Question

Найти производную ctg(∛x/1)+(sin²21x÷7cos7x)

Answer 1

$y=ctg(\sqrt[3]{x}/1)+\frac{sin^221x}{7cos7x}\; ;\; \; \; \; ( \frac{ \sqrt[3]{x} }{1}=\sqrt[3]{x}) \\\\y'= -\frac{1}{sin^2\sqrt[3]{x}} \cdot \frac{1}{3}\cdot \frac{x^{-\frac{2}{3}}}{-\frac{2}{3}}+ \frac{2sin21x\cdot cos21x\cdot 21\cdot 7cos7x-sin^221x\cdot 7\cdot (-sin7x)\cdot 7}{49cos^27x} =\\\\= \frac{}{2\sqrt[3]{x^2}\cdot sin^2\sqrt[3]{x}}+ \frac{7\cdot sin42x\cdot cos7x+49\cdot sin^221x\cdot sin7x}{49\cdot cos^27x}$