Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Lösung 1.2:4a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives
We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}
\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]
&= {}\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]
Zeile 9: Zeile 9:
We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule
We determine the derivative <math>\bigl(\sqrt{1-x^2}\bigr)'</math> by using the chain rule
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
\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\bigl(1-x^2\bigr)'}{1-x^2}\\[5pt]
Zeile 17: Zeile 17:
We simplify the result as far as possible, so as to make the second differentiation easier,
We simplify the result as far as possible, so as to make the second differentiation easier,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\phantom{\frac{d}{dx}\,\frac{x}{\sqrt{1-x^2}}}{}
\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]
&= {}\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]
Zeile 27: Zeile 27:
The second derivative is
The second derivative is
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d^2}{dx^2}\,\frac{x}{\sqrt{1-x^2}}
\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}\,\frac{1}{(1-x^2)^{3/2}}\\[5pt]

Version vom 12:54, 10. Mär. 2009

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

ddxx1−x2=1−x22(x)1−x2−x1−x2=1−x211−x2−x1−x2.

We determine the derivative 1−x2  by using the chain rule

=1−x21−x2−x121−x21−x2=1−x21−x2−x121−x2(−2x).

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

=1−x21−x2+x21−x2=1−x21−x21−x22+x21−x2=1−x21−x21−x2+x2=1(1−x2)32.

The second derivative is

d2dx2x1−x2=ddx1(1−x2)32=ddx1−x2−32=−231−x2−32−11−x2=−231−x2−52(−2x)=3x1−x2−52=3x1−x252.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:4a“