=dfrac (x)(sqrt {1-{x)^2}}，则=dfrac (x)(sqrt {1-{x)^2}}=_________.

步骤 1：定义函数
令$f(x)=x$，$g(x)=\sqrt {1-{x}^{2}}$。
步骤 2：求导
$f'(x)=(x)'=1$，
$g'(x)=(\sqrt {1-{x}^{2}})'=[ {(1-{x}^{2})}^{\dfrac {1}{2}}] '=\dfrac {1}{2}$ .${(1-{x}^{2})}^{\dfrac {1}{2}-1}\cdot (1-{x}^{2})'=\dfrac {1}{2}$ .${(1-{x}^{2})}^{-\dfrac {1}{2}}\cdot (2x)=\dfrac {x}{{(1-{x}^{2})}^{\dfrac {1}{2}}}=\dfrac {x}{\sqrt {1-{x}^{2}}}$。
步骤 3：应用商的导数公式
$[ \dfrac {f(x)}{g(x)}] '=\dfrac {f'(x)g(x)-f(x)g'(x)}{{[ g(x)] }^{2}}$，
代入$f(x)$，$f'(x)$，$g(x)$，$g'(x)$，得
$y'=\dfrac {1\cdot \sqrt {1-{x}^{2}}-x\cdot \dfrac {x}{\sqrt {1-{x}^{2}}}}{{(1-{x}^{2})}^{2}}=\dfrac {\sqrt {1-{x}^{2}}-\dfrac {{x}^{2}}{\sqrt {1-{x}^{2}}}}{{(1-{x}^{2})}^{2}}=\dfrac {1-{x}^{2}-{x}^{2}}{{(1-{x}^{2})}^{\dfrac {3}{2}}}=\dfrac {1-2{x}^{2}}{{(1-{x}^{2})}^{\dfrac {3}{2}}}$。