二、解答题 (本题满分 20 分, 每小题 4 分)
(1) 已知 $y=\arcsin \mathrm{e}^{-\sqrt{x}}$, 求 $y^{\prime}$.
$$
y^{\prime}=\left(\arcsin e^{-\sqrt{x}}\right)_{x}^{\prime}=
$$
$$
\sqrt{1-e^{-2 \sqrt{x}}}\left(e^{-\sqrt{x}}\right)_{x}^{\prime}=
$$
$$
\frac{e^{-\sqrt{x}}}{\sqrt{1-e^{-2 \sqrt{x}}}} \cdot\left(-\frac{1}{2} \cdot \frac{1}{\sqrt{x}}\right)=
$$
$$
\frac{-e^{-\sqrt{x}}}{\sqrt[2]{x\left(1-e^{-2 \sqrt{x}}\right)}}
$$
(2) 求 $\int \frac{\mathrm{d} x}{x \ln ^{2} x}$.
由于 $\left(\frac{1}{\Delta}\right)^{\prime}$ 会产生 $\left(\frac{1}{\Delta^{2}}\right)$ 这样的形式，因此：
$$
\left(\frac{1}{\ln x}\right)^{\prime}=\frac{-\frac{1}{x}}{\ln ^{2} x}=-\frac{1}{x \ln ^{2} x} \Rightarrow
$$
$$
\int \frac{1}{x \ln ^{2} x} \mathrm{~d} x=-\frac{1}{\ln x}+C
$$
或者：
$$
(\ln x)^{\prime}=\frac{1}{x} \Rightarrow \int \frac{1}{x \ln ^{2} x} \mathrm{~d} x=\int \frac{1}{\ln ^{2} x} \mathrm{~d} (\ln x)=
$$
$$
\int \frac{1}{t^{2}} \mathrm{~d} t=\int t^{-2} \mathrm{~d} t=-t^{-1}+C \Rightarrow
$$
$$
\int \frac{1}{x \ln ^{2} x} \mathrm{~d} x=-\frac{1}{\ln x}+C
$$
(3) 求 $\lim \limits_{x \rightarrow 0}(2 \sin x+\cos x)^{\frac{1}{x}}$.
$$
x \rightarrow 0 \Rightarrow
$$
$$
(2 \sin x+\cos x)^{\frac{1}{x}} \Rightarrow 1^{\infty} \Rightarrow
$$
$$
[1+(2 \sin x+\cos x-1)]^{\frac{1}{2 \sin x+\cos x-1} \cdot \frac{1}{x} \cdot \frac{2 \sin x+\cos x-1}{1}}
$$
$$
=e^{\frac{2 \sin x+\cos x-1}{x}}
$$
其中：
$$
\frac{2 \sin x+\cos x-1}{x} \Rightarrow \text { 洛必达 } \Rightarrow
$$
$$
\frac{2 \cos x-\sin x}{1}=2
$$
或者：
$$
\frac{2 \sin x+\cos x-1}{x}= \text { 拆分 } \Rightarrow
$$
$$
\frac{2 \sin x}{x}+\frac{\cos x-1}{x}=
$$
$$
\frac{2 x}{x}+\frac{1-1}{x}=2+0=2
$$
综上：
$$
\lim \limits_{x \rightarrow 0}(2 \sin x+\cos x)^{\frac{1}{x}}=e^{2}
$$
(4) 已知 $\left\{\begin{array}{l}x=\ln \left(1+t^{2}\right), \\ y=\arctan t,\end{array}\right.$ 求 $\frac{\mathrm{d} y}{\mathrm{~d} x}, \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}$.
$$
\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{\mathrm{~d} y}{\mathrm{~d} t} \cdot \frac{\mathrm{~d} t}{\mathrm{~d} x} \Rightarrow
$$
$$
\frac{\mathrm{~d} y}{\mathrm{~d} t}=\frac{1}{1+t^{2}} \quad \frac{\mathrm{~d} x}{\mathrm{~d} t}=\frac{2 t}{1+t^{2}} \Rightarrow
$$
$$
\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{1}{1+t^{2}} \cdot \frac{1+t^{2}}{2 t}=\frac{1}{2 t}
$$
$$
\frac{\mathrm{~d}^{2} y}{\mathrm{~d} x^{2}}=\frac{ \mathrm{~d} }{\mathrm{~d} x}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)=\frac{ \mathrm{~d} }{\mathrm{~d} t}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right) \cdot \frac{\mathrm{~d} t}{\mathrm{~d} x} \Rightarrow
$$
$$
\frac{ \mathrm{~d} }{\mathrm{~d} t}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)=\frac{ \mathrm{~d} }{\mathrm{~d} t}\left(\frac{1}{2 t}\right)=\frac{-2}{4 t^{2}} \Rightarrow
$$
$$
\frac{\mathrm{~d}^{2} y}{\mathrm{~d} x^{2}}=\frac{-2}{4 t^{2}} \cdot \frac{1+t^{2}}{2 t}=\frac{-\left(1+t^{2}\right)}{4 t^{3}}
$$
(5) 已知 $f(2)=\frac{1}{2}, f^{\prime}(2)=0$ 及 $\int_{0}^{2} f(x) \mathrm{d} x=1$, 求 $\int_{0}^{1} x^{2} f^{\prime \prime}(2 x) \mathrm{d} x$.
$$
\int_{0}^{1} x^{2} f^{\prime \prime}(2 x) \mathrm{~d} x=\frac{1}{2} \int_{0}^{1} x^{2} \cdot \mathrm{~d} \left[f^{\prime}(2 x)\right]=
$$
$$
\frac{1}{2}\left[\left.x^{2} f^{\prime}(2 x)\right|_{0} ^{1}-\int_{0}^{1} f^{\prime}(2 x) \cdot 2 x \mathrm{~d} x\right] \Rightarrow
$$
$$
\int_{0}^{1} f^{\prime}(2 x) \cdot 2 x \mathrm{~d} x=2 \int_{0}^{1} x f^{\prime}(2 x) \mathrm{~d} x=
$$
$$
\int_{0} \frac{1}{2} \cdot 2 \int_{0}^{1} x d[f(2 x)]=
$$
$$
\left.x f(2 x)\right|_{0} ^{1}-\int_{0}^{1} f(2 x) \mathrm{~d} x=
$$
$$
\frac{1}{2}-\frac{1}{2} \int_{0}^{1} f(2 x) \mathrm{~d} (2 x)=\frac{1}{2}-\frac{1}{2} \int_{0}^{2} f(x) \mathrm{~d} x=0
$$
$$
\int_{0}^{1} x^{2} f^{\prime \prime}(2 x) \mathrm{~d} x=\frac{1}{2}[0-0]=0
$$