Aus Online Mathematik Brückenkurs 2

We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives

\displaystyle \begin{align} & \frac{d}{dx}\frac{x}{\sqrt{1-x^{2}}}=\frac{\left( x \right)^{\prime }\sqrt{1-x^{2}}-x\left( \sqrt{1-x^{2}} \right)^{\prime }}{\left( \sqrt{1-x^{2}} \right)^{2}} \\ & \\ & =\frac{1\centerdot \sqrt{1-x^{2}}-x\left( \sqrt{1-x^{2}} \right)^{\prime }}{1-x^{2}} \\ \end{align}

We determine the derivative \displaystyle \left( \sqrt{1-x^{2}} \right)^{\prime } by using the chain rule

\displaystyle \begin{align} & =\frac{\sqrt{1-x^{2}}-x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\centerdot \left( 1-x^{2} \right)^{\prime }}{1-x^{2}} \\ & \\ & =\frac{\sqrt{1-x^{2}}-x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\centerdot \left( -2x \right)}{1-x^{2}} \\ \end{align}

We simplify the result as far as possible, so as to make the second differentiation easier:

\displaystyle \begin{align} & =\frac{\sqrt{1-x^{2}}+\frac{x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}} \\ & \\ & =\frac{\frac{\left( \sqrt{1-x^{2}} \right)^{2}}{\sqrt{1-x^{2}}}+\frac{x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}} \\ & \\ & =\frac{\frac{1-x^{2}+x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}} \\ & \\ & =\frac{1}{\left( 1-x^{2} \right)^{{3}/{2}\;}} \\ \end{align}

The second derivative is:

\displaystyle \begin{align} & \frac{d^{^{2}}}{dx^{^{2}}}\frac{x}{\sqrt{1-x^{2}}}=\frac{d}{dx}\frac{1}{\left( 1-x^{2} \right)^{{3}/{2}\;}} \\ & \\ & =\frac{d}{dx}\left( 1-x^{2} \right)^{-{3}/{2}\;}=-\frac{3}{2}\left( 1-x^{2} \right)^{-\frac{3}{2}-1}\centerdot \left( 1-x^{2} \right)^{\prime } \\ & \\ & =-\frac{3}{2}\left( 1-x^{2} \right)^{{-5}/{2}\;}\centerdot \left( -2x \right)=3x\left( 1-x^{2} \right)^{{-5}/{2}\;} \\ & \\ & =\frac{3x}{\left( 1-x^{2} \right)^{{5}/{2}\;}} \\ \end{align}