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