《高等数学》习题2.2选做
Elegia
2021-06-14 23:57:25
1. 下列各题的计算是否正确？若不正确，指出错误并加以改正：
(1) $\displaystyle \left(\cos\sqrt{x}\right) '=-\sin\sqrt{x}$
错误，复合时仅考虑了外层求导，实际应为 $\displaystyle \frac{-\sin\sqrt{x}}{2\sqrt{x}}$。
(2) $\displaystyle [\ln( 1-x)] '=\frac{1}{1-x}$
错误，原因同上，此外定义域不同。实际为 $\displaystyle \frac{1}{x-1} \quad ( x< 1)$
(3) $\displaystyle \left[ x^{2}\sqrt{1+x^{2}}\right] '=\left( x^{2}\right) '\left[\sqrt{1+x^{2}}\right] '=2x\frac{x}{\sqrt{1+x^{2}}}$
第一步就不对，应为
$$
\begin{aligned}
& \quad \left[ x^{2}\sqrt{1+x^{2}}\right] '\\
& =\left( x^{2}\right) '\sqrt{1+x^{2}} +x^{2}\left[\sqrt{1+x^{2}}\right] '\\
& =2x\sqrt{1+x^{2}} +x^{2}\frac{x}{\sqrt{1+x^{2}}}
\end{aligned}
$$
(4) $\displaystyle \left[\ln\left| x+2\sin^{2} x\right| \right] '=\frac{1}{x+2\sin^{2} x}( 1+4\sin x)\cos x$
应为 $\displaystyle \frac{1}{x+2\sin^{2} x}( 1+4\sin x\cos x)$
3. 求下列函数的导函数：
(1) $\displaystyle y=\frac{2}{x^{3} -1}$
解：$\displaystyle y'=\frac{-2}{\left( x^{3} -1\right)^{2}} \cdotp 3x^{2} =\frac{-6x^{2}}{\left( x^{3} -1\right)^{2}}$
(2) $\displaystyle y=\sec x$
解：$\displaystyle y'=\left(\frac{1}{\cos x}\right) '=\frac{-(\cos x) '}{\cos^{2} x} =\frac{\sin x}{\cos^{2} x} =\tan x\cdotp \sec x$。
(4) $\displaystyle y=\sin^{3} x\cdot \cos 3x$
解：$\displaystyle y'=3\sin^{2} x\cos x\cdot \cos 3x-3\sin^{3} x\sin 3x=3\sin^{2} x(\cos x\cos 3x-\sin x\sin 3x) =3\sin^{2} x\cos 4x$
(5) $\displaystyle y=\frac{1+\sin^{2} x}{\cos\left( x^{2}\right)}$
解：$\displaystyle y'=\frac{2\sin x\cdot \cos x\cdot \cos\left( x^{2}\right) +\left( 1+\sin^{2} x\right) \cdot \sin x^{2} \cdot 2x}{\cos^{2} x^{2}}$
(6) $\displaystyle y=\frac{1}{3}\tan^{3} x-\tan x+x$
解：$\displaystyle y'=\left(\tan^{2} x-1\right)\sec^{2} x+1=\tan^{2} x\sec^{2} x+1-\sec^{2} x=\tan^{2} x\sec^{2} x-\tan^{2} x=\tan^{4} x$
(8) $\displaystyle y=\cos^{5}\sqrt{1+x^{2}}$
解：$\displaystyle y'=5\cos^{4}\sqrt{1+x^{2}} \cdot \left( -\sin\sqrt{1+x^{2}}\right) \cdot \frac{x}{\sqrt{1+x^{2}}}$
(9) $\displaystyle y=\ln\left| \tan\left(\frac{x}{2} +\frac{\pi }{4}\right)\right|$
解：$\displaystyle y'=\cot\left(\frac{x}{2} +\frac{\pi }{4}\right) \cdot \sec^{2}\left(\frac{x}{2} +\frac{\pi }{4}\right) \cdot \frac{1}{2} =\frac{1}{2\cos\left(\frac{x}{2} +\frac{\pi }{4}\right)\sin\left(\frac{x}{2} +\frac{\pi }{4}\right)} =\csc\left( x+\frac{\pi }{2}\right) =\sec x$
4. 求下列函数的导函数
(1) $\displaystyle y=\arcsin\frac{x}{a}$
解：$\displaystyle y'=\frac{1}{a\sqrt{1-x^{2}}}$
(3) $\displaystyle y=x^{2}\arccos x$
解：$\displaystyle y'=-\frac{x^{2}}{\sqrt{1-x^{2}}} +2x\arccos x$
(4) $\displaystyle y=\arctan\frac{1}{x}$
解：$\displaystyle y'=\frac{1}{1+( 1/x)^{2}} \cdot \left( -x^{-2}\right) =\frac{-1}{1+x^{2}}$。
(5) $\displaystyle y=\frac{x}{2}\sqrt{a^{2} -x^{2}} +\frac{a^{2}}{2}\arcsin\frac{x}{a} \quad ( a >0)$
解：$\displaystyle y'=\frac{1}{2}\sqrt{a^{2} -x^{2}} +\frac{x}{2} \cdot \frac{-2x}{2\sqrt{a^{2} -x^{2}}} +\frac{a}{2} \cdot \frac{1}{\sqrt{1-( x/a)^{2}}} =\frac{\left( a^{2} -x^{2}\right) -x^{2} +a^{2}}{2\sqrt{a^{2} -x^{2}}} =\sqrt{a^{2} -x^{2}}$
(6) $\displaystyle y=\frac{x}{2}\sqrt{x^{2} +a^{2}} +\frac{a^{2}}{2}\ln\frac{x+\sqrt{x^{2} +a^{2}}}{a} \quad ( a >0)$
解：
$$
\begin{aligned}
y' & =\frac{1}{2}\sqrt{x^{2} +a^{2}} +\frac{x}{2} \cdot \frac{x}{\sqrt{x^{2} +a^{2}}} +\frac{a^{2}}{2} \cdot \frac{1}{x+\sqrt{x^{2} +a^{2}}} \cdot \left( 1+\frac{x}{\sqrt{x^{2} +a^{2}}}\right)\\
& =\frac{x^{2} +a^{2} +x^{2} +a^{2}}{2\sqrt{x^{2} +a^{2}}}\\
& =\sqrt{x^{2} +a^{2}}
\end{aligned}
$$
(7) $\displaystyle y=\arcsin\frac{2x}{x^{2} +1} \quad ( x\neq \pm 1)$
解：
$$
\begin{aligned}
y' & =\frac{1}{\sqrt{1-\left(\frac{2x}{x^{2} +1}\right)^{2}}} \cdot \frac{2\left( x^{2} +1\right) -4x^{2}}{\left( x^{2} +1\right)^{2}}\\
& =\frac{2\left( 1-x^{2}\right)}{\left( x^{2} +1\right)\sqrt{\left( x^{2} +1\right)^{2} -4x^{2}}}\\
& =\frac{2\left( 1-x^{2}\right)}{\left( x^{2} +1\right)\sqrt{\left( x^{2} -1\right)^{2}}}\\
& =\frac{2\left( 1-x^{2}\right)}{\left( x^{2} +1\right)\left| 1-x^{2}\right| }\\
& =\frac{2\operatorname{sgn}\left( 1-x^{2}\right)}{1+x^{2}} \quad ( x\neq \pm 1)
\end{aligned}
$$
(8) $\displaystyle y=\frac{2}{\sqrt{a^{2} -b^{2}}}\arctan\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{x}{2}\right) \quad ( a >b\geq 0)$
解：
$$
\begin{aligned}
y' & =\frac{2}{\sqrt{a^{2} -b^{2}}} \cdot \frac{1}{1+\frac{a-b}{a+b}\tan^{2}\frac{x}{2}} \cdot \sqrt{\frac{a-b}{a+b}} \cdot \frac{1}{2}\sec^{2}\frac{x}{2}\\
& =\frac{1}{\sqrt{a^{2} -b^{2}}} \cdot \frac{1}{( a+b) +( a-b)\tan^{2}\frac{x}{2}} \cdot \sqrt{\frac{\left( a^{2} -b^{2}\right)( a+b)}{a+b}} \cdot \sec^{2}\frac{x}{2}\\
& =\frac{1}{( a+b) +( a-b)\tan^{2}\frac{x}{2}} \cdot \sec^{2}\frac{x}{2}\\
& =\frac{1}{( a+b)\cos^{2}\frac{x}{2} +( a-b)\sin^{2}\frac{x}{2}}\\
& =\frac{1}{a+b\cos x}
\end{aligned}
$$
(9) $\displaystyle y=\left( 1+\sqrt{x}\right)\left( 1+\sqrt{2x}\right)\left( 1+\sqrt{3x}\right)$
解：
$$
\begin{aligned}
\frac{y'}{y} & =\frac{\left( 1+\sqrt{x}\right) '}{1+\sqrt{x}} +\frac{\left( 1+\sqrt{2x}\right) '}{1+\sqrt{2x}} +\frac{\left( 1+\sqrt{3x}\right) '}{1+\sqrt{3x}}\\
& =\frac{x^{-1/2}}{2\left( 1+\sqrt{x}\right)} +2\frac{( 2x)^{-1/2}}{2\left( 1+\sqrt{2x}\right)} +3\frac{( 3x)^{-1/2}}{2\left( 1+\sqrt{3x}\right)}\\
y' & =\left( 1+\sqrt{x}\right)\left( 1+\sqrt{2x}\right)\left( 1+\sqrt{3x}\right)\left[\frac{x^{-1/2}}{2\left( 1+\sqrt{x}\right)} +2\frac{( 2x)^{-1/2}}{2\left( 1+\sqrt{2x}\right)} +3\frac{( 3x)^{-1/2}}{2\left( 1+\sqrt{3x}\right)}\right]
\end{aligned}
$$
(13) $\displaystyle y=\ln\left( x+\sqrt{x^{2} +a^{2}}\right)$
解：
$$
\begin{aligned}
y' & =\frac{1}{x+\sqrt{x^{2} +a^{2}}} \cdot \left( 1+\frac{x}{\sqrt{x^{2} +a^{2}}}\right)\\
& =\frac{1}{x+\sqrt{x^{2} +a^{2}}} \cdot \frac{x+\sqrt{x^{2} +a^{2}}}{\sqrt{x^{2} +a^{2}}}\\
& =\frac{1}{\sqrt{x^{2} +a^{2}}}
\end{aligned}
$$
(15) $\displaystyle y=e^{x} +e^{e^{x}}$
解：$\displaystyle y'=e^{x} +e^{e^{x}} e^{x} =e^{x}\left( e^{e^{x}} +1\right)$
(16) $\displaystyle y=x^{a^{a}} +a^{x^{a}} +a^{a^{x}}$
解：$\displaystyle y'=a^{a} x^{a^{a} -1} +a^{x^{a}} \cdot \ln a\cdot ax^{a-1} +a^{a^{x}}\ln a\cdot a^{x}\ln a=a^{a} x^{a^{a} -1} +a\ln a\cdot a^{x^{a}} x^{a-1} +\ln^{2} a\cdot a^{a^{x} +x}$
5. 一雷达探测器瞄准一枚安置在发射台上的火箭，它与发射台之间的水平距离是 $\displaystyle 400\ \mathrm{m}$，设 $\displaystyle t=0$ 时垂直向上发射火箭，初速度为 $\displaystyle 0$，以 $\displaystyle 8\ \mathrm{m/s^{2}}$ 的加速度垂直向上运动，问：发射 $\displaystyle 10$ 秒时，探测器的仰角 $\displaystyle \theta ( t)$ 变化速率是多少？
设 $\displaystyle t$ 的单位为妙，那么 $\displaystyle \theta ( t) =\arctan\frac{4t^{2}}{400}$，因此有 $\displaystyle \theta '( t) =\frac{1}{1+t^{2} /100} \cdot \frac{2t}{100}$，带入得 $\displaystyle \theta '( 10) =0.1$，也就是此时变化速率是 $\displaystyle 0.1\ \mathrm{rad/s}$。