Помогите с этим задание (сатан не мое) Найдите производную функции y=f(x) в точках х=а f(x)=корень2x-3 при а=2 f(x)=sin 2x при a=пи/6 f(x)=2x cos x при а=0 f(x)=e^x+5 при а=In5 f(x)=x/1+x^2 при а=0

1. f(x) = sqrt(2x - 3) f'(x) = 2/(2*sqrt(2x - 3)) = 1/sqrt(2x - 3) =  [x = 2] = 1/sqrt(1) = 1 2. f(x) = sin(2x) f'(x) = 2*cos(2x) = [x = pi/6] = 2 * 1/2 = 1 3. f(x) = 2*x * cos(x) f'(x) = 2 * cos(x) - 2*x*sin(x) = [x = 0] = 2 4. f(x)=e^x+5 f'(x) = e^x = [x = ln5] = 5 5 .f(x)=x/1+x^2 f'(x) = ((1 + x^2) - 2*x*x)/(1 + x^2)^2 = [x = 0] = 1

[latex]f(x)= \sqrt{2x-3} , a = 2 \\ f'(x) = \frac{1}{\sqrt{2x-3}}\\ f'(2)=1 \\ \\ f(x)=sin 2x,a= \frac{ \pi }{6} \\ f'(x) = 2cos2x \\ f'(\frac{ \pi }{6})=2cos \frac{ \pi }{3} =1 \\ \\ f(x)=2x*cos x, a=0 \\ f'(x) = 2cosx-2x*sinx \\ f'(0) = 2*cos0 -2*0*sin0=2 \\ \\ f(x)=e^x+5, a=ln5 \\f'(x) =e^x \\f'(ln5)=e^{ln5} =5 \\ \\ f(x)= \frac{x}{1+x^2},a=0 \\ f'(x)= \frac{1+x^2-x*2x}{(1+x^2)^2} = \frac{1-x^2}{(1+x^2)^2} \\f'(0)=1[/latex]