\mathrm{e}^{\arctan\sqrt{x}}$
$$
y'=\mathrm{e}^{\arctan\sqrt{x}}\cdot\frac{1}{1+(\sqrt{x})^{2}}\cdot\frac{1}{2\sqrt{x}}
=\frac{\mathrm{e}^{\arctan\sqrt{x}}}{2\sqrt{x}(1+x)}.
$$

(5) $y=\sin^{n}x\cos nx$
$$
y'=n\sin^{n-1}x\cos x\cos nx+\sin^{n}x(-\sin nx)\cdot n
=n\sin^{n-1}x\bigl(\cos x\cos nx-\sin x\sin nx\bigr)
=n\sin^{n-1}x\cos(x+nx)=n\sin^{n-1}x\cos[(n+1)x].
$$

(6) $y=\arctan\frac{x+1}{x-1}$
$$
y'=\frac{1}{1+\left(\frac{x+1}{x-1}\right)^{2}}\cdot\frac{(x-1)-(x+1)}{(x-1)^{2}}
=\frac{(x-1)^{2}}{(x-1)^{2}+(x+1)^{2}}\cdot\frac{-2}{(x-1)^{2}}
=\frac{-2}{2x^{2}+2}=-\frac{1}{x^{2}+1}.
$$

(7) $y=\frac{\arcsin x}{\arccos x}$
$$
y'=\frac{\frac{1}{\sqrt{1-x^{2}}}\arccos x-\arcsin x\left(-\frac{1}{\sqrt{1-x^{2}}}\right)}{(\arccos x)^{2}}
=\frac{\arccos x+\arcsin x}{\sqrt{1-x^{2}}(\arccos x)^{2}}
=\frac{\frac{\pi}{2}}{\sqrt{1-x^{2}}(\arccos x)^{2}}.
$$

(8) $y=\ln\ln\ln x$
$$
y'=\frac{1}{\ln\ln x}\cdot\frac{1}{\ln x}\cdot\frac{1}{x}
=\frac{1}{x\ln x\ln\ln x}.
$$

(9) $y=\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$
先化简：分子分母同乘共轭
$$
y=\frac{(\sqrt{1+x}-\sqrt{1-x})^{2}}{(1+x)-(1-x)}=\frac{2-2\sqrt{1-x^{2}}}{2x}=\frac{1-\sqrt{1-x^{2}}}{x}.
$$
求导：
$$
y'=\frac{-\frac{-x}{\sqrt{1-x^{2}}}\cdot x-(1-\sqrt{1-x^{2}})}{x^{2}}
=\frac{\frac{x^{2}}{\sqrt{1-x^{2}}}-1+\sqrt{1-x^{2}}}{x^{2}}
=\frac{x^{2}-\sqrt{1-x^{2}}+(1-x^{2})}{x^{2}\sqrt{1-x^{2}}}
=\frac{1-\sqrt{1-x^{2}}}{x^{2}\sqrt{1-x^{2}}}.
$$

(10) $y=\arcsin\sqrt{\frac{1-x}{1+x}}$
$$
y'=\frac{1}{\sqrt{1-\frac{1-x}{1+x}}}\cdot\frac{1}{2\sqrt{\frac{1-x}{1+x}}}\cdot\frac{-(1+x)-(1-x)}{(1+x)^{2}}
=\frac{1}{\sqrt{\frac{2x}{1+x}}}\cdot\frac{1}{2\sqrt{\frac{1-x}{1+x}}}\cdot\frac{-2}{(1+x)^{2}}
=\frac{-1}{(1+x)^{2}}\cdot\frac{\sqrt{1+x}}{\sqrt{2x}}\cdot\frac{\sqrt{1+x}}{2\sqrt{1-x}}
=-\frac{1}{(1+x)\sqrt{2x(1-x)}}.
$$

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