2.4 Lösning 4b - Versionshistorik

Taifun den 22 november 2018 kl. 12.01

2018-11-22T12:01:46Z

2018-11-22T11:59:09Z

2018-11-22T11:58:51Z

2014-10-18T10:56:10Z

2014-10-17T14:26:58Z

2014-10-17T14:26:29Z

2011-05-12T18:31:23Z

2011-05-12T18:28:24Z

2011-05-12T18:24:30Z

2011-05-12T18:17:46Z

← Äldre version		Versionen från 22 november 2018 kl. 12.01
Rad 4:		Rad 4:


−	Se [[Potenser#Lagen_om_rationell_exponent_.5C.28_.5Cquad_.5Cdisplaystyle_a.5E.7Bm_.5Cover_n.7D_.5C.3B_.3D_.5C.3B_.5Csqrt.5Bn.5D.7Ba.5Em.7D_.5C.29\|Lagen om rationell exponent]].	+	Se [[Potenser#Lagen_om_rationell_exponent_.5C.28_.5Cquad_.5Cdisplaystyle_a.5E.7Bm_.5Cover_n.7D_.5C.3B_.3D_.5C.3B_.5Csqrt.5Bn.5D.7Ba.5Em.7D_.5C.29\|<b><span style="color:blue">Lagen om rationell exponent</span></b>]].

← Äldre version		Versionen från 22 november 2018 kl. 11.59
Rad 4:		Rad 4:


−	Se [[Potenser#Lagen_om_rationell_exponent_.5C.28_.5Cquad_.5Cdisplaystyle_a.5E.7Bm_.5Cover_n.7D_.5C.3B_.3D_.5C.3B_.5Csqrt.5Bn.5D.7Ba.5Em.7D_.5C.29 Lagen om rationell exponent]].	+	Se [[Potenser#Lagen_om_rationell_exponent_.5C.28_.5Cquad_.5Cdisplaystyle_a.5E.7Bm_.5Cover_n.7D_.5C.3B_.3D_.5C.3B_.5Csqrt.5Bn.5D.7Ba.5Em.7D_.5C.29\|Lagen om rationell exponent]].

← Äldre version		Versionen från 22 november 2018 kl. 11.58
Rad 2:		Rad 2:

	:<math> y\,' = {5 \over 2}\cdot {1 \over 5}\cdot x^{{5 \over 2}-1} = {1 \over 2}\cdot x^{3 \over 2} = {1 \over 2}\cdot \sqrt{x^3} = {1 \over 2}\cdot x\,\sqrt{x} = {x\,\sqrt{x}\over 2} </math>		:<math> y\,' = {5 \over 2}\cdot {1 \over 5}\cdot x^{{5 \over 2}-1} = {1 \over 2}\cdot x^{3 \over 2} = {1 \over 2}\cdot \sqrt{x^3} = {1 \over 2}\cdot x\,\sqrt{x} = {x\,\sqrt{x}\over 2} </math>
		+
		+
		+	Se [[Potenser#Lagen_om_rationell_exponent_.5C.28_.5Cquad_.5Cdisplaystyle_a.5E.7Bm_.5Cover_n.7D_.5C.3B_.3D_.5C.3B_.5Csqrt.5Bn.5D.7Ba.5Em.7D_.5C.29 Lagen om rationell exponent]].

← Äldre version		Versionen från 18 oktober 2014 kl. 10.56
Rad 1:		Rad 1:
	:<math> y = {x^2\,\sqrt{x}\over 5} = {1 \over 5}\cdot x^2\,\sqrt{x} = {1 \over 5}\cdot x^2\cdot x^{1 \over 2} = {1 \over 5}\cdot x^{2+{1 \over 2}} = {1 \over 5}\cdot x^{5 \over 2} </math>		:<math> y = {x^2\,\sqrt{x}\over 5} = {1 \over 5}\cdot x^2\,\sqrt{x} = {1 \over 5}\cdot x^2\cdot x^{1 \over 2} = {1 \over 5}\cdot x^{2+{1 \over 2}} = {1 \over 5}\cdot x^{5 \over 2} </math>
−

	:<math> y\,' = {5 \over 2}\cdot {1 \over 5}\cdot x^{{5 \over 2}-1} = {1 \over 2}\cdot x^{3 \over 2} = {1 \over 2}\cdot \sqrt{x^3} = {1 \over 2}\cdot x\,\sqrt{x} = {x\,\sqrt{x}\over 2} </math>		:<math> y\,' = {5 \over 2}\cdot {1 \over 5}\cdot x^{{5 \over 2}-1} = {1 \over 2}\cdot x^{3 \over 2} = {1 \over 2}\cdot \sqrt{x^3} = {1 \over 2}\cdot x\,\sqrt{x} = {x\,\sqrt{x}\over 2} </math>

← Äldre version		Versionen från 17 oktober 2014 kl. 14.26
Rad 1:		Rad 1:
	:<math> y = {x^2\,\sqrt{x}\over 5} = {1 \over 5}\cdot x^2\,\sqrt{x} = {1 \over 5}\cdot x^2\cdot x^{1 \over 2} = {1 \over 5}\cdot x^{2+{1 \over 2}} = {1 \over 5}\cdot x^{5 \over 2} </math>		:<math> y = {x^2\,\sqrt{x}\over 5} = {1 \over 5}\cdot x^2\,\sqrt{x} = {1 \over 5}\cdot x^2\cdot x^{1 \over 2} = {1 \over 5}\cdot x^{2+{1 \over 2}} = {1 \over 5}\cdot x^{5 \over 2} </math>
		+

	:<math> y\,' = {5 \over 2}\cdot {1 \over 5}\cdot x^{{5 \over 2}-1} = {1 \over 2}\cdot x^{3 \over 2} = {1 \over 2}\cdot \sqrt{x^3} = {1 \over 2}\cdot x\,\sqrt{x} = {x\,\sqrt{x}\over 2} </math>		:<math> y\,' = {5 \over 2}\cdot {1 \over 5}\cdot x^{{5 \over 2}-1} = {1 \over 2}\cdot x^{3 \over 2} = {1 \over 2}\cdot \sqrt{x^3} = {1 \over 2}\cdot x\,\sqrt{x} = {x\,\sqrt{x}\over 2} </math>

@@ Rad 1: / Rad 1: @@
-<math> y = {x^2\,\sqrt{x}\over 5} = {1 \over 5}\cdot x^2\,\sqrt{x} = {1 \over 5}\cdot x^2\cdot x^{1 \over 2} = {1 \over 5}\cdot x^{2+{1 \over 2}} = {1 \over 5}\cdot x^{5 \over 2} </math>
+:<math> y = {x^2\,\sqrt{x}\over 5} = {1 \over 5}\cdot x^2\,\sqrt{x} = {1 \over 5}\cdot x^2\cdot x^{1 \over 2} = {1 \over 5}\cdot x^{2+{1 \over 2}} = {1 \over 5}\cdot x^{5 \over 2} </math>
+:<math> y\,' = {5 \over 2}\cdot {1 \over 5}\cdot x^{{5 \over 2}-1} = {1 \over 2}\cdot x^{3 \over 2} = {1 \over 2}\cdot \sqrt{x^3} = {1 \over 2}\cdot x\,\sqrt{x} = {x\,\sqrt{x}\over 2} </math>
 <math> y\,' = {5 \over 2}\cdot {1 \over 5}\cdot x^{{5 \over 2}-1} = {1 \over 2}\cdot x^{3 \over 2} = {1 \over 2}\cdot \sqrt{x^3} = {1 \over 2}\cdot x\,\sqrt{x} = {x\,\sqrt{x}\over 2} </math>

@@ Rad 2: / Rad 2: @@
-<math> y\,' = {5 \over 2}\cdot {1 \over 5}\cdot x^{{5 \over 2}-1} = {1 \over 2}\cdot x^{3 \over 2}</math>
+<math> y\,' = {5 \over 2}\cdot {1 \over 5}\cdot x^{{5 \over 2}-1} = {1 \over 2}\cdot x^{3 \over 2} = {1 \over 2}\cdot \sqrt{x^3} = {1 \over 2}\cdot x\,\sqrt{x} = {x\,\sqrt{x}\over 2} </math>
-−
-−
-<math> y\,' = {x\,\sqrt{x}\over 2} </math>

← Äldre version		Versionen från 12 maj 2011 kl. 18.24
Rad 1:		Rad 1:
−	<math> y = {x^2\,\sqrt{x}\over 5} = </math>	+	<math> y = {x^2\,\sqrt{x}\over 5} = {1 \over 5}\cdot x^2\,\sqrt{x} = {1 \over 5}\cdot x^2\cdot x^{1 \over 2} = {1 \over 5}\cdot x^{2+{1 \over 2}} = {1 \over 5}\cdot x^{5 \over 2} = </math>