미분법 전체 문제

$$\begin{aligned} f(x)&=\sqrt[3]{\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }}\\ &=\left({\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }}\right)^{\frac{1}{3}} \\ \text{To begin the differentiation, we can choose our ``outside" function to be $g(x)=x^{\frac{1}{3}}$, and our ``inside" function to be $h(x)=\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }$. Then $f(x)=g(h(x))$, so $f'(x)=g'(h(x))\cdot h'(x)=\frac{1}{3}(h(x))^{-\frac{2}{3}}h'(x)$:} \\ f'(x)&=\frac{1}{3}\left( \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right)^{\frac{-2}{3}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right\}\\ &=\frac{1}{3} \left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right\} \\ \text{This leads us to use the quotient rule:} \\ &=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \left(\frac{ \sqrt{x^3-9}\tan x \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}-e^{\csc x^2}\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\tan x \right\}}{(\tan^2 x)(x^3-9) }\right) \\ \text{Let's figure out those two derivatives on their own, then plug them in. Using the chain rule twice:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}&=e^{ \csc x^2}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \csc x^2\right\}=e^{\csc x^2}\cdot (-\csc( x^2)\cot( x^2))\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ x^2\}\\&=-2xe^{\csc x^2}\csc(x^2)\cot(x^2) \\ \text{For the other derivative, we start with the product rule, then chain:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \sqrt{x^3-9} \tan x \right\}&= \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\right\}\cdot\tan x+\sqrt{x^3-9}\sec^2 x\\ &=\frac{1}{2\sqrt{ x^3-9}}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^3-9\right\}\cdot \tan x+\sqrt{x^3-9}\sec^2 x\\ &=\frac{3x^2 \tan x}{2\sqrt{{x^3-9}}}+\sqrt{x^3-9}\sec^2 x \\ \text{Now, we plug these into our equation for $f'(x)$:} \\ f'(x)&=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \left(\frac{ \sqrt{x^3-9}\tan x \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}-e^{\csc x^2} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\tan x \right\}}{(\tan^2 x)(x^3-9) }\right)\\ &=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}}\cdot\\ &~\left(\frac{ \sqrt{x^3-9}\tan x (-2x)e^{\csc x^2}\csc(x^2)\cot(x^2)-e^{\csc x^2} \left(\frac{3x^2\tan x}{2\sqrt{{x^3-9}}}+\sqrt{x^3-9}\sec^2 x\right)}{(\tan^2 x)(x^3-9) }\right) \end{aligned}$$

Let $g(x):=\log(f(x))$. Notice $g'(x)=\frac{f'(x)}{f(x)}$.

In order to show that the two curves have horizontal tangent lines at the same values of $x$, we will show two things: first, that if $f(x)$ has a horizontal tangent line at some value of $x$, then also $g(x)$ has a horizontal tangent line at that value of $x$. Second, we will show that if $g(x)$ has a horizontal tangent line at some value of $x$, then also $f(x)$ has a horizontal tangent line at that value of $x$.

Suppose $f(x)$ has a horizontal tangent line where $x=x_0$ for some point $x_0$. This means $f'(x_0)=0$. Then $g'(x_0)=\frac{f'(x_0)}{f(x_0)}$. Since $f(x_0) \neq 0$, $\frac{f'(x_0)}{f(x_0)}=\frac{0}{f(x_0)}=0$, so $g(x)$ also has a horizontal tangent line when $x=x_0$. This shows that whenever $f$ has a horizontal tangent line, $g$ has one too.

Now suppose $g(x)$ has a horizontal tangent line where $x=x_0$ for some point $x_0$. This means $g'(x_0)=0$. Then $g'(x_0)=\frac{f'(x_0)}{f(x_0)}=0$, so $f'(x_0)$ exists and is equal to zero. Therefore, $f(x)$ also has a horizontal tangent line when $x=x_0$. This shows that whenever $g$ has a horizontal tangent line, $f$ has one too.

Remark: if we were not told that $f(x)$ gives only positive numbers, it would not necessarily be true that $f(x)$ and $\log(f(x))$ have horizontal tangent lines at the same values of $x$. If $f(x)$ had a horizontal tangent line at an $x$-value where $f(x)$ were negative, then $\log(f(x))$ would not exist there, let alone have a horizontal tangent line.

• First we find the $x$-ordinates where $y=0$.

$$\begin{aligned} x^2+(1)e^0 &= 5 \\ x^2 +1 &=5\\ x^2&=4 \end{aligned}$$

So $x=2,-2$.

• Now we use implicit differentiation to get $y'$ in terms of $x,y$:

$$\begin{aligned} 2x+(y+1)e^y\frac{\mathrm{d}y}{\mathrm{d}x}+e^y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \end{aligned}$$

• Now set $y=0$ to get

$$\begin{aligned} 2x+(0+1)e^0\frac{\mathrm{d}y}{\mathrm{d}x}+e^0\frac{\mathrm{d}y}{\mathrm{d}x}&=0\\ 2x+\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\mathrm{d}y}{\mathrm{d}x}&=0\\ 2x&=-2\frac{\mathrm{d}y}{\mathrm{d}x}\\ x&=-\frac{\mathrm{d}y}{\mathrm{d}x} \end{aligned}$$

• So at $(x,y)=(2,0)$ we have $y' = -2$,
• and at $(x,y)=(-2,0)$ we have $y' = 2$.

The slope of the tangent line is, of course, given by the derivative, so let's start by finding $\frac{\mathrm{d}y}{\mathrm{d}x}$ of both shapes.

$$\begin{aligned} \\ \text{For the circle, we differentiate implicitly} \\ 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \\ \text{and solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{y} \\ \text{For the ellipse, we also differentiate implicitly:} \\ 2x+6y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \\ \text{and solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{3y} \\ \text{What we want is a value of $x$ where both derivatives are equal. However, they might have different values of $y$, so let's let $y_1$ be the $y$-values associated with $x$ on the circle, and let $y_2$ be the $y$-values associated with $x$ on the ellipse. That is, $x^2+ y_1^2=1$ and $x^2+3y_2^2=1$. For the slopes at $(x,y_1)$ on the circle and $(x,y_2)$ on the ellipse to be equal, we need:} \\ -\frac{x}{y_1}&=-\frac{x}{3y_2}\\ x\left(\frac{1}{y_1}-\frac{1}{3y_2}\right)&=0 \\ \text{So $x=0$ or $y_1=3y_2$. Let's think about which $x$-values will have a $y$-coordinate of the circle be three times as large as a $y$-coordinate of the ellipse. If $y_1=3y_2$, $(x,y_1)$ is on the circle, and $(x,y_2)$ is on the ellipse, then $x^2+y_1^2 =x^2+(3y_2)^2=1$ and $x^2+3y_2^2=1$. In this case:} \\ x^2+9y_2^2&=x^2+3y_2^2\\ 9y_2^2&=3y_2^2\\ y_2&=0\\ x&=\pm 1 \end{aligned}$$

We need to be a tiny bit careful here: when $y=0$, $y'$ is not defined for either curve. For both curves, when $y=0$, the tangent lines are vertical (and so have no real-valued slope!). Two vertical lines are indeed parallel.

So, for $x=0$ and for $x=\pm1$, the two curves have parallel tangent lines.

Since $f'(x)<0$, we need a decreasing function. This only applies to (ii), (iii), and (v). Since $f''(x)>0$, that means $f'(x)$ is increasing, so the slope of the function must be increasing. In (v), the slope is constant, so $f''(x)=0$–therefore, it's not (v). In (iii), the slope is decreasing, because near $a$ the curve is quite flat ($f'(x)$ near zero) but near $b$ the curve is very steeply decreasing ($f'(x)$ is a large negative number), so (iii) has a negative second derivative. By contrast, in (ii), the line starts out as steeply decreasing ($f'(x)$ is a strongly negative number) and becomes flatter and flatter ($f'(x)$ nears 0), so $f'(x)$ is increasing–in other words, $f''(x)>0$. So, (ii) is the only curve that has $f'(x)<0$ and $f''(x)>0$.

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.