이계도함수와 오목·볼록

$$\begin{aligned} f(x)&=x\log x -x\\ f'(x)&=\log x +x\cdot\frac{1}{x}-1\\ &=\log x\\ f''(x)&=\frac{1}{x} \end{aligned}$$

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\{\arctan x\}&=\frac{1}{1+x^2}\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{1+x^2}\right\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{(1+x^2)^{-1}\right\}\\ &=(-1)(1+x^2)^{-2}(2x)\\&=\frac{-2x}{(1+x^2)^2} \end{aligned}$$

We use implicit differentiation, twice.

$$\begin{aligned} 2x+2yy'&=0\\ 2+(2y)y''+(2y')y'&=0\\ y''&=-\frac{(y')^2+1}{y} \\ \text{So, we need an expression for $y'$. We use the equation $2x+2yy'=0$ to conclude $y'=-\dfrac{x}{y}$:} \\ y''&=-\frac{\left(-\frac{x}{y}\right)^2+1}{y}\\ &=-\frac{\frac{x^2}{y^2}+1}{y}\\ &=-\frac{x^2+y^2}{y^3}\\ &=-\frac{1}{y^3} \end{aligned}$$

We differentiate using the chain rule.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\{\log(5x^2-12)\}&=\frac{10x}{5x^2-12} \\ \text{Using the quotient rule:} \\ \frac{\mathrm{d}^2}{\mathrm{d}{x}^2}\{\log(5x^2-12)\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{10x}{5x^2-12}\right\}\\ &=\frac{(5x^2-12)(10)-10x(10x)}{(5x^2-12)^2}\\ &=\frac{-10(5x^2+12)}{(5x^2-12)^2} \\ \text{Using the quotient rule one last time:} \\ \frac{\mathrm{d}^3}{\mathrm{d}{x}^3}\{\log(5x^2-12)\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \frac{-10(5x^2+12)}{(5x^2-12)^2} \right\}\\ &=\frac{(5x^2-12)^2(-10)(10x)+10(5x^2+12)(2)(5x^2-12)(10x)}{(5x^2-12)^4}\\ &=\frac{(5x^2-12)(-100x)+(200x)(5x^2+12)}{(5x^2-12)^3}\\ &=\frac{100x(-5x^2+12+10x^2+24)}{(5x^2-12)^3}\\ &=\frac{100x(5x^2+36)}{(5x^2-12)^3} \end{aligned}$$

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

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).

For (a) and (b), notice the following:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \sin x &= \cos x\\ \frac{\mathrm{d}}{\mathrm{d}x} \cos x &= -\sin x\\ \frac{\mathrm{d}}{\mathrm{d}x} \{-\sin x\} &= -\cos x\\ \frac{\mathrm{d}}{\mathrm{d}x} \{-\cos x\} &= \sin x\\ \frac{\mathrm{d}}{\mathrm{d}x} \sin x &= \cos x \\ \text{The fourth derivative is $\sin x$ is $\sin x$, and the fourth derivative of $\cos x$ is $\cos x$, so (a) and (b) are true.} \\ \frac{\mathrm{d}}{\mathrm{d}x}\tan x &=\sec^2 x\\ \frac{\mathrm{d}}{\mathrm{d}x}\sec^2 x &=2\sec x (\sec x \tan x)=2\sec^2x\tan x\\ \frac{\mathrm{d}}{\mathrm{d}x}\{2\sec^2x\tan x\}&=(4\sec x \cdot \sec x \tan x)\tan x+2\sec^2x\sec^2x\\ &=4\sec^2x\tan^2x+2\sec^4x\\ \frac{\mathrm{d}}{\mathrm{d}x}\{4\sec^2x\tan^2x+2\sec^4x\}&=(8\sec x \cdot \sec x \tan x)\tan^2x+4\sec^2x(2\tan x \cdot\sec^2x)\\&\quad+8\sec^3x\cdot\sec x \tan x\\ &=8\sec^2x\tan^3x+16\sec^4x\tan x \end{aligned}$$

So, $\displaystyle \frac{\mathrm{d}^4}{\mathrm{d}{x}^4} \tan x =8\sec^2x\tan^3x+16\sec^4x\tan x$. It certainly seems like this is not the same as $\tan x$, but remember that sometimes trig identities can fool you: $\tan^2x+1=\sec^2x$, and so on. So, to be absolutely sure that these are not equal, we need to find a value of $x$ so that the output of one is not the same as the output of the other. When $x=\frac{\pi}{4}$: $$8\sec^2x\tan^3x+16\sec^4x\tan x = 8\left({\sqrt{2}}\right)^2(1)^3+16\left({\sqrt{2}}\right)^4(1)=80\neq 1=\tan x.$$ So, (c) is false.