Question

Помогите, я вас умоляю ​

Answer 1

$\displaystyle\bf\\1)\\\\y=7x^{5} +3x^{4} -\frac{5}{7x} +4\\\\\\y'=7\cdot(x^{5})'+3\cdot(x^{4} )'-\frac{5}{7} \cdot\Big(\frac{1}{x} \Big)'+4'=\\\\\\=7\cdot 5x^{4} +3\cdot4x^{3} -\frac{5}{7}\cdot\Big(-\frac{1}{x^{2} }\Big)+0=35x^{4} +12x^{3}+\frac{5}{7x^{2} } \\\\\\2)\\\\y=-3\sqrt{x} +\frac{1}{3} Cosx-\frac{1}{2} Ctgx\\\\\\y'=-3\cdot(\sqrt{x} )'+\frac{1}{3} \cdot(Cosx)'-\frac{1}{2} \cdot(Ctgx)'=$

$\displaystyle\bf\\=-3\cdot\frac{1}{2\sqrt{x} } +\frac{1}{3} \cdot(-Sinx)-\frac{1}{2} \cdot\Big(-\frac{1}{Sin^{2} x} \Big)=-\frac{3}{2\sqrt{x} } -\frac{1}{3} Sinx+\frac{1}{2Sin^{2}x } \\\\\\3)\\\\y=\sqrt{x} (-2x+1)\\\\\\y'=(\sqrt{x} )'\cdot(-2x+1)+\sqrt{x} \cdot(-2x+1)'=\\\\\\=\frac{1}{2\sqrt{x} }\cdot(-2x+1)+\sqrt{x} \cdot(-2)=\frac{1-2x}{2\sqrt{x} } -2\sqrt{x} =\\\\\\=\frac{1-2x-4x}{2\sqrt{x} }=\frac{1-6x}{2\sqrt{x} }$