다항함수의 미분법

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.)

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

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 $2015\cdot 2^{2014}$.