Question

Найти производную функции
Дам 33 б

Answer 1

$y=ln\sqrt{\frac{1-x^5}{1+2x}}=ln\Big (\frac{1-x^5}{1+2x}\Big )^{\frac{1}{2}}=\frac{1}{2}\cdot (ln(1-x^5)-ln(1+2x))\\\\\\y'=\frac{1}{2}\cdot \Big (\frac{1}{1-x^5}\cdot (-5x^4)-\frac{1}{1+2x}\cdot 2\Big )=\frac{1}{2}\cdot \Big (-\frac{5x^4}{1-x^5}-\frac{2}{1+2x}\Big )=\\\\=\frac{1}{2}\cdot \frac{-5x^4-10x^5-2+2x^5}{(1-x^5)(1+2x)}=\frac{-5x^4-8x^5-2}{2\, (1-x^5)(1+2x)}=-\frac{5x^4+8x^5+2}{2\, (1-x^5)(1+2x)}$