Question

Решите задания и получите легкие была

Answer 1

$1)\ \ y=\dfrac{sin^35x}{cos^2\frac{x}{3}}\\\\y'=\dfrac{3sin^25x\cdot cos5x\cdot 5\, cos^2\frac{x}{3}-sin^35x\cdot 2cos\frac{x}{3}\cdot (-sin\frac{x}{3})\cdot \frac{1}{3}}{cos^4\frac{x}{3}}=\\\\=\dfrac{15\, sin^25x\cdot cos5x\cdot cos\frac{x}{3}+\frac{2}{3}\cdot sin^35x\cdot sin\frac{x}{3}}{cos^3\frac{x}{3}}$

$2)\ \ y=\dfrac{x^8}{8(1-x^2)}\\\\y'=\dfrac{1}{8}\cdot \dfrac{8x^7(1-x^2)-x^8\cdot (-2x)}{(1-x^2)^2}=\dfrac{4x^7(1-x^2)+x^9}{4(1-x^2)^2}\\\\\\3)\ \ y=\dfrac{x^3}{3\cdot \sqrt{(1+x^2)^3}}\\\\y'=\dfrac{1}{3}\cdot \dfrac{3x^2\cdot \sqrt{(1+x^2)^3}-x^3\cdot \frac{3}{2}\cdot (1+x^2)^{\frac{1}{2}}\cdot 2x}{(1+x^2)^3}=\\\\=\dfrac{x^2\cdot \sqrt{(1+x^2)^3}-x^4\cdot \sqrt{1+x^2}}{(1+x^2)^3}=\dfrac{x^2\cdot (1+x^2)-x^4}{\sqrt{(1+x^2)^5}}=\dfrac{x^2}{\sqrt{(1+x^2)^5}}$

$4)\ \ y=\dfrac{\sqrt{2x^2-2x+1}}{x}\ \ ,\ \ y=\dfrac{1}{x}\cdot \sqrt{2x^2-2x+1}\\\\y'=-\dfrac{1}{x^2}\cdot \sqrt{2x^2-2x+1}+\dfrac{1}{x}\cdot \dfrac{4x-2}{2\sqrt{2x^2-2x+1}}=\\\\=-\dfrac{\sqrt{2x^2-2x+1}}{x^2}+\dfrac{2x-1}{x\cdot \sqrt{2x^2-2x+1}}\\\\\\5)\ \ y=\Big (\frac{3}{3}\sqrt[3]{(x+1)^2}\Big):\Big(\dfrac{1}{2}\, x\cdot \sqrt[6]{x}\Big)\ \ ,\ \ y=\sqrt[3]{(x+1)^2}\cdot 2x^{-7/6}\\\\y'=\dfrac{2}{3}\cdot (x+1)^{-1/3}\cdot 2x^{-7/6}+(x+1)^{2/3}\cdot \Big(-\dfrac{7}{3}\cdot x^{-13/6}\Big)=$

$=\dfrac{4}{3}\cdot \dfrac{1}{\sqrt[6]{x^7}\cdot \sqrt[3]{x+1}}-\dfrac{7\sqrt[3]{(x+1)^2}}{3\cdot \sqrt[6]{x^{13}}}$