Question

Найдите f″(0), если f(x)=x/√x2+1

Answer 1

Ответ:

Пошаговое объяснение:

$y=\frac{x}{\sqrt{x^2+1}}\\y'=\frac{\sqrt{x^2+1}-x*\frac{1}{2}(x^2+1)^-^\frac{1}{2}*2x}{x^2+1}=\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}= \frac{x^2+1-x^2}{(x^2+1)*\sqrt{x^2+1} } =\frac{1}{(x^2+1)^\frac{3}{2}}\\ y''=-\frac{\frac{3}{2}(x^2+1)^\frac{1}{2}*2x  }{(x^2+1)^3}=-3x(x^2+1)^-^\frac{5}{2}\\ y''(0)=0$

Answer 2

$f(x)=\frac{x}{\sqrt{x^2+1}}\\\\\\f'(x)=\frac{1\cdot \sqrt{x^2+1}-x\cdot \frac{2x}{2\sqrt{x^2+1}}}{x^2+1}=\frac{x^2+1-x^2}{\sqrt{x^2+1}\cdot (x^2+1)}=\frac{1}{\sqrt{(x^2+1)^3}}=(x^2+1)^{-\frac{3}{2}}\\\\f''(x)=-\frac{3}{2}\cdot (x^2+1)^{-\frac{5}{2}}\cdot 2x=-\frac{3x}{\sqrt{(x^2+1)^5}}\\\\f''(0)=0$