Question

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
f(x)=√3x^2 - 4x+9
f(x)=(3x-2)(7x+5)
f(x)=2x+4/7x-3

Answer 1

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

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

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

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

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

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

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