已知$f'(x)=\frac {2x}{\sqrt {1-{x}^{2}}}$，则$\frac {df(\sqrt {1-{x}^{2})}}{dx}=$$\left(\,\,\,\,\,\right)$
$A、-2$；
$B、-\dfrac{2x}{\left|x\right|}$；
$C、-\dfrac{x}{\sqrt{1-{x}^{2}}}$；
$D、\dfrac{2}{\sqrt{1-{x}^{2}}}$.
=dfrac (x)(sqrt {1-{x)^2}}，则=dfrac (x)(sqrt {1-{x)^2}}=_________.，则=_________.
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(int )_(1)^2dfrac (dx)(x(1+sqrt {2x))}=( )A.(int )_(1)^2dfrac (dx)(x(1+sqrt {2x
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