《高等数学》习题2.3选做
Elegia
2021-06-15 09:13:57
1. 当 $\displaystyle x\rightarrow 0$ 时，下列各函数是 $\displaystyle x$ 的几阶无穷小量？
(1) $\displaystyle y=x+10x^{2} +100x^{3}$
解：$\displaystyle \lim _{x\rightarrow 0}\frac{y}{x} =\lim _{x\rightarrow 0}\left( 1+10x+100x^{2}\right) =1$，是 $\displaystyle 1$ 阶无穷小量。
(2) $\displaystyle y=\left(\sqrt{x+2} -\sqrt{2}\right)\sin x$
解：$\displaystyle \sqrt{x+2} -\sqrt{2} =\frac{x}{\sqrt{x+2} +\sqrt{2}} =x+o( x)$，而 $\displaystyle \sin x=x+o( x)$，因此 $\displaystyle y=x^{2} +o\left( x^{2}\right)$ 是二阶无穷小量。
(3) $\displaystyle y=x( 1-\cos x)$
解：$\displaystyle 1-\cos x=\frac{x^{2}}{2} +o\left( x^{2}\right)$，因此 $\displaystyle y=\frac{x^{3}}{2} +o\left( x^{3}\right)$ 是三阶无穷小量。
2. 已知 $\displaystyle x\rightarrow 0$ 时，$\displaystyle \alpha ( x) =o\left( x^{2}\right)$，证明 $\displaystyle \alpha ( x) =o( x)$。
解：题设即 $\displaystyle \lim _{x\rightarrow 0}\frac{\alpha ( x)}{x^{2}} =0$，那么 $\displaystyle \lim _{x\rightarrow 0}\frac{\alpha ( x)}{x} =\lim _{x\rightarrow 0}\left(\frac{\alpha ( x)}{x^{2}} \cdot x\right) =0\cdot 0=0$。也即 $\displaystyle \alpha ( x) =o( x)$
3. 设 $\displaystyle x\rightarrow 0$ 时 $\displaystyle \alpha ( x) =o( x) ,\beta ( x) =o( x)$，证明 $\displaystyle \alpha ( x) +\beta ( x) =o( x)$
解：$\displaystyle \lim _{x\rightarrow 0}\frac{\alpha ( x) +\beta ( x)}{x} =\lim _{x\rightarrow 0}\left(\frac{\alpha ( x)}{x} +\frac{\beta ( x)}{x}\right) =0$。也即 $\displaystyle \alpha ( x) +\beta ( x) =o( x)$。
4. 计算下列函数在指定点 $\displaystyle x_{0}$ 处的微分
(1) $\displaystyle x\sin x,x_{0} =\pi /4$
解：$\displaystyle \mathrm{d}( x\sin x) =\sin x\mathrm{d} x+x\cos x\mathrm{d} x=(\sin x+x\cos x)\mathrm{d} x$，带入为 $\displaystyle \frac{\sqrt{2}}{2}( 1+\pi /4)\mathrm{d} x$。
(2) $\displaystyle ( 1+x)^{a} ,x_{0} =0\quad ( a >0)$
解：$\displaystyle \mathrm{d}\left[( 1+x)^{a}\right] =a( 1+x)^{a-1}\mathrm{d} x$，带入有 $\displaystyle a\mathrm{d} x$。
7. 试计算 $\displaystyle \sqrt[5]{32.16}$ 的近似值
解：$\displaystyle \sqrt[5]{32.16} =2\cdot \sqrt[5]{1+\frac{1}{200}} \approx 2\cdot \left( 1+\frac{1}{5} \cdot \frac{1}{200}\right) =2.002$。
8. 求下列方程确定的隐函数的导函数：
(1) $\displaystyle x^{2/3} +y^{2/3} =a^{2/3} \quad ( a >0)$
解：
$$
\begin{aligned}
\mathrm{d}\left( x^{2/3}\right) & =-\mathrm{d}\left( y^{2/3}\right)\\
\frac{2}{3} x^{-1/3}\mathrm{d} x & =-\frac{2}{3} y^{-1/3}\mathrm{d} y\\
\frac{\mathrm{d} y}{\mathrm{d} x} & =-\frac{y^{1/3}}{x^{1/3}}
\end{aligned}
$$
(2) $\displaystyle ( x-a)^{2} +( y-b)^{2} =c^{2}$
解：
$$
\begin{aligned}
2( x-a)\mathrm{d} x & =-2( y-b)\mathrm{d} y\\
\frac{\mathrm{d} y}{\mathrm{d} x} & =-\frac{x-a}{y-b}
\end{aligned}
$$
(3) $\displaystyle \arctan\frac{y}{x} =\ln\sqrt{x^{2} +y^{2}}$
解：
$$
\begin{aligned}
\arctan\frac{y}{x} & =\frac{1}{2}\ln\left( x^{2} +y^{2}\right)\\
\frac{1}{1+( y/x)^{2}} \cdot \frac{x\mathrm{d} y-y\mathrm{d} x}{x^{2}} & =\frac{2x\mathrm{d} x+2y\mathrm{d} y}{2\left( x^{2} +y^{2}\right)}\\
x\mathrm{d} y-y\mathrm{d} x & =x\mathrm{d} x+y\mathrm{d} y\\
( x-y)\mathrm{d} y & =( x+y)\mathrm{d} x\\
\frac{\mathrm{d} y}{\mathrm{d} x} & =\frac{x+y}{x-y}
\end{aligned}
$$
(4) $\displaystyle y\sin x-\cos( x-y) =0$
解：
$$
\begin{aligned}
\sin x\mathrm{d} y+y\cos x\mathrm{d} x+\sin( x-y) \cdot (\mathrm{d} x-\mathrm{d} y) & =0\\
(\sin x-\sin( x-y))\mathrm{d} y & =-( y\cos x+\sin( x-y))\mathrm{d} x\\
\frac{\mathrm{d} y}{\mathrm{d} x} & =\frac{\sin( x-y) +y\cos x}{\sin( x-y) -\sin x}
\end{aligned}
$$
9. 求下列隐函数在指定点 $\displaystyle M$ 的导数：
(1) $\displaystyle y^{2} -2xy-x^{2} +2x-4=0,M( 3,7)$
解：
$$
\begin{aligned}
2y\mathrm{d} y-2y\mathrm{d} x-2x\mathrm{d} y-2x\mathrm{d} x+2\mathrm{d} x & =0\\
( y-x)\mathrm{d} y & =( x+y-1)\mathrm{d} x\\
\frac{\mathrm{d} y}{\mathrm{d} x} & =\frac{x+y-1}{y-x}
\end{aligned}
$$
在 $\displaystyle M$ 点处的导数为 $\displaystyle \frac{3+7-1}{4} =\frac{9}{4}$。
(2) $\displaystyle e^{xy} -5x^{2} y=0,M\left(\frac{e^{2}}{10} ,\frac{20}{e^{2}}\right)$
解：
$$
\begin{aligned}
e^{xy}( y\mathrm{d} x+x\mathrm{d} y) -5x^{2}\mathrm{d} y-10xy\mathrm{d} x & =0\\
\left( e^{xy} x-5x^{2}\right)\mathrm{d} y & =\left( 10xy-e^{xy} y\right)\mathrm{d} x\\
\frac{\mathrm{d} y}{\mathrm{d} x} & =\frac{10xy-e^{xy} y}{e^{xy} x-5x^{2}}
\end{aligned}
$$
由 $\displaystyle xy=2$，带入有 $\displaystyle =\frac{20-e^{2} y}{e^{2} x-5x^{2}} =\frac{20-20}{e^{4} /10-e^{4} /20} =0$
10. 设 $\displaystyle y=f( x)$ 由下列参数方程给出，求 $\displaystyle y'=\frac{\mathrm{d} y}{\mathrm{d} x}$：
(1) $\displaystyle \begin{cases}
x=2t-t^{2}\\
y=3t-t^{3}
\end{cases}$
解：$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{3-3t^{2}}{2-2t} =\frac{3( 1+t)}{2}$
(2) $\displaystyle \begin{cases}
x=t\ln t\\
y=e^{t}
\end{cases}$
解：$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{e^{t}}{1+\ln t}$
(3) $\displaystyle \begin{cases}
x=\arccos\frac{1}{\sqrt{1+t^{2}}}\\
y=\arcsin\frac{t}{\sqrt{1+t^{2}}}
\end{cases}$
解：
$$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{d} x} & =\frac{\frac{1}{\sqrt{1-\frac{t^{2}}{1+t^{2}}}} \cdot \left[\frac{1}{\sqrt{1+t^{2}}} -t^{2}\left( 1+t^{2}\right)^{-3/2}\right]}{-\frac{1}{\sqrt{1-\frac{1}{1+t^{2}}}} \cdot \left[ -t\left( 1+t^{2}\right)^{-3/2}\right]}\\
& =\frac{\sqrt{t^{2}}}{1}\frac{\left[ 1-\frac{t^{2}}{1+t^{2}}\right]}{\frac{t}{1+t^{2}}}\\
& =\frac{|t|}{t}\\
& =\operatorname{sgn} t\quad ( t\neq 0)
\end{aligned}
$$