Question

Срочно!!! Найти производные функций

Answer 1

$1)\; \; y=6x^2-\frac{5}{x^2}-\sqrt[3]{x^2}\\y'=12x-5\cdot \frac{-2}{x^3}-\frac{2}{3}\cdot x^{-\frac{1}{3}}\\\\2)\; \; y=5^{x}(x^5-10x)\\y'=5^{x}\, ln5(x^5-10x)+5^{x}\cdot (5x^4-10)\\\\3)\; \; y=8^{4x}\cdot \sqrt {e^{x}}\\y'=8^{4x}\cdot 4ln8\cdot \sqrt{e^{x}}+8^{4x}\cdot \frac{1}{2\sqrt{e^{x}}}\cdot e^{x}\\\\4)\; \; y=\frac{5x^2}{x-3}\\y'=\frac{10x(x-3)-5x^2\cdot 1}{(x-3)^2}=\frac{5x^2-30x}{(x-3)^2}=\frac{5x(x-6)}{(x-3)^2}\\\\5)\; \; y=ln\sqrt[6]{cosx}\\y'=\frac{1}{\sqrt[6]{cosx}}\cdot \frac{1}{6}\cdot (cosx)^{-\frac{5}{6}}\cdot (-sinx)$

$6)\; \; y=cos^4x\cdot 2\, lnx\\y'=4cos^3x\cdot (-sinx)\cdot 2\, lnx+cos^4x\cdot \frac{2}{x}\\\\7)\; \; y=arccos\frac{1-\sqrt{x}}{1+\sqrt{x}}\\\\y'=\frac{-1}{\sqrt{1-(\frac{1-\sqrt{x}}{1+\sqrt{x}}})^2}}\cdot \frac{-\frac{1}{2\sqrt{x}}\cdot (1+\sqrt{x})-(1-\sqrt{x})\cdot \frac{1}{2\sqrt{x}}}{(1+\sqrt{x})^2}=\\\\=-\frac{1+\sqrt{x}}{\sqrt{(1+\sqrt{x})^2-(1-\sqrt{x})^2}}\cdot \frac{-1-\sqrt{x}-1+\sqrt{x}}{2\sqrt{x}(1+\sqrt{x})^2}=\frac{1}{2\sqrt[4]{x^3}\cdot (1+\sqrt{x})}$