Question

производная
$y = \sin(2x) \cos(x)$

$y = \frac{ \sin(x) }{2 + \cos(x) }$

решить:

$\frac{2}{ \sqrt{7} + \sqrt{5} }$

$\frac{2x - 2y}{ \sqrt{13x} + \sqrt{13y} }$

$\frac{x}{ \sqrt{19} }$

Answer 1

$1)\; \; y=sin2x\cdot cosx\\\\y'=2\, cos2x\cdot cosx+sin2x\cdot (-sinx)=2\, cos2x\cdot cosx-sin2x\cdot sinx\\\\2)\; \; y=\frac{sinx}{2+cosx}\\\\y'=\frac{cosx(2+cosx)-sinx\cdot (-sinx)}{(2+cosx)^2}=\frac{cosx(2+cosx)+sin^2x}{(2+cosx)^2}=\frac{2cosx+1}{(2+cosx)^2}\\\\3)\; \; \frac{2}{\sqrt7+\sqrt5}=\frac{2(\sqrt7-\sqrt5)}{(\sqrt7+\sqrt5)(\sqrt7-\sqrt5)}=\frac{2(\sqrt7-\sqrt5)}{7-5}=\sqrt7-\sqrt5\\\\4)\; \; \frac{2x-2y}{\sqrt{13x}+\sqrt{13y}}=\frac{2(x-y)}{\sqrt{13}\, (\sqrt{x}+\sqrt{y})}=\frac{2(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{13}\, (\sqrt{x}+\sqrt{y})}=\frac{2\, (\sqrt{x}-\sqrt{y})}{\sqrt{13}}$

$5)\; \; \frac{x}{\sqrt{19}}=\frac{x\sqrt{19}}{19}$