이계도함수와 오목·볼록

Concave up on all $x,$ no inflection points

a. Increasing for all $x$ where defined b. No local minima or maxima c. Concave up for $x<1;$ concave down for $x>1$ d. No inflection points in domain

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

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

We use implicit differentiation, twice.

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

The question asks for $s''(1)$. We start our differentiation using the quotient rule:

$$\begin{aligned} s'(t)&=\frac{e^t(t^2+1)-e^t(2t)}{(t^2+1)^2}\\ &=\frac{e^t(t^2-2t+1)}{(t^2+1)^2} \\ \text{Using the quotient rule again,} \\ s''(t)&=\frac{(t^2+1)^2 \frac{\mathrm{d}}{\mathrm{d}t}\{e^t(t^2-2t+1)\}-e^t(t^2-2t+1) \frac{\mathrm{d}}{\mathrm{d}t}\{(t^2+1)^2\}}{(t^2+1)^4}\\ &=\frac{(t^2+1)^2\cdot\left[ e^t(2t-2)+e^t(t^2-2t+1)\right]- e^t(t^2-2t+1) \cdot 2(t^2+1)(2t)}{(t^2+1)^4}\\ &=\frac{e^t(t^2+1)^2(t^2-1)-4te^t(t-1)^2(t^2+1)}{(t^2+1)^4}\\ s''(1)&=0 \end{aligned}$$

We differentiate using the chain rule.

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

The velocity of the particle is given by $h'(t)=\sin t$. Note $0<1<\pi$, so $h'(1)>0$–the particle is rising (moving in the positive direction, in this case “up"). The acceleration of the particle is $h''(t)=\cos t$. Since $0<1<\frac{\pi}{2}$, $h''(t)>0$, so $h'(t)$ is increasing: the particle is moving up, and it's doing so at an increasing rate. So, the particle is speeding up.

For this problem, remember that velocity has a sign indicating direction, while speed does not.

The velocity of the particle is given by $h'(t)=3t^2-2t-5$. At $t=1$, the velocity of the particle is $-4$, so the particle is moving downwards with a speed of 4 units per second. The acceleration of the particle is $h''(t)=6t-2$, so when $t=1$, the acceleration is (positive) $4$ units per second per second. That means the velocity (currently $-4$ units per second) is becoming a bigger number–since the velocity is negative, a bigger number is closer to zero, so the speed of the particle is getting smaller. (For instance, a velocity of $-3$ represents a slower motion than a velocity of $-4$.) So, the particle is slowing down at $t=1$.

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

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.

Inflection points occur where $f''(x)$ changes sign. Since $f(x)$ is a polynomial, its first and second derivatives exist everywhere, and are themselves polynomials. In particular,

$$\begin{aligned}f(x)&=3x^5-5x^4+13x\\ f'(x)&=15x^4-20x^3+13\\ f''(x)&=60x^3-60x^2=60x^2(x-1) \end{aligned}$$

The second derivative is negative for $x<1$ and positive for $x>1$. Thus the concavity changes between concave up and concave down at $x=1,\ y=11$.

This is the only inflection point. It is true that $f''(0)=0$, but for values of $x$ both a little larger than and a little smaller than 0, $f''(x)<0$, so the concavity does not change at $x=0$.