How to find the intervals on which $f$ is concave up? $f(x) = \arctan(\sin x )$ \begin{align*}f(x) &= \arctan(\sin x )\\[4pt]
f'(x) &= \frac{\cos x }{1+\sin^2 x }\\[4pt]
f''(x) &= -\frac{(\sin x )(2+\cos^2 x )}{(1+ \sin^2 x )^2}
\end{align*}
So I need to use these three equations to find the intervals on which $f$ is concave up.
Can you reason why it is?
 A: For f to concave up, it's second derivative needs to be positive, so we just need to look at when $f^{\prime\prime}>0$.
Since $(2+\cos^2(x))$ and the denominator $(1+\sin^2(x))^2$ are always positive due to the squares, what we need to look at is simply when $\sin(x)$ is negative.
$$f^{\prime\prime}>0\longrightarrow\sin(x)<0\longrightarrow x\in((2n-1)\pi,(2n)\pi)\{n\in\mathbb{N} \}$$
Hope this is helpful.
A: A function is concave up when its second derivative is positive.
$$f''(x)=-\frac{(\sin x )(2+\cos^2 x )}{(1+ \sin^2 x )^2}$$
The negative sign can be taken into the numerator to give
$$f''(x)=\frac{-(\sin x )(2+\cos^2 x )}{(1+ \sin^2 x )^2}$$
The denominator is always positive and the $(2+\cos^2 x)$ in the numerator is always positive.

This means that the sign of $f''(x)$ is determined by the sign of $-(\sin x)$.

When $\sin x$ is negative, $-(\sin x)$ is positive and hence $f''(x)$ is positive.

$\sin x$ is negative in the 3rd and 4th quadrants.

Hence, $f''(x)$ is positive when $x$ is in the 3rd and 4th quadrants.
Hence, $f(x)=\arctan(\sin x)$ is concave up when $x$ is in the 3rd and 4th quadrants, such as when $x\in(-5\pi,-4\pi)\cup(-3\pi,-2\pi)\cup(-\pi,0)\cup(\pi,2\pi)\cup(3\pi,4\pi)\cup(5\pi,6\pi)$ and so on.
