Skillnad mellan versioner av "2.3a Lösning 10"
Från Mathonline
Taifun (Diskussion | bidrag) (Skapade sidan med '<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)}...') |
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| − | <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> | ::<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> | ||
Versionen från 4 september 2014 kl. 10.40
\[ f(x+h) - f(x) = {1 \over x+h} - {1 \over x} = {x \over x\,(x+h)} - {x+h \over x\,(x+h)} = {x - (x+h) \over x\,(x+h)} = {x - x - h \over x\,(x+h)} = {- h \over x\,(x+h)} \]
\[ f\,'(x) = \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} {{- h \over x\,(x+h)} \over h} = \lim_{h \to 0} {{- h \over x\,(x+h)} \over {h \over 1}} = \lim_{h \to 0} \; {- h\cdot 1 \over h\cdot x\,(x+h)} = \]
- \[ = \lim_{h \to 0} \; {- 1 \over x\,(x+h)} = \lim_{h \to 0} \; {- 1 \over x^2 + x\,h} = - \, {1 \over x^2} \]
