Skillnad mellan versioner av "2.5 Lösning 3f"
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Taifun (Diskussion | bidrag) m (Created page with "<math> y = {e\,^x + e\,^{-x} \over 2} = {e\,^x \over 2} + {e\,^{-x} \over 2} = {1 \over 2}\cdot e\,^x + {1 \over 2}\cdot e\,^{-x}</math> <math> y\,' = {1 \over 2}\cdot e\,^x - ...") |
Taifun (Diskussion | bidrag) m |
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| + | <math> y = y = {3\,^x + 3\,^{-x} \over 3} </math> | ||
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<math> y = {e\,^x + e\,^{-x} \over 2} = {e\,^x \over 2} + {e\,^{-x} \over 2} = {1 \over 2}\cdot e\,^x + {1 \over 2}\cdot e\,^{-x}</math> | <math> y = {e\,^x + e\,^{-x} \over 2} = {e\,^x \over 2} + {e\,^{-x} \over 2} = {1 \over 2}\cdot e\,^x + {1 \over 2}\cdot e\,^{-x}</math> | ||
<math> y\,' = {1 \over 2}\cdot e\,^x - {1 \over 2}\cdot e\,^{-x} = {e\,^x - e\,^{-x} \over 2} </math> | <math> y\,' = {1 \over 2}\cdot e\,^x - {1 \over 2}\cdot e\,^{-x} = {e\,^x - e\,^{-x} \over 2} </math> | ||
Versionen från 15 maj 2011 kl. 14.01
\( y = y = {3\,^x + 3\,^{-x} \over 3} \)
\( y = {e\,^x + e\,^{-x} \over 2} = {e\,^x \over 2} + {e\,^{-x} \over 2} = {1 \over 2}\cdot e\,^x + {1 \over 2}\cdot e\,^{-x}\)
\( y\,' = {1 \over 2}\cdot e\,^x - {1 \over 2}\cdot e\,^{-x} = {e\,^x - e\,^{-x} \over 2} \)
