2.5 Fördjupning till Deriveringsregler - Versionshistorik

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← Äldre version		Versionen från 2 maj 2020 kl. 20.22
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−	[[Matte:Copyrights\|Copyright]] © ~~2011-2018 Math Online Sweden~~ AB. All Rights Reserved.	+	[[Matte:Copyrights\|Copyright]] © 2020 [https://www.techpages.se <b><span style="color:blue">TechPages AB</span></b>]. All Rights Reserved.

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-[[Media: Lektion 17 Deriveringsregler I Ruta.pdf|<b><span style="color:blue">Lektion 17 Deriveringsregler I</span></b>]]
+[[Media: Lektion 18 Deriveringsregler II Ruta.pdf|<b><span style="color:blue">Lektion 18 Deriveringsregler II</span></b>]] -->
 == <b><span style="color:#931136">Bevis av deriveringsreglerna</span></b> ==
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-[[Matte:Copyrights|Copyright]] © 2011-2017 Math Online Sweden AB. All Rights Reserved.
+[[Matte:Copyrights|Copyright]] © 2011-2018 Math Online Sweden AB. All Rights Reserved.

← Äldre version		Versionen från 23 november 2017 kl. 08.43
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@@ Rad 93: / Rad 93: @@
 För funktionen <math> f(x) = -8\,x + 9 </math> blir derivatan:
-:<math> f\,'(x) = \lim_{h \to 0} \, {f(x+h) \, - \, f(x) \over h} = \lim_{h \to 0} \, {-8\, (x+h) + 9 - (-8\,x + 9) \over h} = \lim_{h \to 0} \, {-8\, x -8\, h + 9 + 8\, x - 9 \over h} = \lim_{h \to 0} \, {-8\, h \over h} = \lim_{h \to 0} \, (-8) = -8 </math>
+:<math> f\,'(x) = \lim_{h \to 0} \, {f(x+h) \, - \, f(x) \over h} = \lim_{h \to 0} \, {-8\, (x+h) + 9 - (-8\,x + 9) \over h} = \lim_{h \to 0} \, {-8\, x -8\, h + 9 + 8\, x - 9 \over h} = \lim_{h \to 0} \, {-8\,{\color{Red} {\cancel{{\color{Black} h}}}} \over {\color{Red} {\cancel{{\color{Black} h}}}}} = \lim_{h \to 0} \, (-8) = -8 </math>
 Att <math> f(x+h) = -8\, (x+h) + 9 </math> inser man när man i funktionen <math> f(x)= -8\,x + 9 </math> ersätter <math> x\, </math> med <math> x+h\, </math>.

@@ Rad 84: / Rad 84: @@
 Om vi tillämpar derivatans definition på <math> f(x) = k\cdot x + m </math> kan vi skriva:
-<math> f\,'(x) = \lim_{h \to 0} \, {f(x+h) - f(x) \over h} = \lim_{h \to 0} \, {k\cdot (x+h) + m - (k\cdot x + m) \over h} = \lim_{h \to 0} \, {k\cdot x + k\cdot h + m - k\cdot x - m \over h} = \lim_{h \to 0} \, {k\cdot {\color{Red} {\cancel{{\color{Black} h}}}} \over \cancel{h}} = \lim_{h \to 0} \, k = k </math>
+<math> f\,'(x) = \lim_{h \to 0} \, {f(x+h) - f(x) \over h} = \lim_{h \to 0} \, {k\cdot (x+h) + m - (k\cdot x + m) \over h} = \lim_{h \to 0} \, {k\cdot x + k\cdot h + m - k\cdot x - m \over h} = \lim_{h \to 0} \, {k\cdot {\color{Red} {\cancel{{\color{Black} h}}}} \over {\color{Red} {\cancel{{\color{Black} h}}}}} = \lim_{h \to 0} \, k = k </math>
 Att <math> f(x+h) = k\cdot (x+h) + m </math> inser man när man i funktionen <math> f(x)= k\cdot x + m </math> ersätter <math> x\, </math> med <math> x+h\, </math>.