Skillnad mellan versioner av "2.3 Lösning 3b"
Från Mathonline
Taifun (Diskussion | bidrag) m (Created page with "<math> {\Delta y \over \Delta x} = {f(1+h) - f(1) \over h} = {5+10\,h+5\,h^2 - 5\cdot 1 \over h} = {5+10\,h+5\,h^2 - 5 \over h} = </math> <math> = {10\,h+5\,h^2 \over h} = {h\cd...") |
Taifun (Diskussion | bidrag) m |
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| Rad 1: | Rad 1: | ||
| − | <math> {\Delta y \over \Delta x} = {f(1+h) - f(1) \over h} = {5+10\,h+5\,h^2 - 5\cdot 1 \over h} = { | + | <math> {\Delta y \over \Delta x} = {f(1+h) - f(1) \over h} = {5+10\,h+5\,h^2 - 5\cdot 1 \over h} = {10\,h+5\,h^2 \over h} = </math> |
| − | <math> | + | <math> = {h\cdot (10 + 5\,h) \over h} = 10 + 5\,h </math> |
<math> f\,'(1) \; = \; \lim_{h \to 0} \; 10 + 5\,h \; = \; 10 </math> | <math> f\,'(1) \; = \; \lim_{h \to 0} \; 10 + 5\,h \; = \; 10 </math> | ||
Versionen från 5 maj 2011 kl. 16.16
\( {\Delta y \over \Delta x} = {f(1+h) - f(1) \over h} = {5+10\,h+5\,h^2 - 5\cdot 1 \over h} = {10\,h+5\,h^2 \over h} = \)
\( = {h\cdot (10 + 5\,h) \over h} = 10 + 5\,h \)
\( f\,'(1) \; = \; \lim_{h \to 0} \; 10 + 5\,h \; = \; 10 \)
