미분 전체 문제

The Mean Value Theorem does not apply.

True, examine derivative

The function is differentiable at $x=0$ if the following limit: $$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)-0}{x}=\lim_{x\to 0} \frac{f(x)}{x}$$ exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$). We start by computing the left limit. For this computation, recall that if $x<0$ then $\sqrt{x^2}=|x|=-x$. $$\lim_{x\to 0^-}\frac{f(x)}{x}=\lim_{x\to 0^-}\frac{\sqrt{x^2+x^4}}{x}=\lim_{x\to 0^-} \frac{\sqrt{x^2}\sqrt{1+x^2}}{x}=\lim_{x \rightarrow 0}\frac{-x\sqrt{1+x^2}}{x}=-1$$ Now, from the right: $$\lim_{x\to 0^+}\frac{x\cos x}{x}=\lim_{x\to 0^+}\cos x = 1.$$ Since the limit from the left does not equal the limit from the right, the derivative does not exist at $x=0$.

The function is differentiable at $x=0$ if the following limit: $$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)-0}{x}=\lim_{x\to 0} \frac{f(x)}{x}$$ exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$).

We start by computing the left limit.

$$\begin{aligned} \lim_{x\to 0^-}\frac{f(x)}{x}=\lim_{x\to 0^-} \frac{x\cos x}{x} =\lim_{x\to 0^-} \cos x = 1. \end{aligned}$$

Now, from the right:

$$\begin{aligned} \lim_{x\to 0^+}\frac{\sqrt{1+x}-1}{x} &= \lim_{x\to 0^+}\frac{\sqrt{1+x}-1}{x} \cdot \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1} \\ &= \lim_{x\to 0^+} \frac{1+x-1}{x(\sqrt{1+x}+1)} = \lim_{x\to 0^+} \frac{1}{\sqrt{1+x}+1} = \frac{1}{2} \end{aligned}$$

Since the limit from the left does not equal the limit from the right, the derivative does not exist at $x=0$.

The function is differentiable at $x=0$ if the following limit:

$$\begin{aligned} \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)-0}{x}=\lim_{x\to 0} \frac{f(x)}{x} \end{aligned}$$

exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$). We compute left and right limits; so

$$\begin{aligned} \lim_{x\to 0^-}\frac{f(x)}{x}=\lim_{x\to 0^-}\frac{x^3-7x^2}{x}=\lim_{x\to 0^-} x^2-7x=0 \end{aligned}$$

and

$$\begin{aligned} \lim_{x\to 0^+}\frac{x^3\cos\left(\frac{1}{x}\right)}{x}=\lim_{x\to 0^+}x^2\cdot \cos\left(\frac{1}{x}\right). \end{aligned}$$

This last limit equals $0$ by the Squeeze Theorem since

$$\begin{aligned} -1\le \cos\left(\frac{1}{x}\right)\le 1 \end{aligned}$$

and so,

$$\begin{aligned} -x^2\le x^2\cdot \cos\left(\frac{1}{x}\right)\le x^2, \end{aligned}$$

where in these inequalities we used the fact that $x^2\ge 0$. Finally, since $\lim_{x\to 0^+}-x^2=\lim_{x\to 0^+}x^2=0$, the Squeeze Theorem yields that also $\lim_{x\to 0^+}x^2\cos\left(\frac{1}{x}\right) =0$, as claimed.

Since the left and right limits match (they're both equal to $0$), we conclude that indeed $f(x)$ is differentiable at $x=0$ (and its derivative at $x=0$ is actually equal to $0$).

The function is differentiable at $x=1$ if the following limit: $$\lim_{x\to 1}\frac{f(x)-f(1)}{x-1} = \lim_{x\to 1}\frac{f(x)-0}{x-1}=\lim_{x\to 1} \frac{f(x)}{x-1}$$ exists (note that we used the fact that $f(1)=0$ as per the definition of the first branch which includes the point $x=0$). We compute left and right limits; so $$\lim_{x\to 1^-}\frac{f(x)}{x-1}=\lim_{x\to 1^-}\frac{4x^2-8x+4}{x-1}=\lim_{x\rightarrow 1^-}\frac{4(x-1)^2}{x-1} =\lim_{x\to 1^-} 4(x-1)=0$$ and $$\lim_{x\to 1^+}\frac{(x-1)^2\sin\left(\frac{1}{x-1}\right)}{x-1}=\lim_{x\to 1^+}(x-1)\cdot \sin\left(\frac{1}{x-1}\right).$$ This last limit equals $0$ by the Squeeze Theorem since $$-1\le \sin\left(\frac{1}{x-1}\right)\le 1$$ and so, $$-(x-1)\le (x-1)\cdot \sin\left(\frac{1}{x-1}\right)\le x-1,$$ where in these inequalities we used the fact that $x\to 1^+$ yields positive values for $x-1$. Finally, since $\lim_{x\to 1^+}-x+1=\lim_{x\to 1^+}x-1=0$, the Squeeze Theorem yields that also $\lim_{x\to 1^+}(x-1)\sin\left(\frac{1}{x-1}\right) =0$, as claimed.

Since the left and right limits match (they're both equal to $0$), we conclude that indeed $f(x)$ is differentiable at $x=1$ (and its derivative at $x=1$ is actually equal to $0$).

Many answers are possible; here is one.

The key is to realize that the few points you're given suggest a pattern, but don't guarantee it. You only know nine points; anything can happen in between.

We know that $y'=2x$. So, if we choose a point $(\alpha,\alpha^2)$ on the curve $y=x^2$, then the tangent line to the curve at that point has slope $2\alpha$. That is, the tangent line has equation

$$\begin{aligned} (y-\alpha^2)&=2\alpha(x-\alpha)\\ \text{simplified, } \qquad y&=(2\alpha)x-\alpha^2 \\ \text{So, if $(1,-3)$ is on the tangent line, then} \\ -3&=(2\alpha)(1)-\alpha^2\\ \iff\qquad 0&=\alpha^2-2\alpha-3\\ \iff\qquad 0&=(\alpha-3)(\alpha+1)\\ \iff\qquad \alpha&=3, \quad\text{or}\qquad \alpha=-1. \\ \text{So, the tangent lines $y=(2\alpha)x-\alpha^2$ are} \\ y&=6x-9 \quad\text{and}\quad y=-2x-1. \end{aligned}$$

$R'(t)$ is the rate at which the wheel is rotating measured in rotations per second. To convert to degrees, we multiply by 360: $\boxed{360R'(t)}$.

Population growth is rate of change of population. Population in year $2000+t$ is given by $P(t)=P_0+b(t)-d(t)$, where $P_0$ is the initial population of the town. Then $P'(t)$ is the expression we're looking for, and $P'(t)=b'(t)-d'(t)$.

It is interesting to note that the initial population does not obviously show up in this calculation. It would probably affect $b(t)$ and $d(t)$, but if we know these we do not need to know $P_0$ to answer our question.

We already know that $\frac{\mathrm{d}}{\mathrm{d}x}x^2=2x$. So the slope of $y=3x^2$ at $x=a$ is $6a$. The tangent line to $y=3x^2$ at $x=a, y=3a^2$ is $y-3a^2=6a(x-a)$. This tangent line passes through $(2,9)$ if

$$\begin{aligned} 9-3a^2&=6a(2-a)\\ 3a^2-12a+9&=0\\ a^2-4a+3&=0\\ (a-3)(a-1)&=0\\ \implies~~a&=1,3 \end{aligned}$$

The points are $(1,3),\ (3,27)$.

This limit represents the derivative computed at $x=100180$ of the function $f(x)=\sqrt{x}$. Since the derivative of $f(x)$ is $\dfrac{1}{2\sqrt{x}}$, then its value at $x=100180$ is exactly $\dfrac{1}{2\sqrt{100180}}$.

Let $w(t)$ and $l(t)$ be the width and length of the rectangle. Given in the problem is that $w'(t)=2$ and $l'(t)=5$. Since both functions have constant slopes, both must be lines. Their slopes are given, and their intercepts are $w(0)=l(0)=1$. So, $w(t)=2t+1$ and $l(t)=5t+1$.

The area of the rectangle is $A(t)=w(t)\cdot l(t)$, so using the product rule, the rate at which the area is increasing is $A'(t)=w'(t)l(t)+w(t)l'(t)=2(5t+1)+5(2t+1)=20t+7$ square metres per second.

Using the product rule, $f'(x)=(2x)g(x)+x^2g'(x)$, so $f'(0)=0\cdot g(x)+0\cdot g'(x)=0$. (Since $g$ is differentiable, $g'$ exists.)