Processing Math: Done

Aus Online Mathematik Brückenkurs 2

Version vom 14:26, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 1.2:3c moved to Solution 1.2:3c: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 11:57, 15. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	We can write the expression as
-	<center> [[Image:1_2_3c-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+	<math>\frac{1}{x\sqrt{1-x^{2}}}=\left( x\sqrt{1-x^{2}} \right)^{-1}</math>,
-	<center> [[Image:1_2_3c-2(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	and then we see that we have "something raised to
		+	<math>-\text{1}</math>", which can be differentiated one step by using the chain rule:
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}\left( \left\{ \left. x\sqrt{1-x^{2}} \right\} \right. \right)^{-1}=-1\centerdot \left( \left\{ \left. x\sqrt{1-x^{2}} \right\} \right. \right)\centerdot \left( \left\{ \left. x\sqrt{1-x^{2}} \right\} \right. \right)^{\prime } \\
		+	& =-\frac{1}{\left( x\sqrt{1-x^{2}} \right)^{2}}\centerdot \left( x\sqrt{1-x^{2}} \right)^{\prime } \\
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( x\sqrt{1-x^{2}} \right)^{\prime } \\
		+	\end{align}</math>
		+
		+
		+	The next step is to differentiate the product
		+	<math>x\centerdot \sqrt{1-x^{2}}</math>
		+	using the product rule:
		+
		+
		+	<math>\begin{align}
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( \left( x \right)^{\prime }\sqrt{1-x^{2}}+x\left( \sqrt{1-x^{2}} \right)^{\prime } \right) \\
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( 1\sqrt{1-x^{2}}+x\left( \sqrt{1-x^{2}} \right)^{\prime } \right) \\
		+	\end{align}</math>
		+
		+
		+	The expression
		+	<math>\sqrt{1-x^{2}}</math>
		+	is of the type "root of something", so we use the chain rule to differentiate,
		+
		+
		+	<math>\begin{align}
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}+x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\left( 1-x^{2} \right)^{\prime } \right) \\
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}+x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\centerdot \left( -2x \right) \right) \\
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}-\frac{x^{2}}{\sqrt{1-x^{2}}} \right) \\
		+	\end{align}</math>
		+
		+
		+	We write the expression on the right over a common denominator:
		+
		+	<math>\begin{align}
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( \frac{\left( \sqrt{1-x^{2}} \right)^{2}-x^{2}}{\sqrt{1-x^{2}}} \right) \\
		+	& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( \frac{1-x^{2}-x^{2}}{\sqrt{1-x^{2}}} \right) \\
		+	& =-\frac{1-2x^{2}}{x^{2}\left( 1-x^{2} \right)^{{3}/{2}\;}} \\
		+	\end{align}</math>
		+
		+
		+	NOTE: When we make simplifications of the form
		+	<math>\left( \sqrt{1-x^{2}} \right)^{2}=1-x^{2}</math>, we assume that both sides are well defined (i.e. in this case that
		+	<math>x</math>
		+	lies between
		+	<math>-\text{1}</math>
		+	and
		+	<math>\text{1}</math>
		+	).

---

Version vom 11:57, 15. Okt. 2008

We can write the expression as

1x1−x2=x1−x2−1 ,

and then we see that we have "something raised to −1", which can be differentiated one step by using the chain rule:

ddxx1−x2−1=−1x1−x2x1−x2=−1x1−x22x1−x2=−1x21−x2x1−x2

The next step is to differentiate the product x1−x2  using the product rule:

=−1x21−x2x1−x2+x1−x2=−1x21−x211−x2+x1−x2

The expression 1−x2  is of the type "root of something", so we use the chain rule to differentiate,

=−1x21−x21−x2+x121−x21−x2=−1x21−x21−x2+x121−x2−2x=−1x21−x21−x2−x21−x2

We write the expression on the right over a common denominator:

=−1x21−x21−x21−x22−x2=−1x21−x21−x21−x2−x2=−1−2x2x21−x232

NOTE: When we make simplifications of the form 1−x22=1−x2 , we assume that both sides are well defined (i.e. in this case that x lies between −1 and 1 ).