I need to do the following: show that the equation $x \cdot $ cos $(x \cdot y)=0$ has a unique solution in the neighborhood of the point $(1,\frac{\pi}{2})$ (we will denote it $y(x)$), and that $y(x)$ is a convex function in that neighborhood.
For the first part, I used the implicit function theorem to show that there is a unique $y(x)$. Now I want to show it is convex; I have found that the derivative is: $$y'(x)=\frac{\cos\bigl(x \cdot y(x)\bigr)-x \cdot y(x) \cdot \sin\bigl(x \cdot y(x)\bigr)}{x^2 \cdot \sin\bigl(x \cdot y(x)\bigr )}$$ Now in order to show it is convex, I thought about showing that $y'(x)$ is increasing in that neighborhood, but I don't know how to do it since I don't know much about that neighborhood nor about $y(x)$. Any ideas?
Suppose $x \cdot \cos (x \cdot y) = 0$. Then $x = 0$ or $\cos(x \cdot y) = 0$. In the latter case, this is equivalent to $x \cdot y = \frac{\pi}{2} + 2 k \pi$ for some $k$. Note that the vertical line $x = 0$ can never contribute to any of these continuous function curves, so literally, your solution must be of the form,
$$y = \frac{\frac{\pi}{2} + 2 k \pi}{x}.$$
Given that it passes through the point $(1, \pi/2)$, we're looking at $k = 0$. So, the function in question is,
$$y = \frac{\pi}{2x}$$
which is convex for $x > 0$.