Question

y=(ctgx)^√2x-5 найти производную​

Answer 1

Ответ:

$y=ctgx^{\sqrt{2x-5}}\\\\lny=ln\Big(ctgx^{\sqrt{2x-5}}\Big)\\\\lny=\sqrt{2x-5}\cdot ln(ctgx)\\\\\\\dfrac{y'}{y}=\dfrac{2}{2\sqrt{2x-5}}\cdot ln(ctgx)+\sqrt{2x-5}\cdot \dfrac{1}{ctgx}\cdot \dfrac{-1}{sin^2x}\\\\\\y'=y\cdot \Big(\dfrac{1}{\sqrt{2x-5}}\cdot ln(ctgx)+\sqrt{2x-5}\cdot \dfrac{sinx}{cosx}\cdot \dfrac{-1}{sin^2x}\Big)\\\\\\y'=(ctgx)^{\sqrt{2x-5}}\cdot \Big(\ \dfrac{ln(ctgx)}{\sqrt{2x-5}}-\dfrac{2\sqrt{2x-5}}{sin2x}\ \Big)$