Question

Решение произдводных

Answer 1

$1)\; \; y=sinx\cdot cosx=\frac{1}{2}\cdot sin2x\\\\y'=\frac{1}{2}\cdot cos2x\cdot 2=cos2x\\\\2)\; \; y=\frac{ctg(4x-1)}{8}\\\\y'=\frac{1}{8}\cdot \frac{-4}{sin^2(4x-1)}=-\frac{1}{2\, sin^2(4x-1)} \\\\3)\; \; y=sinax\cdot cosbx\\\\y'=a\, cosax\cdot cosbx-b\, sinbx\cdot sinax\\\\4)\; \; y=\frac{\sqrt{1-cosx}}{1-cosx}=\frac{1}{\sqrt{1-cosx}}\\\\y'=-\frac{(\sqrt{1-cosx})'}{(1-cosx)^2}=-\frac{sinx}{2\sqrt{1-cosx}\; \cdot \, (1-cosx)}}=-\frac{sinx}{2\sqrt{(1-cosx)^3}}$