Question

Помогите!!! Прикрепил скриншот

Answer 1

f'(x)=sqrt(x^2+1)-x^2/sqrt(x^2+1)/(x^2+1)=

=(x^2+1-x^2)/(x^2+1)^(3/2)=1/(x^2+1)^3/2

f''(x)=-3/2*(x^2+1)^(-5/2)*2x=-3x*(x^2+1)^(-5/2)

f''(0)=0

Answer 2

Ответ:

$f(x)=\frac{x}{\sqrt{x^2+1}}\\ f'(x)=\frac{1*\sqrt{x^2+1}-x*\frac{1}{2}(x^2+1)^-^\frac{1}{2}*2x}{x^2+1}=\frac{\sqrt{x^2+1}-\frac{2x^2}{\sqrt{x^2+1} }}{x^2+1}=\frac{x^2+1-2x^2}{(x^2+1)^\frac{3}{2}}=\frac{1-x^2}{(x^2+1)^\frac{3}{2}}\\f''(x)=\frac{-2x*(x^2+1)^\frac{3}{2}-(1-x^2)*\frac{3}{2}(x^2+1)^\frac{1}{2}*2x}{(x^2+1)^3}=\frac{(x^2+1)^\frac{1}{2}(-2x^3-2x-3x+3x^3)}{(x^2+1)^3}=\frac{x^3-5x}{(x^2+1)^\frac{5}{2} }$

f''(0)=0