How can I prove the monotonicity of a function? Given is the function $$f(x) =\frac{x}{\sqrt{1-x^2}}$$
How can I prove that this function is monotonic and thus injective?
 A: On the interval $[0,1)$, the numerator $x$ is increasing and the denominator $\sqrt{1-x^2}$ is decreasing. So $f(x)$ is increasing on the interval.
For $-1\le x\le 0$, use the fact that $f(x)$ is an odd function, so has symmetry across the origin. We conclude that $f(x)$ is monotone on $(-1,1)$, the natural domain of definition.
Remark: For other functions, a different approach, such as looking at the sign of the derivative, may be the appropriate one. The derivative should not be the automatic go-to technique, since the derivative may be complicated, and difficult to analyze. 
A: We observe that $f(x) =\tan({\arcsin (x))}$  so as a composition of increasing functions is increasing.

Notice in that triangle $$\arcsin(\frac{x}{1}) =\theta$$
and
$$\tan({\theta})= \frac{x}{\sqrt{1-x^2}}$$
$\sin(x)$ is increasing on $[ -\frac{\pi}{2},\frac{\pi}{2}]$ so then is 
$\arcsin{x}$ as his inverse.
A: Generally approach is to take the derivative of the function and check if the derivative is positive or negative. If positive, the function is increasing. If negative, the function is decreasing.
For this function, $f'(x)=\frac{1}{(1-x^2)^{3/2}}$ which is apparently positive on interval $(-1,1)$. So $f(x)$ is an increasing function on $(-1, 1)$.
