问题及解答

求 $y=\arcsin x$ 的高阶导数.

验证其满足等式 $xy'=(1-x^2)y''$, 并求 $y^{(n)}(0)$.

References:

梅加强, 《数学分析》 P.129, Exer 4.2 (10)

对于 $x\in(-1,1)$, 

\[
\begin{aligned}
y'&=\frac{1}{\sqrt{1-x^2}}\\
y''&=-\frac{1}{1-x^2}\cdot\frac{-2x}{2\sqrt{1-x^2}}=\frac{x}{(1-x^2)^{3/2}}=x\cdot(\frac{1}{\sqrt{1-x^2}})^3=x\cdot(y')^3
\end{aligned}
\]

记 $t=y'$, 则 $t'=y''=xt^3$. 于是有

\[
\begin{aligned}
y'&=t\\
y''&=xt^3\\
y'''&=(xt^3)'=1\cdot t^3+x\cdot 3t^2\cdot t'\\
&=t^3+3xt^2\cdot xt^3\\
&=t^3+3x^2 t^5
\end{aligned}
\]

\[
\begin{aligned}
y^{(4)}&=(t^3+3x^2 t^5)'_x=3t^2\cdot t'+3(2x\cdot t^5+x^2\cdot 5t^4\cdot t')\\
&=3t^2\cdot xt^3+6x\cdot t^5+15x^2 t^4\cdot xt^3\\
&=9xt^5+15x^3 t^7
\end{aligned}
\]

\[
\begin{aligned}
y^{(5)}&=(9xt^5+15x^3 t^7)'_x\\
&=9(1\cdot t^5+x\cdot 5t^4\cdot t')+15(3x^2\cdot t^7+x^3\cdot 7t^6\cdot t')\\
&=9t^5+45xt^4\cdot xt^3+45x^2 t^7+105x^3\cdot t^6\cdot xt^3\\
&=9t^5+90x^2 t^7+105x^4 t^9
\end{aligned}
\]

\[
\begin{aligned}
y^{(6)}&=(9t^5+90x^2 t^7+105x^4 t^9)'_x\\
&=9\cdot 5t^4\cdot t'+90(2x\cdot t^7+x^2\cdot 7t^6\cdot t')+105(4x^3\cdot t^9+x^4\cdot 9t^8\cdot t')\\
&=45t^4\cdot xt^3+180xt^7+630x^2\cdot t^6\cdot xt^3+420x^3\cdot t^9+945x^4\cdot t^8\cdot xt^3\\
&=225xt^7+1050x^3 t^9+945x^5 t^{11}
\end{aligned}
\]

\[
\begin{aligned}
y^{(7)}&=(225xt^7+1050x^3 t^9+945x^5 t^{11})'_x\\
&=225(1\cdot t^7+x\cdot 7t^6\cdot t')+1050(3x^2\cdot t^9+x^3\cdot 9t^8\cdot t')+945(5x^4\cdot t^{11}+x^5\cdot 11t^{10}\cdot t')\\
&=225t^7+1575xt^6\cdot xt^3+3150x^2 t^9+9450x^3 t^8\cdot xt^3+4725x^4\cdot t^{11}+10395x^5\cdot t^{10}\cdot xt^3\\
&=225t^7+4725x^2 t^9+14175x^4 t^{11}+10395x^6 t^{13}
\end{aligned}
\]

将以上关于 $y$ 的前七阶导数的表达式系数整理为下表

这里系数之间的关系可以由下图解释

对于 $x\in(-1,1)$, 

\[
\begin{aligned}
y'&=\frac{1}{\sqrt{1-x^2}}\\
y''&=-\frac{1}{1-x^2}\cdot\frac{-2x}{2\sqrt{1-x^2}}=\frac{x}{(1-x^2)^{3/2}}=x\cdot(\frac{1}{\sqrt{1-x^2}})^3=x\cdot(y')^3
\end{aligned}
\]

\[
(1-x^2)y''=(1-x^2)\cdot\frac{x}{(1-x^2)^{3/2}}=\frac{x}{\sqrt{1-x^2}}=xy'
\]