미분 전체 문제

We've already seen that $\frac{\mathrm{d}}{\mathrm{d}x}\{\sqrt{x}\}=\frac{1}{2\sqrt{x}}$, but if you forget this formula it is easy to figure out: $\sqrt{x}=x^{1/2}$, so $\frac{\mathrm{d}}{\mathrm{d}x}\{\sqrt{x}\}=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$.

Using the quotient rule:

$$\begin{aligned} T(x) &= \dfrac{\sqrt{x}+1}{x^2+3}\\ T'(x)&=\frac{(x^2+3)\left(\frac{1}{2\sqrt{x}}\right)-(\sqrt{x}+1)(2x)}{(x^2+3)^2} \end{aligned}$$

We use quotient rule:

$$\begin{aligned} \frac{(x^2+3)\cdot 7 - 2x\cdot (7x+2)}{(x^2+3)^2}=\frac{21 - 4x - 7x^2}{(x^2+3)^2} \end{aligned}$$

Instead of multiplying to get our usual form of this polynomial, we can use the product rule. If $f_1(x)=3x^3+4x^2+x+1$ and $f_2(x)=2x+5$, then $f_1'(x)=9x^2+8x+1$ and $f_2'(x)=2$. Then

$$\begin{aligned} f'(0)&=f_1'(0)f_2(0)+f_1(0)f_2'(0)\\ &=(1)(5)+(1)(2)=7\end{aligned}$$

Using the quotient rule, $$f'(x) = \frac{(x^2+5x)(9x^2)-(3x^3+1)(2x+5)}{(x^2+5x)^2} = \frac{3x^4+30x^3-2x-5}{(x^2+5x)^2}$$

We use quotient rule:

$$\begin{aligned} \frac{(2-x)(6x)-(3x^2+5)(-1)}{(2-x)^2}=\frac{-3x^2+12x+5}{(x-2)^2} \end{aligned}$$

We use quotient rule:

$$\begin{aligned} \frac{(3x^2+5)(-2x) - (2-x^2)(6x)}{(3x^2+5)^2}=\frac{-22x}{(3x^2+5)^2} \end{aligned}$$

We use quotient rule:

$$\begin{aligned} \frac{6x^2\cdot (x+2)-(2x^3+1)\cdot 1}{(x+2)^2}=\frac{4x^3+12x^2-1}{(x+2)^2} \end{aligned}$$

The derivative of the function is

$$\begin{aligned} \frac{(1-x^2)\cdot\frac{1}{2\sqrt{x}} - \sqrt{x} \cdot (-2x)}{(1-x^2)^2} &= \frac{(1-x^2) - 2x \cdot (-2x)}{2\sqrt{x}(1-x^2)^2} \end{aligned}$$

The derivative is undefined if either $x<0$ or $x = 0,\pm 1$ (since the square-root is undefined for $x<0$ and the denominator is zero when $x=0,1,-1$. Putting this together — the derivative exists for $x>0, x\neq 1$.

Using the product rule seems faster than expanding.

$$\begin{aligned} f'(x)&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{3\sqrt[5]{x}+15\sqrt[3]{x}+8\right\}\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\frac{\mathrm{d}}{\mathrm{d}x}\left\{3x^2+8x-5\right\}\\ &=\frac{\mathrm{d}}{\mathrm{d}x}\left\{3{x}^{\frac{1}{5}}+15{x}^{\frac{1}{3}}+8\right\}\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\frac{\mathrm{d}}{\mathrm{d}x}\left\{3x^2+8x-5\right\}\\ &=\left(\frac{3}{5}{x}^{\frac{-4}{5}}+5{x}^{\frac{-2}{3}}\right)\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\left(6x+8\right) \end{aligned}$$

To avoid the quotient rule, we can divide through the denominator:

$$\begin{aligned} f(x)&=\dfrac{(x^2+5x+1)(\sqrt{x}+\sqrt[3]{x})}{x} =(x^2+5x+1)\dfrac{(\sqrt{x}+\sqrt[3]{x})}{x}\\ &=(x^2+5x+1)(x^{-1/2}+x^{-2/3}) \\ \text{Now, product rule:} \\ f'(x)&=(2x+5)(x^{-1/2}+x^{-2/3})+(x^2+5x+1)\left(\frac{-1}{2}x^{-3/2}-\frac{2}{3}x^{-5/3}\right) \end{aligned}$$

(If you simplified differently, or used the quotient rule, you probably came up with a different-looking answer. There is only one derivative, though, so all correct answers will look the same after sufficient algebraic manipulation.)

Denote by $m$ the slope of the common tangent, by $(x_1,y_1)$ the point of tangency with $y=x^2$, and by $(x_2,y_2)$ the point of tangency with $y=x^2-2x+2$. Then we must have $$y_1=x_1^2\qquad y_2=x_2^2-2x_2+2\qquad m=2x_1=2x_2-2=\frac{y_2-y_1}{x_2-x_1}$$ From the “$m$” equations we get $x_1=\frac{m}{2}$, $x_2=\frac{m}{2}+1$ and

$$\begin{aligned} m&=\frac{y_2-y_1}{x_2-x_1}\\ &=y_2-y_1\\ &=x_2^2-2x_2+2-x_1^2\\ \\& =(x_2-x_1)(x_2+x_1)-2(x_2-1)\\ &=\left(\frac{m}{2}+1-\frac{m}{2}\right)\left(\frac{m}{2}+1+\frac{m}{2}\right) -2\left(\frac{m}{2}+1-1\right) \\ &=(1)(m+1)-2\frac{m}{2}\\ &=1\\ \text{So,}~~~m&=1,\qquad x_1=\tfrac{1}{2},\qquad y_1=\frac{1}{4},\qquad x_2=\frac{3}{2},\qquad y_2=\frac{9}{4}-3+2=\frac{5}{4} \end{aligned}$$

An equation of the common tangent is $y=x-\frac{1}{4}$.

The line $y=mx+b$ is tangent to $y=x^2$ at $x=\alpha$ if $$2\alpha=m\text{ and }\alpha^2=m\alpha+b \iff m=2\alpha\text{ and }b=-\alpha^2$$ The same line $y=mx+b$ is tangent to $y=-x^2+2x-5$ at $x=\beta$ if

$$\begin{aligned} -2\beta+2=m&\text{ and }-\beta^2+2\beta-5=m\beta+b\\ \iff m=2-2\beta&\text{ and }b=-\beta^2+2\beta-5-(2-2\beta)\beta=\beta^2-5 \end{aligned}$$

For the line to be simultaneously tangent to the two parabolas we need $$m=2\alpha=2-2\beta\text{ and }b=-\alpha^2=\beta^2-5$$ Substituting $\alpha=1-\beta$ into $-\alpha^2=\beta^2-5$ gives $-(1-\beta)^2=\beta^2-5$ or $2\beta^2-2\beta-4=0$ or $\beta=-1,2$. The corresponding values of the other parameters are $\alpha=2,-1$, $m=4,-2$ and $b=-4,-1$. The two lines are $y=4x-4$ and $y=-2x-1$.

This limit represents the derivative computed at $x=2$ of the function $f(x)=x^{2015}$. To see this, simply use the definition of the derivative at $a=2$ with $f(x)=x^{2015}$:

$$\begin{aligned} \left.\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)\}\right|_{a} &= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}\\ \left.\frac{\mathrm{d}}{\mathrm{d}x}\{x^{2015}\}\right|_{2} &=\lim_{x\to2} \frac{x^{2015}-2^{2015}}{x-2} \end{aligned}$$

Since the derivative of $f(x)$ is $2015\cdot x^{2014}$, then its value at $x=2$ is exactly $2015\cdot 2^{2014}$.

We know the top speed of the caribou, so we can use this to give the minimum possible number of hours the caribou spent travelling during its migration. If the caribou travels at 70kph, it will take $(5000 \mathrm{km})\left(\dfrac{1 \mathrm{hr}}{70 \mathrm{km}}\right) \approx 71.4 \mathrm{hrs}$ to travel 5000 kilometres. Probably the caribou wasn't sprinting the whole time, so probably it took it longer than that, but we can only say for sure that the caribou spent at least about 71.4 hours migrating.

If $f(x)$ is the position of the crane at time $x$, measured in hours, then (if we let $x=0$ be the beginning of the day) we know that $f(24)-f(0)=240$. Since $f(x)$ is the position of the bird, $f(x)$ is continuous and differentiable. So, the MVT says there is a $c$ in $(0,24)$ such that $f'(x)=\dfrac{f(24)-f(0)}{24-0}=\dfrac{240}{24}=10$. That is, at some point $c$ during the day, the speed of the crane was exactly 10 kph.

Since $f(x)$ is differentiable for all $x \in (0,10)$, then $f(x)$ is also continuous for all $x \in (0,10)$. If $f(x)$ were continuous on the closed interval $[0,10]$, then the MVT would guarantee $f'(x)=\dfrac{f(10)-f(0)}{10-0}=1$ for some $c \in (0,10)$; however, this is not the case. So, it must be that $f(x)$ is continuous for all $x \in (0,10)$, but not for all $x \in [0,10]$.

Since $f'(c)=0$ for $c \in (0,10)$, that means $f$ is constant on that interval. So, $f(x)$ is a function like this:

where the height of the constant function can be anything.

So, one possible answer is $f(x) = \left\{\begin{array}{lr} 0&x \neq 10\\ 10&x=10 \end{array}\right.$.

(a) No such function is possible: Rolle's Theorem guarantees $f'(c)=0$ for at least one point $c \in (1,2)$.

For the other functions, examples are below, but many answers are possible.

The function $f(x)$ is continuous over all real numbers, but it is only differentiable when $x \neq 0$. So, if we want to apply the MVT, our interval must consist of only positive numbers or only negative numbers: the interval $(-4,13)$ is not valid.

It is possible to use the mean value theorem to prove what we want: if $a=1$ and $b=144$, then $f(x)$ is differentiable over the interval $(1,144)$ (since 0 is not contained in that interval), and $f(x)$ is continuous everywhere, so by the mean value theorem there exists some point $c$ where $f'(x)=\dfrac{\sqrt{|144|}-\sqrt{|1|}}{144-1}=\dfrac{11}{143}=\dfrac{1}{13}$.

That being said, an easier way to prove that a point exists is to simply find it–without using the MVT. When $x>0$, $f(x)=\sqrt{x}$, so $f'(x)=\dfrac{1}{2\sqrt {x}}$. Then $f'\left(\dfrac{169}{4}\right)=\dfrac{1}{13}$.