Skillnad mellan versioner av "2.4 Lösning 4c"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 1: | Rad 1: | ||
| − | <math> y = {2 \over 3}\,x\,\sqrt{x} - {1 \over x^2} = {2 \over 3}\,x\cdot x^{1 \over 2} - {1 \over x^2} = {2 \over 3}\, x^{1+{1 \over 2}} - {1 \over x^2} = {2 \over 3}\, x^{3 \over 2} - {1 \over x^2} = {2 \over 3}\, x^{3 \over 2} - x^{-2} </math> | + | :<math> y = {2 \over 3}\,x\,\sqrt{x} - {1 \over x^2} = {2 \over 3}\,x\cdot x^{1 \over 2} - {1 \over x^2} = {2 \over 3}\, x^{1+{1 \over 2}} - {1 \over x^2} = {2 \over 3}\, x^{3 \over 2} - {1 \over x^2} = {2 \over 3}\, x^{3 \over 2} - x^{-2} </math> |
| − | <math> y\,' = {3 \over 2}\cdot {2 \over 3}\, x^{{3 \over 2}-1} - (-2)\cdot x^{-2-1} = x^{1 \over 2} + 2\cdot x^{-3} = \sqrt{x} + 2\cdot {1 \over x^3} = \sqrt{x} + {2 \over x^3}</math> | + | :<math> y\,' = {3 \over 2}\cdot {2 \over 3}\, x^{{3 \over 2}-1} - (-2)\cdot x^{-2-1} = x^{1 \over 2} + 2\cdot x^{-3} = \sqrt{x} + 2\cdot {1 \over x^3} = \sqrt{x} + {2 \over x^3}</math> |
Versionen från 17 oktober 2014 kl. 14.28
\[ y = {2 \over 3}\,x\,\sqrt{x} - {1 \over x^2} = {2 \over 3}\,x\cdot x^{1 \over 2} - {1 \over x^2} = {2 \over 3}\, x^{1+{1 \over 2}} - {1 \over x^2} = {2 \over 3}\, x^{3 \over 2} - {1 \over x^2} = {2 \over 3}\, x^{3 \over 2} - x^{-2} \]
\[ y\,' = {3 \over 2}\cdot {2 \over 3}\, x^{{3 \over 2}-1} - (-2)\cdot x^{-2-1} = x^{1 \over 2} + 2\cdot x^{-3} = \sqrt{x} + 2\cdot {1 \over x^3} = \sqrt{x} + {2 \over x^3}\]
