Найдите значение производной функции в точке x0: а) f(x)=2tgx, x0= -3Pi/4 б)...

A)f'(x)=(2tgx)'=2*(1/cos²x), f'(-3π/4)=2*(1/cos²(-3π/4)=2*(-2/√2)²=4
б) $f'(x)=( \frac{4x+1}{x+3})'= \frac{(4x+1)'(x+3)-(4x+1)(x+3)'}{(x+3)^2}= \frac{4x+12-4x-1}{(x+3)^2}= \frac{11}{(x+3)^2}$ , f'(-2)=11/1=11
в) $f'(x)= (\sqrt{4x-7})'= \frac{1}{2 \sqrt{4x-7} }$ ,
$f'(2)= \frac{1}{2 \sqrt{2*4-7} }= \frac{1}{2}$
г)f'(x)=(sin(3x-π/4))'=3*cos(3x-π/4), f'(π/4)=3*cos(3*(π/4)-π/4)=3*cos(π/2)=0
д)f'(x)=(tg6x)'=6*(1/cos²(6x)), f'(π/24)=6*(1/cos²(6*π/24)=6*(1/cos²π/4)=6*(2/√2)²=6*4/2=12