How to find lipschitz constant of the gradiant of $\sqrt{1+x^2}$?

Question

I know $\left|\frac{x}{\sqrt{1+x^2}}\right| \le 1$. But I don't know how to find $L$ that $\left|\frac{x}{\sqrt{1+x^2}}-\frac{y}{\sqrt{1+y^2}}\right| \le L|x-y|$.

Would you please explain it? Thank you.

Answer 1

Hint: the function $f(x):=\sqrt{1+x^2}$ has second derivative $\dfrac{1}{(1+x^2)^{3/2}}$, so $f$ is in $\mathrm C^2$.

Clue: Lagrange's theorem will apply.

Clue: Lagrange gives here that $|f'(x) -f'(y)| \leq |f''(z_0)||x-y|$ where $z_0$ is somewhere in $(x, y)$.