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Решите задачу:

$3)\; \; y=\frac{6\, arcsinx}{log_5x}\\\\y'=\frac{6\cdot \frac{1}{\sqrt{1-x^2}}\cdot log_5x-6\, arcsinx\cdot \frac{1}{x\, ln5}}{log_5^2x}=\frac{6\cdot (x\cdot ln5\cdot log_5x-\sqrt{1-x^2}\cdot arcsinx)}{x\cdot ln5\cdot \sqrt{1-x^2}\cdot log_5^2x}\\\\4)\; \; y=\sqrt8\, arctgx\cdot \sqrt{3x}\\\\y'=\sqrt8\cdot \frac{1}{1+x^2}\cdot \sqrt{3x}+\sqrt8\, arctgx\cdot \frac{3}{2\sqrt{3x}}\\\\5)\; \; y=cos(arcsin\sqrt[3]{x})\\\\y'=-sin(arcsin\sqrt[3]{x})\cdot \frac{1}{\sqrt{1-\sqrt[3]{x^2}}}\cdot \frac{1}{3}\cdot x^{-\frac{2}{3}}$

$II.\; \; y=\frac{x^2}{10}+3\; \; ,\; \; x_0=2\\\\y(2)=\frac{4}{10}+3=\frac{17}{5}\\\\y'(x)=\frac{1}{10}\cdot 2x=\frac{x}{5}\; \; ,\; \; y'(2)=\frac{2}{5}\\\\y=\frac{17}{5}+\frac{2}{5}\cdot (x-2)\\\\y=\frac{2}{5}\cdot x+\frac{13}{5}\\\\y=0,4x+2,6$