Question

срочно!!! помогите хотя бы с одним!!​

Answer 1

$1)y=6-\sqrt{x}\\\\y'=6' (\sqrt{x})'=0-\frac{1}{2\sqrt{x}}=-\frac{1}{2\sqrt{x}} \\\\y'(x_{0})=y'(25)=-\frac{1}{2\sqrt{25} } =-\frac{1}{2*5}= \boxed{-0,1} \\\\\\2)y=6Sinx\\\\y'=6(Sinx)'=6Cosx\\\\y'(x_{0} )=y'(-\frac{\pi }{6})=6Cos(-\frac{\pi }{6})=6Cos\frac{\pi }{6}=6*\frac{\sqrt{3} }{2}=\boxed{3\sqrt{3}}$

Answer 2

Ответ:

$1)\ \ y=6-\sqrt{x} \ \ ,\ \ \ y'=-\dfrac{1}{2\sqrt{x} }\ \ ,\ \ \ y'(25)=-\dfrac{1}{2\sqrt{25} }=-\dfrac{1}{2\cdot 5}=-\dfrac{1}{10}\\\\\\2)\ \ y=6\, sinx\ \ ,\ \ y'=6\, cosx\ \ ,\ \ y'\Big(-\dfrac{\pi}{6}\Big)=6cos\Big(-\frac{\pi}{6}\Big)=6cos\dfrac{\pi}{6}=6\cdot \dfrac{\sqrt3}{2}=3\sqrt3$