3.5 教材习题

解

(1)

$y^{\prime }=-\frac{9}{2}{x}^2\sqrt{1-x^3}$。

(2)

$y^{\prime }=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}(1+\frac{1}{2\sqrt{x+\sqrt{x}}}(1+\frac{1}{2\sqrt{x}}))$。

(3)

$y^{\prime }=\frac{1}{1+x^2}$。

解 利用隐函数求导法则可得$$\begin{array}{}\text{(1)}& y+x{y}^{\prime }+\frac{y^{\prime }}{y}=0\Rightarrow y^{\prime }=-\frac{y^2}{xy+1}\Rightarrow y^{\prime }{|}_{x=1}=-\frac{1}{2}\end{array}$$ 因此切线方程为$$\begin{array}{}\text{(2)}& y=-\frac{1}{2}(x-1)+1=-\frac{1}{2}x+\frac{3}{2}\end{array}$$ $◻$

解 (1) 利用链式法则可得$$\begin{array}{}\text{(3)}& \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y^{\prime }(t)}{x^{\prime }(t)}=\frac{3t^2}{1-2t^3}\end{array}$$ (2) 视极坐标方程为关于$\theta$的参数方程$x=\rho \cos \theta ,y=\rho \sin \theta$，利用链式法则可得$$\begin{array}{}\text{(4)}& \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y^{\prime }(\theta )}{x^{\prime }(\theta )}=\frac{(a{\mathrm{e}}^{m\theta }\cos \theta {)}^{\prime }}{(a{\mathrm{e}}^{m\theta }\sin \theta {)}^{\prime }}=\frac{m-\tan \theta }{1+m\tan \theta }\end{array}$$ $◻$

解 (1) 利用Leibniz公式可得$$\begin{array}{}\text{(5)}& \begin{array}{rl}{y}^{(n)}& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k}){\left(\frac{1}{x}\right)}^{(n-k)}(\ln x{)}^{(k)}\\ & =\frac{(-1{)}^{n}n!}{x^{n+1}}\ln x+\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{(-1{)}^{n-k}(n-k)!}{x^{n-k+1}}\frac{(-1{)}^{k-1}(k-1)!}{x^k}\\ & =\frac{(-1{)}^{n-1}n!}{x^{n+1}}(-\ln x+\sum _{k=1}^n\frac{(n-k)!(k-1)!}{(n-k)!k!})=\frac{(-1{)}^{n-1}n!}{x^{n+1}}(-\ln x+\sum _{k=1}^n\frac{1}{k})\end{array}\end{array}$$ 故$$\begin{array}{}\text{(6)}& y^{(5)}=\frac{274-120\ln x}{x^6}\end{array}$$

(2) 因式分解可得$$\begin{array}{}\text{(7)}& y^{(n)}=\frac{1}{3}{(\frac{1}{x+2}-\frac{1}{x-1})}^{(n)}=\frac{(-1{)}^{n}n!}{3}[\frac{1}{(x+2{)}^{n+1}}-\frac{1}{(x-1{)}^{n+1}}]\end{array}$$ 故$$\begin{array}{}\text{(8)}& y^{(20)}=\frac{20!}{3}[\frac{1}{(x+2{)}^{20}}-\frac{1}{(x-1{)}^{20}}]\end{array}$$

(3) 利用Leibniz公式可得$$\begin{array}{}\text{(9)}& \begin{array}{rl}{y}^{(n)}& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(x^3{)}^{(k)}({\mathrm{e}}^x{)}^{(n-k)}=\sum _{k=0}^{max\{n,3\}}(\genfrac{}{}{0ex}{}{n}{k})\frac{3!}{(3-k)!}{x}^{3-k}{\mathrm{e}}^x\\ & ={\mathrm{e}}^x[x^3+3n{x}^2+3n(n-1)x+n(n-1)(n-2)]\end{array}\end{array}$$ $◻$

解 根据链式法则可得$$\begin{array}{}\text{(10)}& \begin{array}{rl}\frac{\mathrm{d}y}{\mathrm{d}x}& =\frac{y^{\prime }(t)}{x^{\prime }(t)}=\frac{\sin t}{1-\cos t}\\ \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =\frac{1}{x^{\prime }(t)}\frac{\mathrm{d}}{\mathrm{d}t}\frac{\sin t}{1-\cos t}=-\frac{1}{a(1-\cos t{)}^2}\\ \frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}& =\frac{1}{x^{\prime }(t)}\frac{\mathrm{d}}{\mathrm{d}t}(-\frac{1}{a(1-\cos t{)}^2})=\frac{2\sin t}{a^2(1-\cos t{)}^4}\end{array}\end{array}$$ $◻$