Question

Найдите значение производной функции в точке x0: а) f(x)=2tgx, x0= -3pi/4 б) f(x)=(4x+1)/(x+3), x0= -2 в) f(x)=корень из 4x-7, x0= 2 г) f(x)=sin(3x-pi/4), x0= pi/4 д) f(x)=tg6x, x0= pi/24

Answer 1

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