Упр.41.20 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

Найдём производные по правилу производной частного:

1. $$y=\frac{x^9-3}{x^3}$$
   $$y’=\frac{(x^9-3)’ \cdot x^3-(x^9-3)\cdot (x^3)’}{(x^3)^2}$$
   $$y’=\frac{9x^8\cdot x^3-(x^9-3)\cdot 3x^2}{x^6}$$
   $$y’=\frac{9x^{11}-3x^{11}+9x^2}{x^6}=\frac{6x^{11}+9x^2}{x^6}=6x^5+\frac{9}{x^4}.$$
2. $$y=\frac{x^{15}}{x^{10}+1}$$
   $$y’=\frac{(x^{15})’\cdot (x^{10}+1)-x^{15}\cdot (x^{10}+1)’}{(x^{10}+1)^2}$$
   $$y’=\frac{15x^{14}(x^{10}+1)-x^{15}\cdot 10x^9}{(x^{10}+1)^2}$$
   $$y’=\frac{15x^{24}+15x^{14}-10x^{24}}{(x^{10}+1)^2}=\frac{5x^{24}+15x^{14}}{(x^{10}+1)^2}=\frac{5x^{14}(x^{10}+3)}{(x^{10}+1)^2}.$$
3. $$y=\frac{x^5+x}{x^5-1}$$
   $$y’=\frac{(x^5+x)’\cdot (x^5-1)-(x^5+x)\cdot (x^5-1)’}{(x^5-1)^2}$$
   $$y’=\frac{(5x^4+1)(x^5-1)-(x^5+x)\cdot 5x^4}{(x^5-1)^2}$$
   $$y’=\frac{5x^9-5x^4+x^5-1-5x^9-5x^5}{(x^5-1)^2}$$
   $$y’=\frac{-4x^5-5x^4-1}{(x^5-1)^2}=-\frac{4x^5+5x^4+1}{(x^5-1)^2}.$$
4. $$y=\frac{x^{13}}{x^4-2}$$
   $$y’=\frac{(x^{13})’\cdot (x^4-2)-x^{13}\cdot (x^4-2)’}{(x^4-2)^2}$$
   $$y’=\frac{13x^{12}(x^4-2)-x^{13}\cdot 4x^3}{(x^4-2)^2}$$
   $$y’=\frac{13x^{16}-26x^{12}-4x^{16}}{(x^4-2)^2}=\frac{9x^{16}-26x^{12}}{(x^4-2)^2}=\frac{x^{12}(9x^4-26)}{(x^4-2)^2}.$$

Ответ:

а) $$y’=6x^5+\frac{9}{x^4}$$;
б) $$y’=\frac{5x^{14}(x^{10}+3)}{(x^{10}+1)^2}$$;
в) $$y’=-\frac{4x^5+5x^4+1}{(x^5-1)^2}$$;
г) $$y’=\frac{x^{12}(9x^4-26)}{(x^4-2)^2}$$.