이계도함수와 오목·볼록

$$\begin{aligned} x^2+x+y&=\sin(xy) \\ \text{We differentiate implicitly. For ease of notation, we write $y'$ for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$.} \\ 2x+1+y'&=\cos(xy)(y+xy') \\ \text{We're interested in $y''$, so we implicitly differentiate again.} \\ 2+y''&=-\sin(xy)(y+xy')^2+\cos(xy)(2y'+xy'') \\ \text{We want to know what $y''$ is when $x=y=0$. Plugging these in yields the following:} \\ 2+y''&=2y' \\ \text{So, we need to know what $y'$ is when $x=y=0$. We can get this from the equation $2x+1+y'=\cos(xy)(y+xy')$, which becomes $1+y'=0$ when $x=y=0$. So, at the origin, $y'=-1$, and} \\ 2+y''&=2(-1)\\ y''&=-4 \end{aligned}$$

Remark: a common mistake is to stop at the equation $2x+1+y'=\cos(xy)(y+xy')$, plug in $x=y=0$, find $y'=-1$, and decide $y''=\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\{-1\}=0$. This is due to a slight sloppiness in the usual notation. When we wrote $y'=1$, what we meant is that at the point $(0,0)$, $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-1$. More properly written: $\left.\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=0,~y=0}=-1$. This is not the same as saying $y'=1$ everywhere (in which case, indeed, $y''$ would be 0 everywhere).