Question

Помогите по алгебре чем сможете. Даю 100 баллов.

Answer 1

$1)f(x)=2\sqrt{3} Cos4x-2Cosx\\f'(x)=2\sqrt{3}(Cos4x)'- 2(Cosx)'=2 \sqrt{3}*(-Sin4x)*(4x)'+2Sinx=-8 \sqrt{3}Sin4x+2Sinx\\\\\\f'( \pi/6)=-8 \sqrt{3}Sin(4* \pi/6)+2Sin \pi/6=-8 \sqrt{3}Sin2 \pi/3+2Sin \pi/6=-8 \sqrt{3}* \frac{\sqrt{3}} {2}+2* \frac{1}{2}=-12+1=-11$

2) y = (- 10x - 5)⁷

y' = [(- 10x- 5)⁷]' = 7(- 10x - 5)⁶ * (- 10x - 5)' = 7(- 10x - 5)⁶ * (- 10) =

= - 70(- 10x - 5)⁶

$y=lntgx\\y' = (lntgx)' = \frac{1}{tgx} *(tgx)' = \frac{1}{tgx} *\frac{1}{Cos^{2}x}= \frac{Cosx}{Sinx}* \frac{1}{Cos^{2}x}= \frac{1}{SinxCosx}= \frac{2}{Sin2x} \\y'(\pi/12)= \frac{2}{Sin(2*\pi/12)}= \frac{2}{Sin\pi/6}= \frac{2}{\frac{1}{2}} =4$