Skillnad mellan versioner av "2.4 Lösning 4c"
Från Mathonline
Taifun (Diskussion | bidrag) m (Created page with "<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 \ov...") |
Taifun (Diskussion | bidrag) m |
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| − | <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} = </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> |
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| + | <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} = </math> | ||
Versionen från 12 maj 2011 kl. 18.43
\( y = {2 \over 3}\,x\,\sqrt{x} - {1 \over x^2} = {2 \over 3}\,x\cdot x^{1 \over 2} - {1 \over x^2} = {2 \over 3}\, x^{1+{1 \over 2}} - {1 \over x^2} = {2 \over 3}\, x^{3 \over 2} - {1 \over x^2} = {2 \over 3}\, x^{3 \over 2} - x^{-2} \)
\( y\,' = {3 \over 2}\cdot {2 \over 3}\, x^{{3 \over 2}-1} - (-2)\cdot x^{-2-1} = x^{1 \over 2} + 2\cdot x^{-3} = \)
