미분 전체 문제

A good approximation from the graph is $f(5)=0.5$. We want to find a secant line whose endpoints are both very close to $x=5$, but that also give us clear $y$-values. It looks like $f(5.25) \approx 1$, and $f(4.75)\approx \frac{1}{8}$. The secant line from $x=5$ to $x=5.25$ has approximate slope $\dfrac{f(5.25)-f(5)}{5.25-5}\approx \dfrac{1-.5}{.25}=2$. The secant line from $x=5$ to $x=4.75$ has approximate slope $\dfrac{0.5-\frac{1}{8}}{5-4.75}=\dfrac{3}{2}$.

The graph increases more and more quickly (gets steeper and steeper), so it makes sense that the secant line to the left of $x=5$ has a smaller slope than the secant line to the right of $x=5$. Also, if you're taking secant lines that have endpoints farther out from $x=5$, you'll notice that the slopes of the secant lines change quite dramatically. You have to be very, very close to $x=5$ to get any kind of accuracy.

If we split the difference, we might approximate the slope of the secant line to be the average of $\frac{3}{2}$ and $2$, which is $\frac{7}{4}$.

Another way to try to figure out the tangent line is by carefully drawing it in with a ruler. This is shown here in blue:

It's much easier to take the slope of a line than a curve, and this one looks like it has slope about 1.5. However, we drew this with a computer: by hand it's much harder to draw an accurate tangent line. (That's why we need calculus!)

The actual slope of the tangent line to the function at $x=5$ is about $1.484$. This is extremely hard to figure out just from the graph–by hand, a guess between $1.25$ and $1.75$ would be very accurate.

There is only one tangent line to $f(x)$ at $P$ (shown in blue), but there are infinitely many choices of $Q$ and $R$ (one possibility shown in red). One easy way to sketch the secant line on paper is to draw any line parallel to the tangent line, and choose two intercepts with $y=f(x)$.

Any place the graph looks flat (if you imagine zooming in) is where the tangent line has slope 0. This occurs three times.

Notice that two of the indicated points are at a low point and a high point, respectively. Later, we'll use these places where the tangent line has slope zero to find where a graph achieves its biggest and smallest values.

We can use point-slope form to get the equation of the line, if we have a point and its slope. The point is given: $(1,6)$. The slope is the derivative:

$$\begin{aligned} y'(1)&=\lim_{h \rightarrow 0}\frac{y(1+h)-y(1)}{h}\\ &=\lim_{h \rightarrow 0}\frac{[(1+h)^3+5]-[1^3+5]}{h}\\ &=\lim_{h \rightarrow 0}\frac{[1+3h+3h^2+h^3+5]-[1+5]}{h}\\ &=\lim_{h \rightarrow 0}\frac{3h+3h^2+h^3}{h}\\ &=\lim_{h \rightarrow 0}{3+3h+h^2}\\ &=3 \end{aligned}$$

So our slope is 3, which gives a line of equation $y-6=3(x-1)$.

We set up the definition of the derivative.

$$\begin{aligned} f'(x)&=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h \rightarrow 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}\\ &=\lim_{h \rightarrow 0}\frac{\frac{x}{x(x+h)}-\frac{x+h}{x(x+h)}}{h}\\ &=\lim_{h \rightarrow 0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}\\ &=\lim_{h \rightarrow 0}\frac{\frac{-h}{x(x+h)}}{h}\\ &=\lim_{h \rightarrow 0}\frac{-1}{x(x+h)}\\ &=\frac{-1}{x^2} \end{aligned}$$

By definition $$f'(0)=\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} =\lim_{h\rightarrow 0}\frac{h|h|}{h} =\lim_{h\rightarrow 0}|h|=0$$ In particular, the limit exists, so the derivative exists (and is equal to zero).

We set up the definition of the derivative.

$$\begin{aligned} f'(x)&=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} =\lim_{h\rightarrow 0}\frac{1}{h}\Big(\frac{2}{x+h+1} -\frac{2}{x+1}\Big) =\lim_{h\rightarrow 0}\frac{2}{h}\ \frac{(x+1)-(x+h+1)}{(x+h+1)(x+1)}\\ &=\lim_{h\rightarrow 0}\frac{2}{h}\ \frac{-h}{(x+h+1)(x+1)} =\lim_{h\rightarrow 0}\frac{-2}{(x+h+1)(x+1)} =\frac{-2}{(x+1)^2} \end{aligned}$$

$$\begin{aligned} f'(x)&=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} =\lim_{h\rightarrow 0}\frac{1}{h}\Big(\frac{1}{(x+h)^2+3} -\frac{1}{x^2+3}\Big) =\lim_{h\rightarrow 0}\frac{1}{h}\frac{x^2-(x+h)^2}{[(x+h)^2+3][x^2+3]}\\ &=\lim_{h\rightarrow 0}\frac{1}{h}\frac{-2xh-h^2}{[(x+h)^2+3][x^2+3]} =\lim_{h\rightarrow 0}\frac{-2x-h}{[(x+h)^2+3][x^2+3]} =\boxed{\frac{-2x}{[x^2+3]^2}} \end{aligned}$$

The slope of the tangent line is the derivative. We set this up using the same definition of the derivative that we always do. This limit is hard to take for general $x$, but easy when $x=0$.

$$\begin{aligned} f'(0)&=\lim_{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\\ &=\lim_{h \rightarrow 0} \frac{h\log_{10}(2h+10)-0}{h}\\ &=\lim_{h \rightarrow 0} \log_{10}(2h+10)=\log_{10}(10)=1 \end{aligned}$$

So, the slope of the tangent line is 1.

$$\begin{aligned} f'(x)&=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} =\lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h} =\lim_{h\rightarrow 0}\frac{x^2-(x+h)^2}{(x+h)^2x^2h} =\lim_{h\rightarrow 0}\frac{-2xh-h^2}{(x+h)^2x^2h}\\ &=\lim_{h\rightarrow 0}\frac{-2x-h}{(x+h)^2x^2}=\frac{-2x}{x^4} =-\frac{2}{x^3} \end{aligned}$$

When $x$ is not equal to 2, then the function is differentiable– the only place we have to worry about is when $x$ is exactly 2.

In order for $f$ to be differentiable at $x=2$, it must also be continuous at $x=2$. This forces $x^2\big|_{x=2}=\big[ax+b\big]_{x=2}$ or $$2a+b=4.$$

In order for a limit to exist, the left- and right-hand limits must exist and be equal to each other. Since a derivative is a limit, in order for $f$ to be differentiable at $x=2$, the left hand derivative of $ax+b$ at $x=2$ must be the same as the right hand derivative of $x^2$ at $x=2$. Since $ax+b$ is a line, its derivative is $a$ everywhere. We've already seen the derivative of $x^2$ is $2x$, so we need $$a=2x\big|_{x=2}=4.$$

So, the values of $a$ and $b$ that makes $f$ differentiable everywhere are $a=4$ and $b=-4$.

We plug in $f(x)$ to the definition of a derivative. To evaluate the limit, we multiply and divide by the conjugate of the numerator, then simplify.

$$\begin{aligned} f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} &=\lim\limits_{h\rightarrow 0}\frac{\sqrt{1+x+h}-\sqrt{1+x}}{h}\\ &=\lim\limits_{h\rightarrow 0}\frac{\sqrt{1+x+h}-\sqrt{1+x}}{h} \left(\frac{\sqrt{1+x+h}+\sqrt{1+x}}{\sqrt{1+x+h}+\sqrt{1+x}}\right)\\ &=\lim\limits_{h\rightarrow 0}\frac{(1+x+h)-(1+x)}{h(\sqrt{1+x+h}+\sqrt{1+x})}\\ &=\lim\limits_{h\rightarrow 0}\frac{h}{h(\sqrt{1+x+h}+\sqrt{1+x})}\\ &=\lim\limits_{h\rightarrow 0}\frac{1}{\sqrt{1+x+h}+\sqrt{1+x}}\\ &=\frac{1}{\sqrt{1+x+0}+\sqrt{1+x}}=\frac{1}{2\sqrt{1+x}} \end{aligned}$$

The domain of the function is $[-1,\infty)$. In particular, $f(x)$ is defined when $x=-1$. However, $f'(x)$ is not defined when $x=-1$, so $f'(x)$ only exists over $(-1,\infty)$.

Remark: $\displaystyle \lim_{x \to -1^+}f'(x)=\infty$, so the tangent line to $f(x)$ at the point $x=-1$ has a vertical slope.

$p'(t) = \displaystyle\lim_{h \rightarrow 0}\frac{p(t+h)-p(t)}{h} \approx \frac{p(t+1)-p(t)}{1} = p(t+1)-p(t)$, or the difference in profit caused by the sale of the $(t+1)^{\mathrm{st}}$ widget. So, $p'(t)$ is the profit from the $(t+1)^{\mathrm{st}}$ widget. That is, $p'(t)$ is the profit per additional widget sold, when $t$ widgets are being sold. This is called the marginal profit per widget, when $t$ widgets are being sold.

How quickly the temperature is changing per unit change of depth, measured in degrees per metre. In an ordinary body of water, the temperature near the surface ($d=0$) is pretty variable, depending on the sun, but deep down it is more stable (unless there are heat sources). So, one might reasonably expect that $|T'(d)|$ is larger when $d$ is near 0.

$C'(w)=\displaystyle\lim_{h \rightarrow 0} \dfrac{C(w+h)-C(w)}{h} \approx \dfrac{C(w+1)-C(w)}{1}=C(w+1)-C(w)$, which is the number of calories in $C(w+1)$ grams minus the number of calories in $C(w)$ grams. This is the number of calories per additional gram, when there are $w$ grams.

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.

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

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