Теория:

Нахождение производной сложной функции:

f(g(x)) = (g(x))' * (f(g(x)) )'

Нахождение производной дроби:

[tex]\displaystyle (\frac{f(x)}{g(x)})' = \frac{(f(x))'*g(x)-(g(x))'*f(x)}{(g(x))^2}[/tex]

Решение:

2) [tex]\displaystyle f(x) = 4*sin(3x)[/tex]

[tex]\displaystyle (f(x))' = 4*(3x)'*(sin(3x))'[/tex]

[tex]\displaystyle (f(x))' = 4*3*cos(3x)[/tex]

[tex]\displaystyle (f(x))' = 12*cos(3x)[/tex]

[tex]\displaystyle (f(\frac{\pi }{12} ))' = 12*cos(3*\frac{\pi }{12})=12*cos(\frac{\pi }{4})=12*\frac{{\sqrt{2} } }{2}=6\sqrt{2}[/tex]

[tex]\displaystyle (f(0))' = 12*cos(3*0)=12*cos(0)=12*1=12[/tex]

3) [tex]\displaystyle f(x) = \frac{5x+4}{3-7x}[/tex]

[tex]\displaystyle (f(x))' = \frac{(5x+4)'*(3-7x)-(3-7x)'*(5x+4)}{(3-7x)^2}[/tex]

[tex]\displaystyle (f(x))' = \frac{5*(3-7x)-(-7)*(5x+4)}{(3-7x)^2}[/tex]

[tex]\displaystyle (f(x))' = \frac{15-35x+35x+28}{(3-7x)^2} =\frac{43}{(3-7x)^2}[/tex]

[tex]\displaystyle (f(-1))' =\frac{43}{(3-7*(-1))^2} =\frac{43}{(3+7)^2}=\frac{43}{100}=0,43[/tex]

[tex]\displaystyle (f(0))' =\frac{43}{(3-7*0)^2} =\frac{43}{3^2}=\frac{43}{9}=4\frac{7}{9}[/tex]

Verified answer

Ответ:

во вложении

Объяснение:

там же