Skillnad mellan versioner av "2.3a Lösning 10"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| (8 mellanliggande versioner av samma användare visas inte) | |||
| 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> | ||
| − | :<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> | |
Nuvarande version från 28 september 2014 kl. 20.59
\[ \begin{array}{rcl} f(x+h) - f(x) & = & {1 \over x+h} - {1 \over x} = {x \over x\,(x+h)} - {x+h \over x\,(x+h)} = {x - (x+h) \over x\,(x+h)} = {x - x - h \over x\,(x+h)} = \\ \\ & = & {- h \over x\,(x+h)} \end{array}\]
\[ {f(x+h) - f(x) \over h} = {- h/h \over x\,(x+h)}= {- 1 \over x\,(x+h)} \]
\[ \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} \; {- 1 \over x\,(x+h)} = {- 1 \over x\,(x+0)} = - \, {1 \over x^2} \]
