Lösung 1.2:4a
Aus Online Mathematik Brückenkurs 2
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Version vom 10:07, 11. Mär. 2009
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}} &= {}\rlap{\frac{(x)'\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{\bigl(\sqrt{1-x^2}\bigr)^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt] &= \frac{1\cdot\sqrt{1-x^2}-x\bigl(\sqrt{1-x^2}\bigr)'}{1-x^2}\,\textrm{.} \end{align} |
We determine the derivative \displaystyle \bigl(\sqrt{1-x^2}\bigr)' by using the chain rule
| \displaystyle \begin{align}
\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{} &= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt] &= \frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot (-2x)}{1-x^2}\,\textrm{.} \end{align} |
We simplify the result as far as possible, so as to make the second differentiation easier,
| \displaystyle \begin{align}
\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{} &= {}\rlap{\frac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}}\phantom{\frac{\sqrt{1-x^2}-x\cdot\dfrac{1}{2\sqrt{1-x^2}}\cdot\bigl(1-x^2\bigr)'}{1-x^2}}\\[5pt] &= \frac{\dfrac{\bigl(\sqrt{1-x^2}\bigr)^2}{\sqrt{1-x^2}}+\dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt] &= \frac{\dfrac{1-x^2+x^2}{\sqrt{1-x^2}}}{1-x^2}\\[5pt] &= \frac{1}{(1-x^2)^{3/2}}\,\textrm{.} \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}{(1-x^2)^{3/2}}\\[5pt] &= \frac{d}{dx}\,\bigl(1-x^2\bigr)^{-3/2}\\[5pt] &= -\tfrac{3}{2}\bigl(1-x^2\bigr)^{-3/2-1}\cdot\bigl(1-x^2\bigr)'\\[5pt] &= -\tfrac{3}{2}\bigl(1-x^2\bigr)^{-5/2}\cdot (-2x)\\[5pt] &= 3x\bigl(1-x^2\bigr)^{-5/2}\\[5pt] &= \frac{3x}{\bigl(1-x^2\bigr)^{5/2}}\,\textrm{.} \end{align} |
