2.3a Lösning 10 - Versionshistorik

Taifun den 28 september 2014 kl. 19.59

2014-09-28T19:59:50Z

2014-09-28T19:58:40Z

2014-09-04T09:50:37Z

2014-09-04T09:48:07Z

2014-09-04T09:47:17Z

2014-09-04T09:46:40Z

2014-09-04T09:45:15Z

2014-09-04T09:44:29Z

2014-09-04T09:43:50Z

2014-09-04T09:40:03Z

@@ Rad 1: / Rad 1: @@
-:<math> \begin{array}{rcl} f(x+h) - f(x) = {1 \over x+h} - {1 \over x} = {x \over x\,(x+h)} - {x+h \over x\,(x+h)} & = & {x - (x+h) \over x\,(x+h)} = {x - x - h \over x\,(x+h)} = \\
+:<math> \begin{array}{rcl} f(x+h) - f(x) & = & {1 \over x+h} - {1 \over x} = {x \over x\,(x+h)} - {x+h \over x\,(x+h)} = {x - (x+h) \over x\,(x+h)} = {x - x - h \over x\,(x+h)} = \\
-                                                                                                                   & = & {- h \over x\,(x+h)}
+                  \\
 \end{array}</math>

← Äldre version		Versionen från 28 september 2014 kl. 19.58
Rad 1:		Rad 1:
−	:<math> f(x+h) - f(x) = {1 \over x+h} - {1 \over x} = {x \over x\,(x+h)} - {x+h \over x\,(x+h)} = {x - (x+h) \over x\,(x+h)} = {x - x - h \over x\,(x+h)} = {- h \over x\,(x+h)} </math>	+	:<math> \begin{array}{rcl} f(x+h) - f(x) = {1 \over x+h} - {1 \over x} = {x \over x\,(x+h)} - {x+h \over x\,(x+h)} & = & {x - (x+h) \over x\,(x+h)} = {x - x - h \over x\,(x+h)} = \\
		+	& = & {- h \over x\,(x+h)}
		+	\end{array}</math>

← Äldre version		Versionen från 4 september 2014 kl. 09.50
Rad 2:		Rad 2:


−	:<math> {f(x+h) - f(x) \over h} = {- 1 \over x\,(x+h)} </math>	+	:<math> {f(x+h) - f(x) \over h} = {- h/h \over x\,(x+h)}= {- 1 \over x\,(x+h)} </math>


	:<math> \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} \; {- 1 \over x\,(x+h)} = {- 1 \over x\,(x+0)} = - \, {1 \over x^2} </math>		:<math> \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} \; {- 1 \over x\,(x+h)} = {- 1 \over x\,(x+0)} = - \, {1 \over x^2} </math>

← Äldre version		Versionen från 4 september 2014 kl. 09.48
Rad 5:		Rad 5:


−	:<math> \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} \; {- 1 \over x\,(x+0)} = - \, {1 \over x^2} </math>	+	:<math> \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} \; {- 1 \over x\,(x+h)} = {- 1 \over x\,(x+0)} = - \, {1 \over x^2} </math>

← Äldre version		Versionen från 4 september 2014 kl. 09.47
Rad 5:		Rad 5:


−	:<math> \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} \; {- 1 \over x\,(x+~~h)} = {- 1 \over x^2 + x \cdot~~ 0} = - \, {1 \over x^2} </math>	+	:<math> \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} \; {- 1 \over x\,(x+0)} = - \, {1 \over x^2} </math>

@@ Rad 5: / Rad 5: @@
-:<math> \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} {{- h \over x\,(x+h)} \over h} = \lim_{h \to 0} {{- h \over x\,(x+h)} \over {h \over 1}} = \lim_{h \to 0} \; {- h\cdot 1 \over h\cdot x\,(x+h)} =  </math>
+:<math> \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} \; {- 1 \over x\,(x+h)} = \lim_{h \to 0} \; {- 1 \over x^2 + x\,h} = - \, {1 \over x^2} </math>
-−
-−
-::<math> = \lim_{h \to 0} \; {- 1 \over x\,(x+h)} = \lim_{h \to 0} \; {- 1 \over x^2 + x\,h} = - \, {1 \over x^2} </math>

@@ Rad 1: / Rad 1: @@
-<math> f(x+h) - f(x) = {1 \over x+h} - {1 \over x} = {x \over x\,(x+h)} - {x+h \over x\,(x+h)} = {x - (x+h) \over x\,(x+h)} = {x - x - h \over x\,(x+h)} = {- h \over x\,(x+h)} </math>
+:<math> f(x+h) - f(x) = {1 \over x+h} - {1 \over x} = {x \over x\,(x+h)} - {x+h \over x\,(x+h)} = {x - (x+h) \over x\,(x+h)} = {x - x - h \over x\,(x+h)} = {- h \over x\,(x+h)} </math>
-<math> f\,'(x) = \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} {{- h \over x\,(x+h)} \over h} = \lim_{h \to 0} {{- h \over x\,(x+h)} \over {h \over 1}} = \lim_{h \to 0} \; {- h\cdot 1 \over h\cdot x\,(x+h)} =  </math>
+:<math> f\,'(x) = \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} {{- h \over x\,(x+h)} \over h} = \lim_{h \to 0} {{- h \over x\,(x+h)} \over {h \over 1}} = \lim_{h \to 0} \; {- h\cdot 1 \over h\cdot x\,(x+h)} =  </math>
 ::<math> = \lim_{h \to 0} \; {- 1 \over x\,(x+h)} = \lim_{h \to 0} \; {- 1 \over x^2 + x\,h} = - \, {1 \over x^2} </math>