TITLE: If $f:\mathbb{R}\to \mathbb{R}$ such that $f(x)=\frac{\sin{x}}{\sqrt{1+x^2}}$ show that $f'(x)$ is bounded.
QUESTION [1 upvotes]: If $f:\mathbb{R}\to \mathbb{R}$ such that $f(x)=\frac{\sin{x}}{\sqrt{1+x^2}}$ show that $f'(x)$ is bounded.
We have $f'(x)=\frac{\cos{x}}{\sqrt{1+x^2}}-\frac{x\sin{x}}{(1+x^2)^{\frac{3}{2}}}$. then $|f'(x)|\leq |\frac{1}{\sqrt{1+x^2}}|+|\frac{x}{(1+x^2)^{\frac{3}{2}}}|$ 
How to show that $|f'(x)|<A$

REPLY [0 votes]: Since $f'(x)$ is continuous, the Boundedness theorem shows it is enough to prove $\lim\limits_{x\to\pm\infty}f'(x)=0$, which is obvious from the formula for $f'(x)$.
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