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Lösung 1.2:2c

Aus Online Mathematik Brückenkurs 2

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When we see this expression, we should think "square root of something",
When we see this expression, we should think "square root of something",
-
{{Displayed math||<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}}}\,\textrm{,}</math>}}
+
{{Abgesetzte Formel||<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}}}\,\textrm{,}</math>}}
and in order to differentiate it, we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression <math>\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x</math>,
and in order to differentiate it, we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression <math>\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x</math>,
-
{{Displayed math||<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,</math>}}
+
{{Abgesetzte Formel||<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,</math>}}
where we have used the differentiation rule
where we have used the differentiation rule
-
{{Displayed math||<math>\frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
Thus, we obtain
Thus, we obtain
-
{{Displayed math||<math>\frac{d}{dx}\,\sqrt{\cos x} = \frac{1}{2\sqrt{\cos x}}\cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\frac{d}{dx}\,\sqrt{\cos x} = \frac{1}{2\sqrt{\cos x}}\cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}\,\textrm{.}</math>}}

Version vom 12:53, 10. Mär. 2009

When we see this expression, we should think "square root of something",

, 

and in order to differentiate it, we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression =cosx,

ddxcosx=12cosxcosx 

where we have used the differentiation rule

ddxx=ddxx12=21x12−1=21x−12=12x.

Thus, we obtain

ddxcosx=12cosx(−sinx)=−sinx2cosx.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2c“