Question

решите пожалуйста срочно надо

Answer 1

$1)\; \; y=\frac{3\sqrt{x}}{\sqrt[3]{x}(3x+1)}=\frac{3x^{\frac{1}{2}}}{3x^{\frac{4}{3}}+x^{\frac{1}{3}}}\\\\y'=\frac{\frac{3}{2}x^{-\frac{1}{2}}\cdot (3x^{\frac{4}{3}}+x^{\frac{1}{3}})-3x^{\frac{1}{2}}\cdot (4x^{\frac{1}{3}}+\frac{1}{3}\cdot x^{-\frac{2}{3}})}{(3x^{\frac{4}{3}}+x^{\frac{1}{3}})^2}=\frac{\frac{9}{2}x^{\frac{5}{6}}+\frac{3}{2}x^{-\frac{1}{6}}-12x^{\frac{5}{6}}-x^{-\frac{1}{6}}}{(3x^{\frac{4}{3}}+x^{\frac{1}{3}})^2}\\\\x_0=1\; ,\; \; y'(1)=\frac{4,5+1,5-12-1}{(3+1)^2}=-\frac{7}{16}$

$2)\; \; y=2sin2x+3cos3x\\\\y'=2\cdot cos2x\cdot 2+3\cdot (-sin3x)\cdot 3=4cos2x-9sin3x\\\\x_0=\frac{11\pi }{6}\; ,\; \; y'(\frac{11\pi }{6})=4\cdot cos\frac{11\pi }{3}-9\cdot sin\frac{11\pi }{2}=\\\\=4\cdot cos(4\pi -\frac{\pi}{3})-9\cdot sin(6\pi-\frac{\pi }{2})=4\cdot cos\frac{\pi }{3}-9\cdot sin(-\frac{\pi }{2})=\\\\=4\cdot \frac{1}{2}-9\cdot (-1)=2+9=11$