Matematické Fórum

ako-na-to

derivace a ako pokracovat,v pripade chyby mi poradite ako na to?
$y&=(\sin x )^{\cos x}\\ [(\sin x)^{\cos x}]'&=\cos x\cdot(-\sin x)=$

jarrro · 26. 01. 2012 21:04 — Editoval jarrro (26. 01. 2012 21:06)

$y=\sin{\left(x\right)}^{\cos{\left(x\right)}}=\mathrm{e}^{\cos{\left(x\right)}\ln{\left(\sin{x}\right)}}\nl y^{\prime}=\mathrm{e}^{\cos{\left(x\right)}\ln{\left(\sin{\left(x\right)}\right)}}\left(-\sin{\left(x\right)}\ln{\left(\sin{\left(x\right)}\right)}+\cos{\left(x\right)\frac{1}{\sin{\left(x\right)}}}\cos{\left(x\right)}\right)=\nl=\sin{\left(x\right)}^{\cos{\left(x\right)}}\left(-\sin{\left(x\right)}\ln{\left(\sin{\left(x\right)}\right)}+\frac{\cos^2{\left(x\right)}}{\sin{\left(x\right)}}\right)=\sin{\left(x\right)}^{\cos{\left(x\right)}+1}\left(\mathrm{cotg}^2{\left(x\right)}-\ln{\left(\sin{\left(x\right)}\right)}\right)$

Matematické Fórum

#1 26. 01. 2012 20:49 — Editoval Pavel Brožek (26. 01. 2012 21:01)

derivace sin x

#2 26. 01. 2012 21:04 — Editoval jarrro (26. 01. 2012 21:06)

Re: derivace sin x

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