Question

срочно надо пж помогите
найти производную

Answer 1

-------------------------------------------------------
Готово™
-------------------------------------------------------

Answer 2

$y = \sqrt{x} \cos \: x$
$y ' = ( \sqrt{x} )'\cos x + \sqrt{x} (\cos x)'=(x^ \frac{1}{2} )'\cos x+ \sqrt{x} ( - \sin x) = \\ = \frac{1}{2} {x}^{ \frac{1}{2} - 1} \cdot \cos x - \sqrt{x} \sin x = \frac{1}{2} {x}^{- \frac{1}{2} } \cos x - \sqrt{x} \sin x = \\ = \frac{\cos x}{2 {x}^{ \frac{1}{2} } } - \sqrt{x} \sin x = \frac{\cos x}{2 \sqrt{x} } - \sqrt{x} \sin x$
$y = x \cos x$
$y = (x \cos \: x)'=x' \cos x + x(\cos x)' = 1\cdot \cos x + \\ + x\cdot(-\sin x) = \cos x - x\sin x$
$y = x \sin6x$
$y' = (x\sin6x)' = x'\cdot \sin6x+x\cdot(\sin6x)' = 1\cdot\sin6x + \\ + x\cdot \cos 6x\cdot(6x)' = \sin6x + x\cdot \cos6x\cdot6 = \\ = \sin6x + 6x\cos6x$