Question

Решите кто нибудь 4,5,6

Answer 1

$y= \frac{ x^{2} }{3-4x} \\\\y'= \frac{( x^{2} )'*(3-4x)- x^{2} *(3-4x)'}{(3-4x) ^{2} } = \frac{2x*(3-4x)+4 x^{2} }{(3-4x) ^{2} } = \frac{6x-8 x^{2} +4 x^{2} }{(3-4x) ^{2} }= \frac{6x-4 x^{2} }{(3-4x) ^{2} }$

$y= \frac{Cosx}{x}\\\\y'= \frac{(Cosx)'*x-Cosx*(x)'}{ x^{2} } = \frac{-xSinx-Cosx}{ x^{2} }$

$y= \sqrt{x} Sinx\\\\y'=( \sqrt{x} )'* Sinx+ \sqrt{x} *(Sinx)' = \frac{1}{2 \sqrt{x} } Sinx+ \sqrt{x} Cosx= \frac{Sinx}{2 \sqrt{x} }$$+ \sqrt{x} Cosx$