Найти производную функции F(x)=11x+2/10x-5 f(x)=x^4/x^2-1 f(x)=1/(3x+6)^3 f(x)=√5x-7...

Решите задачу:

$\displaystyle F(x)=\frac{11x+2}{10x-5};\ f'(x)=\frac{(11x+2)'(10x-5)-(11x+2)(10x-5)'}{(10x-5)^2}=\\=\frac{11(10x-5)-(11x+2)10}{(10x-5)^2}=\frac{110x-55-110x-20}{(10x-5)^2}=\frac{-75}{(10x-5)^2}$

$\displaystyle f(x)=\frac{x^4}{x^2-1};\ f'(x)=\frac{4x^3(x^2-1)-2x*x^4}{(x^2-1)^2}=\\\frac{4x^5-4x^3-2x^5}{(x^2-1)^2}=\frac{2x^5-4x^3}{(x^2-1)^2};$

$\displaystyle f(x)=\frac{1}{(3x+6)^3};\ f'(x)=\frac{-9}{(3x+6)^4};$

$\displaystyle f(x)=\sqrt{5x-7};\ f'(x)=\frac{5}{2\sqrt{5x-7}};$

$\displaystyle f(x)=\sqrt{3x^2-4x+9};\ f'(x)=\frac{6x-4}{2\sqrt{3x^2-4x+9}};$

$\displaystyle f(x)=\frac{3x-2}{7x+5};\ f'(x)=\frac{3(7x+5)-(3x-2)7}{(7x+5)^2}=\\\frac{21x+15-21x+14}{(7x+5)^2}=\frac{29}{(7x+5)^2};$

$\displaystyle f(x)=\frac{2x+4}{7x-3};\ f'(x)=\frac{2(7x-3)-(2x+4)7}{(7x-3)^2}=\\=\frac{14x-6-14x-28}{(7x-3)^2}=\frac{-34}{(7x-3)^2}$