Lösung 1.1:2d

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 1.1:2d moved to Solution 1.1:2d: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
If we write
-
<center> [[Image:1_1_2d.gif]] </center>
+
<math>\sqrt{x}</math>
-
{{NAVCONTENT_STOP}}
+
in power form
 +
<math>x^{{1}/{2}\;}</math>, we see that the square root is a function having the appearance of
 +
<math>x^{n}</math>
 +
and its derivative is therefore equal to
 +
 
 +
 
 +
<math>{f}'\left( x \right)=\frac{d}{dx}\sqrt{x}=\frac{d}{dx}x^{{1}/{2}\;}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}</math>
 +
 
 +
The answer can also be written as
 +
 
 +
 
 +
<math>{f}'\left( x \right)=\frac{1}{2\sqrt{x}}</math>
 +
 
 +
 
 +
because
 +
<math>x^{-\frac{1}{2}}=\left( x^{\frac{1}{2}} \right)^{-1}=\left( \sqrt{x} \right)^{-1}=\frac{1}{\sqrt{x}}</math>

Version vom 11:30, 10. Okt. 2008

If we write \displaystyle \sqrt{x} in power form \displaystyle x^{{1}/{2}\;}, we see that the square root is a function having the appearance of \displaystyle x^{n} and its derivative is therefore equal to


\displaystyle {f}'\left( x \right)=\frac{d}{dx}\sqrt{x}=\frac{d}{dx}x^{{1}/{2}\;}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}

The answer can also be written as


\displaystyle {f}'\left( x \right)=\frac{1}{2\sqrt{x}}


because \displaystyle x^{-\frac{1}{2}}=\left( x^{\frac{1}{2}} \right)^{-1}=\left( \sqrt{x} \right)^{-1}=\frac{1}{\sqrt{x}}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.1:2d“