극한·미분 심화 전체 문제

Using the chain rule,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\arcsin\left(\frac{x}{3}\right)\right\}&=\frac{1}{\sqrt{1-\left(\frac{x}{3}\right)^2}}\cdot \frac{1}{3}\\ &=\frac{1}{3\sqrt{1-\frac{x^2}{9}}}\\ &=\frac{1}{\sqrt{9-x^2}} \end{aligned}$$

Since the domain of arcsine is $[-1,1]$, and we are plugging in $\dfrac{x}{3}$ to arcsine, the values of $x$ that we can plug in are those that satisfy $-1 \le \dfrac{x}{3} \leq 1$, or $-3\leq x \leq 3$. So the domain of $f$ is $[-3,3]$.

Using the quotient rule,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\left\{\dfrac{\arccos t}{t^2-1}\right\}&= \frac{(t^2-1)\left(\frac{-1}{\sqrt{1-t^2}}\right)-(\arccos t)(2t)}{(t^2-1)^2} \end{aligned}$$

The domain of arccosine is $[-1,1]$, and since $t^2-1$ is in the denominator, the domain of $f$ requires $t^2-1 \neq 0$, that is, $t \neq \pm 1$. So the domain of $f(t)$ is $(-1,1)$.

The domain of $\mathop{\mathrm{arcsec}} x$ is $|x| \geq 1$: that is, we can plug into arcsecant only values with absolute value greater than or equal to one. Since $-x^2-2 \leq -2$, every real value of $x$ gives us an acceptable value to plug into arcsecant. So, the domain of $f(x)$ is all real numbers.

To differentiate, we use the chain rule. Remember $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\mathop{\mathrm{arcsec}} x\right\} = \dfrac{1}{|x|\sqrt{x^2-1}}$.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\mathop{\mathrm{arcsec}}(-x^2-2)\right\}&= \dfrac{1}{|-x^2-2|\sqrt{(-x^2-2)^2-1}}\cdot(-2x)\\ &=\dfrac{-2x}{(x^2+2)\sqrt{x^4+4x^2+3}}. \end{aligned}$$

We use the chain rule, remembering that $a$ is a constant.

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

The domain of arctangent is all real numbers, so the domain of $f(x)$ is also all real numbers.

We differentiate using the product and chain rules.

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

The domain of $\arcsin x$ is $[-1,1]$, and the domain of $\sqrt{1-x^2}$ is all values of $x$ so that $1-x^2 \geq 0$, so $x$ in $[-1,1]$. Therefore, the domain of $f(x)$ is $[-1,1]$.

We differentiate using the chain rule:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\{\arctan(x^2)\}&=\frac{2x}{1+x^4} \end{aligned}$$

This is zero exactly when $x=0$.

Using formulas you should memorize from this section, $$\frac{\mathrm{d}}{\mathrm{d}x}\{\arcsin x + \arccos x\}=\frac{1}{\sqrt{1-x^2}}+\frac{-1}{\sqrt{1-x^2}}=0$$

Remark: the only functions with derivative equal to zero everywhere are constant functions, so $\arcsin x + \arccos x$ should be a constant. Since $\sin \theta = \cos \left(\frac{\pi}{2}-\theta\right)$, we can set

$$\begin{aligned} \sin\theta&=x & \cos\left(\frac{\pi}{2}-\theta\right)&=x\\ \text{where $x$ and $\theta$ are the same in both expressions, and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$. Then} \\ \arcsin x &=\theta & \arccos x &= \frac{\pi}{2}-\theta \\ \text{We note here that arcsine is the inverse of the sine function \textit{ restricted to} $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. So, since we restricted $\theta$ to this domain, $\sin \theta=x$ really does imply $\arcsin x = \theta$. (For an example of why this matters, note $\sin(2\pi)=0$, but $\arcsin (0)=0 \neq 2\pi$.) Similarly, arccosine is the inverse of the cosine function restricted to $[0,\pi]$. Since $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, then $0 \leq (\frac{\pi}{2}-\theta) \leq \pi$, so $\cos\left(\frac{\pi}{2}- \theta\right) =x$ really does imply $\arccos x=\frac{\pi}{2}-\theta$.} \\ \end{aligned}$$

So, $$\arcsin x+\arccos x =\theta+\frac{\pi}{2}-\theta =\frac{\pi}{2}$$ which means the derivative we were calculating was actually just $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\dfrac{\pi}{2}\right\}=0$.

Using the chain rule, $$y'=\frac{-{\frac{1}{ x^2}}}{\sqrt{1-\left({\frac{1}{ x}}\right)^2}}=\frac{-1}{x^2\sqrt{1-\frac{1}{x^2}}}.$$

Using the chain rule, $$y'=\frac{-{\frac{1}{ x^2}}}{1+\left({\frac{1}{ x}}\right)^2}=\frac{-1}{x^2+1}.$$

Using the product rule:

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

Let $\theta = \arctan x$. Then $\theta$ is the angle of a right triangle that gives $\tan \theta = x$. In particular, the ratio of the opposite side to the adjacent side is $x$. So, we have a triangle that looks like this:

where the length of the hypotenuse came from the Pythagorean Theorem. Now, $$\sin\left(\arctan x\right) = \sin \theta = \frac{\text{opp}}{\text{hyp}} = \frac{x}{\sqrt{x^2+1}}$$ From here, we differentiate using the quotient rule:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{x}{\sqrt{x^2+1}} \right\}&= \frac{\sqrt{x^2+1}-x\frac{2x}{2\sqrt{x^2+1}}}{x^2+1}\\ &=\left(\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}\right)\cdot\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}\\ &=\frac{(x^2+1)-x^2}{(x^2+1)^{3/2}}\\ &=\frac{1}{(x^2+1)^{3/2}}=(x^2+1)^{-3/2} \end{aligned}$$

Remark: another strategy is to differentiate first, using the chain rule, then draw a triangle to simplify the resulting expression $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sin\left(\arctan x\right)\right\}=\dfrac{\cos(\arctan x)}{1+x^2}$.

Let $\theta = \arcsin x$. Then $\theta$ is the angle of a right triangle that gives $\sin \theta = x$. In particular, the ratio of the opposite side to the hypotenuse is $x$. So, we have a triangle that looks like this:

where the length of the adjacent side came from the Pythagorean Theorem. Now, $$\cot\left(\arcsin x\right) = \cot \theta = \frac{\text{adj}}{\text{opp}} = \frac{\sqrt{1-x^2}}{x}$$ From here, we differentiate using the quotient rule:

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

Remark: another strategy is to differentiate first, using the chain rule, then draw a triangle to simplify the resulting expression $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\cot\left(\arcsin x\right)\right\}=\frac{-\csc^2(\arcsin x)}{\sqrt{1-x^2}}$.

The line $y=2x+9$ has slope $2$, so we must find all values of $x$ between $-1$ and $1$ ($\arcsin x$ is only defined for these values of $x$) for which $\frac{\mathrm{d}}{\mathrm{d}x}\{\arcsin x\}=2$. Evaluating the derivative:

$$\begin{aligned} y&=\arcsin x\\ 2=y'&=\frac{1}{\sqrt{1-x^2}}\\ 4&=\frac{1}{1-x^2}\\ \frac{1}{4}&=1-x^2\\ x^2&=\frac{3}{4}\\ x&=\pm\frac{\sqrt{3}}{2}\\ (x,y)&=\pm\big(\frac{\sqrt{3}}{2},\frac{\pi}{3}\big) \end{aligned}$$

We differentiate using the chain rule:

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

So if $f'(x)=0$, then $\cos x=0$, and this happens when $x=\dfrac{(2n+1)\pi}{2}$ for any integer $n$. We should check that these points are in the domain of $f$. Arctangent is defined for all real numbers, so we only need to check the domain of cosecant; when $x=\dfrac{(2n+1)\pi}{2}$, then $\sin x=\pm1 \neq 0$, so $\csc x = \dfrac{1}{\sin x}$ exists.

We have no idea what $f(0.93)$ is, but 0.93 is pretty close to 1, and we definitely know $f(1)$. The linear approximation of $f(x)$ about $x=1$ is given by

$$\begin{aligned} f(x) &\approx f(1)+f'(1)(x-1) \\ \text{So, we calculate:} \\ f(1)&=\log(1)=0\\ f'(x)&=\frac{1}{x}\\ f'(1)&=\frac{1}{1}=1 \\ \text{Therefore,} \\ f(x) &\approx 0+1(x-1)=x-1 \\ \text{When $x=0.93$:} \\ f(0.93) &\approx 0.93-1=-0.07 \end{aligned}$$

We approximate the function $f(x)=\sqrt{x}$ about $x=4$, since 4 is a perfect square and it is close to 5.

$$\begin{aligned} f(4)&=\sqrt{4}=2\\ f'(x)&=\frac{1}{2\sqrt{x}} \qquad \Rightarrow \qquad f'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4}\\ f(x) &\approx f(4)+f'(4)(x-4)=2+\frac{1}{4}(x-4)\\ f(5)&\approx 2+\frac{1}{4}(5-4)=\frac{9}{4} \end{aligned}$$

We estimate $\sqrt{5}\approx\dfrac{9}{4}$.

Remark: $\left(\dfrac{9}{4}\right)^2=\dfrac{81}{16}$, which is pretty close to $\dfrac{80}{16}=5$. Our approximation seems pretty good.

We approximate the function $f(x)=\sqrt[5]{x}$. We need to centre the approximation about some value $x=a$ such that we know $f(a)$ and $f'(a)$, and $a$ is not too far from $30$.

$$\begin{aligned} f(x)&=\sqrt[5]{x}=x^{\frac{1}{5}}\\ f'(x)&=\frac{1}{5}x^{-\frac{4}{5}}=\frac{1}{5\sqrt[5]{x}^4} \\ \text{$a$ needs to be a number whose fifth root we know. Since $\sqrt[5]{32}=2$, and 32 is reasonably close to 30, $a=32$ is a great choice.} \\ f(32)&=\sqrt[5]{32}=2\\ f'(32)&=\frac{1}{5\cdot 2^4}=\frac{1}{80} \\ \text{The linear approximation of $f(x)$ about $x=32$ is} \\ f(x) &\approx 2+\frac{1}{80}(x-32) \\ \text{When $x=30$:} \\ f(30) &\approx 2+\frac{1}{80}(30-32)=2-\frac{1}{40}=\frac{79}{40} \end{aligned}$$