Ответ:

Находим производные функций, применяя правила дифференцирования и таблицу производных функций .

[tex]\bf 1)\ \ y'=(2x^5-5\, sinx+5)'=10x^4-5\, cosx\\\\2)\ \ y'=(5\, e^{x}-arcsinx)'=5\, e^{x}-\dfrac{1}{\sqrt{1-x^2}}\\\\3)\ \ y'=(3\sqrt{x}-5\, arctg\, x)'=\dfrac{3}{2\sqrt{x}}-\dfrac{5}{1+x^2}\\\\4)\ \ y'=((5x+1)\cdot (arccosx-5))'=5\cdot (arccosx-5)-\dfrac{5x+1}{\sqrt{1-x^2}}\\\\5)\ \ y'=(5\, lnx-log_{571}\, x)'=\dfrac{5}{x}-\dfrac{1}{x\cdot ln\, 571}[/tex]

[tex]\bf 6)\ \ y'=(e^{x^2+5\, cos2x})'=e^{x^2+5\, cos2x}\cdot (2x-10\cdot sin2x)\\\\7)\ \ y'=\Big(\dfrac{5-tg\, x}{ctgx-5\sqrt[3]{\bf x}}\Big)'=\\\\\\=\dfrac{-\dfrac{1}{cos^2x}\cdot (ctgx-5\sqrt[3]{\bf x})-(5-tg\, x)\cdot \Big(-\dfrac{1}{sin^2x}-5\cdot \dfrac{1}{3}\cdot x^{-\frac{2}{3}}\Big)}{(ctgx-5\sqrt[3]{\bf x})^2}=\\\\\\=\dfrac{\dfrac{ctgx-5\sqrt[3]{\bf x}}{cos^2x}+(5-tg\, x)\cdot \Big(\dfrac{1}{sin^2x}+\dfrac{5}{3\sqrt[3]{\bf x^2}}\Big)}{(ctgx-5\sqrt[3]{\bf x})^2}=[/tex]  

[tex]\bf =\dfrac{(ctgx-5\sqrt[3]{\bf x})\cdot 3\sqrt[3]{\bf x^2}\cdot sin^2x+(5-tg\, x)(3\sqrt[3]{\bf x^2}+5sin^2x)\cdot cos^2x}{3\sqrt[3]{\bf x^2}\cdot sin^2x\cdot cos^2x\cdot (ctgx-5\sqrt[3]{\bf x})^2}[/tex]        

[tex]\bf 8)\ \ y'=\Big(arcsin(x^2-5x)\Big)'=\dfrac{1}{\sqrt{1-(x^2-5x)^2}}\cdot (2x-5)\\\\\\9)\ \ y'=\Big((x^5-arccos\, x^5)\cdot (ln\, 5x-\sqrt[3]{\bf x^5})\Big)'=\\\\=\Big(5x^4+\dfrac{1}{\sqrt{1-x^{10}}}\cdot 5x^4\Big)\cdot \Big(ln\, 5x-\sqrt[3]{\bf x^5}\Big)+\Big(x^5-arccos\, x^5\Big)\cdot \Big(\dfrac{5}{5x}-\dfrac{5}{3}\sqrt[3]{\bf x^2}\Big)[/tex]