미분 전체 문제

The rate of change of velocity is acceleration. (If your velocity is increasing, you're accelerating; if your velocity is decreasing, you have negative acceleration.)

The rate of change in this case will be the relationship between the heat added and the temperature change. $\displaystyle\lim_{h \rightarrow 0} \dfrac{T(j+h)-T(j)}{h} \approx \dfrac{T(j+1)-T(j)}{1}=T(j+1)-T(j)$, or the change in temperature after the application of one joule. (This is closely related to heat capacity and to specific heat — there's a nice explanation of this on Wikipedia.)

As usual, it is instructive to think about the definition of the derivative: $$P'(T) = \displaystyle\lim_{h \rightarrow 0}\dfrac{P(T+h)=P(T)}{h} \approx \dfrac{P(T+1)-P(t)}{1} = P(T+1)-P(T).$$ This is the difference in population between two hypothetical populations, raised one degree in temperature apart. So, it is the number of extra individuals that exist in the hotter experiment (with the understanding that this number could be negative, as one would expect in conditions that are hotter than the bacteria prefer). So $P'(T)$ is the number of bacteria added to the colony per degree.

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

True: this is exactly what the Sum Rule states.

False, in general. The product rule tells us $\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)g(x)\}=f'(x)g(x)+f(x)g'(x)$. An easy example of why we can't do it the other way is to take $f(x)=g(x)=x$. Then the equation becomes $\frac{\mathrm{d}}{\mathrm{d}x}\{x^2\}=(1)(1)$, which is false.

True: the quotient rule tells us $$\frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{f(x)}{g(x)}\right\}=\frac{g(x)f'(x)-f(x)g'(x)}{g^2(x)} = \frac{g(x)f'(x)}{g^2(x)}-\frac{f(x)g'(x)}{g^2(x)} = \frac{f'(x)}{g(x)}-\frac{f(x)g'(x)}{g^2(x)}.$$

If you're creative, you can find lots of ways to differentiate!

Constant multiple: $g'(x)=3f'(x)$.

Product rule: $g'(x) = \frac{\mathrm{d}}{\mathrm{d}x}\{3\}f(x)+3f'(x)=0f(x)+3f'(x)=3f'(x)$.

Sum rule: $g'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)+f(x)+f(x)\}=f'(x)+f'(x)+f'(x)=3f'(x)$.

Quotient rule: $g'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{f(x)}{\frac{1}{3}}\right\}= \frac{\frac{1}{3}f'(x)-f(x)(0)}{\frac{1}{9}}=\frac{\frac{1}{3}f'(x)}{\frac{1}{9}}=9\left(\frac{1}{3}\right)f'(x)=3f'(x)$.

All rules give $g'(x)=3f'(x)$.

We already know that $\frac{\mathrm{d}}{\mathrm{d}x}x=1$ and $\frac{\mathrm{d}}{\mathrm{d}x}x^2=2x$, so we can compute the derivative of $x^3$ by writing $x^3=(x)(x^2)$,

$$\frac{\mathrm{d}}{\mathrm{d}x} x^3 =\frac{\mathrm{d}}{\mathrm{d}x} (x)(x^2) = (1)(x^2)+(x)(2x)=3x^2$$

When this is evaluated at $x=\frac{1}{2}$ we get $\frac{3}{4}$. Since we also compute $\left( \frac{1}{2}\right)^3=\frac{1}{8}$, the equation of the tangent line is

$$\begin{aligned} y - \frac{1}{8} = \frac{3}{4}\cdot \left(x-\frac{1}{2}\right). \end{aligned}$$

We can use the quotient rule here.

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

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.)

$$\begin{aligned} \\ \text{First expression, $f(x)=\dfrac{g(x)}{h(x)}$:} \\ f'(x)&=\frac{h(x)g'(x)-g(x)h'(x)}{h^2(x)} \\ \text{Second expresson, $f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)}$:} \\ f'(x)&=\left(\frac{k(x)g'(x)-g(x)k'(x)}{k^2(x)}\right)\left(\frac{k(x)}{h(x)}\right)+\left(\frac{g(x)}{k(x)}\right)\left(\frac{h(x)k'(x)-k(x)h'(x)}{h^2(x)}\right)\\ &=\frac{k(x)g'(x)-g(x)k'(x)}{k(x)h(x)}+ \frac{g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-h(x)g(x)k'(x)}{k(x)h^2(x)}+ \frac{g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-h(x)g(x)k'(x)+g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)g'(x)-g(x)h'(x)}{h^2(x)} \\ \text{and this is exactly what we got from differentiating the first expression.} \\ \end{aligned}$$

In the quotient rule, there is a minus, not a plus. Also, $2(x+1)+2x$ is not the same as $2(x+1)$.

The correct version is:

$$\begin{aligned} f(x)&=\frac{2x}{x+1}\\ f'(x)&=\frac{2(x+1) -2x}{(x+1)^2}\\ &=\frac{2}{(x+1)^2} \end{aligned}$$

$f(x)=\frac{2}{3}x^6+5x^4+12x^2+9$ is a polynomial:

$$\begin{aligned} f'(x)&=4x^5+20x^3+24x\\& =4x(x^4+5x^2+6)\\& = 4x((x^2)^2+5(x^2)+6)\\&=4x(x^2+2)(x^2+3) \end{aligned}$$

We can rewrite slightly to make every term into a power of $t$:

$$\begin{aligned} s(t)&=3t^4+5t^3-t^{-1}\\ s'(t)&=4\cdot 3t^{3}+3\cdot 5t^2-(-1)\cdot t^{-2}\\ &=12t^3+15t^2+\frac{1}{t^2} \end{aligned}$$

We could use the product rule here, but it's easier to simplify first. Don't be confused by the role reversal of $x$ and $y$: $x$ is the name of the function, and $y$ is the variable.

$$\begin{aligned} x(y) &= \left(2y+\frac{1}{y}\right)\cdot y^3\\ &=2y^4+y^2\\ x'(y)&=8y^3+2y \end{aligned}$$