Question

Найти производную четвертого порядка
y=(x^2 +3)ln(x-1)

Answer 1

$y=(x^2+3)ln(x-1)\\y'=2xln(x-1)+frac{x^2+3}{x-1}\\y''=2ln(x-1)+frac{2x}{x-1}+frac{2x(x-1)-(x^2+3)}{(x-1)^2}=\\=2ln(x-1)+frac{2x}{x-1}+frac{x^2-2x-3}{(x-1)^2}\\y'''=frac{2}{x-1}+frac{2(x-1)-2x}{(x-1)^2}+frac{(2x-2)-(x^2-2x-3)cdot 2(x-1)}{(x-1)^4}=\\=frac{2}{x-1}+frac{-1}{(x-1)^2}+frac{2(x-1)(1-x^2+2x+3)}{(x-1)^4}=frac{2}{x-1}-frac{2}{(x-1)^2}+frac{2(4+2x-x^2)}{(x-1)^3}\\y^{(4)}=frac{-2}{(x-1)^2}-frac{2cdot 2(x-1)}{(x-1)^4}+frac{2(2-2x)(x-1)^3-2(4+2x-x^2)3(x-1)^2}{(x-1)^6}$

$=-frac{2}{(x-1)^2}-frac{4}{(x-1)^3}-frac{2cdot (14+2x-x^2)}{(x-1)^4}$