Question

!Помогите решить математику!

Answer 1

а) $y'= \sqrt{x^2-1} - \frac{(x+1)*2x}{2 \sqrt{x^2-1} } = \frac{x^2-1-x(x+1)}{\sqrt{x^2-1}} = \frac{-x-1}{ \sqrt{(x+1)(x-1)} } =- \sqrt{ \frac{x+1}{x-1} }$
y'(2) = -√(3/1) = -√3

б) $y'= \frac{(4x-3)(x+1)-(2x^2-3x-1)}{(x+1)^2} = \frac{4x^2+x-3-2x^2+3x+1}{(x+1)^2} = \frac{2x^2+4x-2}{(x+1)^2}$
y'(1) = (2+4-2)/(2^2) = 4/4 = 1

в) $y'=4sin^3(x)cos(x)+4cos^3(x)(-sin(x))= \\ =4cos(x)sin(x)(sin^2(x)-cos^2(x))=-2sin(2x)cos(2x)=-sin(4x)$