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).

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.

We differentiate a few time to find the pattern.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\{2^x\}&=2^x\log 2\\ \frac{\mathrm{d}^2}{\mathrm{d}{x}^2}\{2^x\}&=2^x\log2 \cdot \log 2 = 2^x(\log2)^2\\ \frac{\mathrm{d}^3}{\mathrm{d}{x}^3}\{2^x\}&=2^x(\log2)^2 \cdot \log 2 = 2^x(\log2)^3 \\ \text{Every time we differentiate, we multiply the original function by another factor of $\log 2$. So, the $n$th derivative is given by:} \\ \frac{\mathrm{d}^n}{\mathrm{d}{x}^n}\{2^x\}&=2^x(\log2)^n \end{aligned}$$

We differentiate using the power rule.

$$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}x}&=3ax^2+2bx+c\\ \frac{\mathrm{d}^2f}{\mathrm{d}{x}^2}&=6ax+2b\\ \frac{\mathrm{d}^3f}{\mathrm{d}{x}^3}&=6a\\ \frac{\mathrm{d}^4f}{\mathrm{d}{x}^4}&=0 \end{aligned}$$

In the above work, remember that $a$, $b$, $c$, and $d$ are all constants. Since they are nonzero constants, $\displaystyle \frac{\mathrm{d}^3f}{\mathrm{d}{x}^3} =6a\neq 0$. So, the fourth derivative is the first derivative to be identically zero: $n=4$.

Since $f(x)$ is differentiable over all real numbers, it is also continuous over all real numbers. Similarly, $f'(x)$ is differentiable over all real numbers, so it is also continuous over all real numbers, and so on for the first $n$ derivatives of $f(x)$.

Rolle's Theorem tells us that if $a$ and $b$ are distinct roots of a function $g$, then $g'(x)=0$ for some $c$ in $(a,b)$. That is, $g'$ has a root strictly between $a$ and $b$. Expanding this idea, if $g$ has $m+1$ distinct roots, then $g'$ must have at least $m$ distinct roots, as in the sketch below.

So, if $f^{(n)}(x)$ has only $m$ roots, then $f^{(n-1)}(x)$ has at most $m+1$ roots. Similarly, since $f^{(n-1)}(x)$ has at most $m+1$ roots, $f^{(n-2)}(x)$ has at most $m+2$ roots. Continuing in this way, we see $f(x)=f^{(n-n)}(x)$ has at most $m+n$ distinct roots.