미분 전체 문제

$4$

$-\frac{3}{4}$

${f}^{'}(x)=3{x}^{2}$

${f}^{'}(x)=\frac{-5}{{x}^{2}}$

$T(x)=4x-5$

a. 5 ft/s b. 9 ft/s

a. $R(x)=10x-0.001{x}^{2}$ b. ${R}^{'}(x)=10-0.002x$ c. $6 per item, $0 per item

If $Q$ is to the left of the $y$ axis, the line through $Q$ and $P$ is increasing, so the secant line has positive slope. If $Q$ is to the right of the $y$ axis, the line through $Q$ and $P$ is decreasing, so the secant line has negative slope.

The slope of the secant line will be $\dfrac{f(2)-f(-2)}{2-(-2)} = \dfrac{f(2)-f(-2)}{4}$, in every part. So, if two lines have the same slope, that means their differences $f(2)-f(-2)$ will be the same.

The graphs in (a),(c), and (e) all have $f(2)-f(-2)=1$, so they all have the same secant line slope. The graphs in (b) and (f) both have $f(2)-f(-2)=-1$, so they both have the same secant line slope. The graph in (d) has $f(2)-f(-2)=0$, and it is the only graph with this property, so it does not share its secant line slope with any of the other graphs.

The function shown is a line, so it has a constant slope–(a) . Since the function is always increasing, $f'$ is always positive, so also (d) holds. Remark: it does not matter that the function itself is sometimes negative; the slope is always positive because the function is always increasing. Also, since the slope is constant, $f'$ is neither increasing nor decreasing: it is the function that is increasing, not its derivative.

The function is always decreasing, so $f'$ is always negative, option (e). However, the function alternates between being more and less steep, so $f'$ alternates between increasing and decreasing several times, and no other options hold.

Remark: $f$ is always positive, but (d) does not hold!

At the left end of the graph, $f$ is decreasing rapidly, so $f'$ is a strongly negative number. Then as we move towards $x=0$, $f$ decreases less rapidly, so $f'$ is a less strongly negative number. As we pass 0, $f$ increases, so $f'$ is a positive number. As we move to the right, $f$ increases more and more rapidly, so $f'$ is an increasing positive number. This description tells us that $f'$ increases for the entire range shown. So (b) holds, but not (a) or (c). Since $f'$ is negative to the left of the $y$ axis, and positive to the right of it, also (d) and (e) do not hold.

By definition, $f(x) = x^3$ is differentiable at $x = 0$ if the limit $$\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} =\lim_{h\rightarrow 0}\frac{h^3-0}{h}$$ exists.

$f'(-1)$ does not exist, because to the left of $x=-1$ the slope is a pretty big positive number (looks like around $+1$) and to the right the slope is $-1/4$. Since the derivative involves a limit, that limit needs to match the limit from the left and the limit from the right. The sharp angle made by the graph at $x=-1$ indicates that the left and right limits do not match, so the derivative does not exist.

$f'(3)$ also does not exist. One way to see this is to notice that the function is discontinuous here. More viscerally, note that $f(3)=1$, so as we take secant lines with one endpoint $(3,1)$, and the other endpoint just to the right of $x=3$, we get slopes that are more and more strongly negative, as shown in the picture below. If we take the limit of the slopes of these secant lines as $x$ goes to $3$ from the right, we get $-\infty$. (This certainly doesn't match the slope from the left, which is $-\frac{1}{4}$.)

At $x=-3$, there is some kind of “change" in the graph; however, it is a smooth curve, so the derivative exists here.

Using the definition of the derivative, $$s'(t)=\lim_{h \to 0}\frac{s(t+h)-s(t)}{h}$$ The units of the numerator are meters, and the units of the denominator are seconds (since the denominator comes from the change in the input of the function). So, the units of $s'(t)$ are metres per second.

Remark: we learned that the derivative of a position function gives velocity. In this example, the position is given in metres, and the velocity is measured in metres per second.

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 $g'(x)=3f'(x)$.

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:

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