이계도함수와 오목·볼록

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.