6.2 Introduction to Derivatives

Average and Instantaneous Velocity, Revisited

We begin this section by revisiting (again!) the notion of average velocity - a concept we first encountered in Example 1.2.8 in Section 1.2 and later revisited in Example 3.1.3 in Section 3.1.

In this scenario, the position function¹ $s(t)=-5t^2+100t$, $0\le t\le 20$ gives the height of a model rocket above the Moon's surface, in feet, $t$ seconds after liftoff. The average rate of change of $s$ over an interval is the average velocity of the rocket over that interval. The average velocity provides two pieces of information: the average speed of the rocket along with the rocket's direction. We formalized the average velocity in Definition 3.5 in Section 3.1:

Interactive figureShrink the step h and the secant line settles onto the tangentDrag the Point a and Step h sliders.

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If we define the change in time, $\mathrm{\Delta }t=t-t_0$, we get $t=t_0+\mathrm{\Delta }t$ which gives:

$$\overline{v}(\mathrm{\Delta }t)=\frac{\mathrm{\Delta }[s(t)]}{\mathrm{\Delta }t}=\frac{s(t_0+\mathrm{\Delta }t)-s(t_0)}{\mathrm{\Delta }t},\mathrm{\Delta }t\ne 0.$$

The above formula measures the average velocity between time $t_0$ and time $t_0+\mathrm{\Delta }t$ as a function of $\mathrm{\Delta }t$.

We now revisit Example 3.1.3 in Section 3.1 using this new formulation.²

The reader is invited to compare Example 3.1.3 in Section 3.1 with Exercise 6.2.1 above. We obtain the exact same information because we are asking the exact same questions - they are just framed differently. We now take the time to formally define instantaneous velocity:

Based on our work in Examples 3.1.3 and 6.2.1, we have $v(15)=-50$. In both of those examples, we've seen what $v(15)$ means on the graph of $\overline{v}$, but there is a more important interpretation when we analyze the graph of $s$. Recall that the average velocity, and, more generally, average rates of change can be visualized as slopes of secant lines.³

Below is a sequence of secant lines along with the graph of $y=s(t)$. In each case, the secant line is graphed between $(15,s(15))=(15,375)$ and another point on the graph.⁴ As the points on the parabola approach $(15,375)$ the secant lines approach what is known as the tangent line.

To find the equation of the tangent line in this case, which we'll call $L(t)$, we refer to the point-slope form of a line, Equation 1.1:

$$L(t)=s\left(t_0\right)+m\left(t-t_0\right)=s(15)+v(15)(t-15)=375-50(t-15)=-50t+1125.$$

The tangent line can best be thought of as `the best linear approximation' to the graph of $y=s(t)$ at $(15,375)$. That is, if we zoom in near $(15,375)$, the graph of $y=s(t)$ and this tangent line become indistinguishable. This geometric property of $y=s(t)$ is called local linearity ⁵ and is foundational to the analysis of functions. Below we graph $y=s(t)=-5t^2+100t$ along with $y=L(t)=-50t+1125$ and observe the local linearity near $(15,375)$.

The tangent line at $(15,375)$.

Zooming in near $(15,375)$.

Zooming in closer to $(15,375)$.

Our next step is to generalize these notions to all functions.

Difference Quotients and Derivatives

Recall in Section 1.2.4 the concept of the average rate of change of a function over the interval $[a,b]$ is the slope between the two points $(a,f(a))$ and $(b,f(b))$ and is given by

$$\frac{\mathrm{\Delta }[f(x)]}{\mathrm{\Delta }x}=\frac{f(b)-f(a)}{b-a}.$$

Geometrically, the average rate of change is the slope of the so-called secant line which `cuts' through the graph of $y=f(x)$ at the points $(a,f(a))$ and $(b,f(b))$:

Consider a function $f$ defined over an interval containing $x$ and $x+h$ where $h\ne 0$. The average rate of change of $f$ over the interval $[x,x+h]$ is thus given by the formula:⁶

$$\frac{\mathrm{\Delta }[f(x)]}{\mathrm{\Delta }x}=\frac{f(x+h)-f(x)}{h},h\ne 0.$$

The above is an example of what is traditionally called the difference quotient or Newton quotient of $f$, since it is the quotient of two differences, namely $\mathrm{\Delta }[f(x)]$ and $\mathrm{\Delta }x$. Another formula for the difference quotient (as seen in Section 5.2.1) keeps with the notation $\mathrm{\Delta }x$ instead of $h$:

$$\frac{\mathrm{\Delta }[f(x)]}{\mathrm{\Delta }x}=\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x},\mathrm{\Delta }x\ne 0.$$

It is important to understand that in this formulation of the difference quotient, the variables `$x$' and `$\mathrm{\Delta }x$' are distinct - that is they do not combine as like terms.

Note that, regardless of which form the difference quotient takes, when $h$, $\mathrm{\Delta }x$, or $\mathrm{\Delta }t$ is $0$, the difference quotient returns the indeterminate form `$\frac{0}{0}$.' As we've seen with rational functions in Section 3.1, when this happens, we can use a limit to help us determine the indeterminate form.

In Section 6.2.1, taking the limit of average velocity as $\mathrm{\Delta }t\to 0$ produced instantaneous velocity. More generally, taking the limit of the average rate of change as the denominator approaches $0$ produces the instantaneous rate of change of the function at that point. The instantaneous rate of change of a function, called the derivative of the function, is defined below.

Using the language of derivatives, Examples 3.1.3 and 6.2.1 have us computing $v(15)=s^{\prime }(15)$. Moreover, since the derivative is a rate of change, it's important to note that the associated units of $f^{\prime }(a)$ are $\frac{\text{units of }f(x)}{\text{units of }x}$. This tracks with the units of $v(15)=s^{\prime }(15)$ being $\frac{\text{feet}}{\text{second}}$, a velocity.

As in Section 6.2.1, $f^{\prime }(a)$ represents the slope of the tangent line at the point $(a,f(a))$. We use this to formally define the tangent line below.

We put these definitions to good use in the following example.

What if we wanted to find the equation of the tangent line to the graph of the function in Example 6.2.2 at $x=0$? $x=1$? $x=5$? We'd ostensibly need to run through difference quotients and limit calculations for each and every input value: $x=0$, $x=1$, and $x=5$. Or we could do a single limit with a generic `$x$', simplify the difference quotient and take the limit once, and substitute in particular values of $x$:

It is worth noting that if we set $h=\mathrm{\Delta }x$, and consider the graph $y=f(x)$, we get:

$$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }[f(x)]}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x},$$

which is why sometimes the derivative is denoted⁷ $\frac{dy}{dx}$.

The astute reader will note that the graph of $f(x)=-x^2+3x-1$ in Example 6.2.3 is a parabola and finding where $f^{\prime }(x)=0$ lead us right back to the vertex. Using a derivative to find the vertex may seem a bit excessive given that we've algebraically derived a handy `vertex formula' in Section 1.4. However, as the functions we aim to analyze become more and more sophisticated, the tools we use to analyze them must also become more sophisticated. The derivative is one such tool that has a near universal application.⁸

Exercises

In Exercises -, find the limit of the following difference quotients.

• $\underset{h\to 0}{lim}\frac{f(2+h)-f(2)}{h}$
• $\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

1. $f(x)=2x-5$
2. $f(x)=-3x+5$
3. $f(x)=6$
4. $f(x)=3x^2-x$
5. $f(x)=-x^2+2x-1$
6. $f(x)=4x^2$

In Exercises -, find:

• $$f^{\prime }(2)=\underset{h\to 0}{lim}\frac{f(2+h)-f(2)}{h}$$
• The equation of the tangent line at $(2,f(2))$. Check your answer graphically.
• $$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$$
• The equation of the tangent line at $(0,f(0))$. Check your answer graphically.
• $f(x)=x-x^2$
• $f(x)=x^3+1$
• Find $$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$$ for $f(x)=mx+b$ where $m\ne 0$
• 1. Find $$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$$ for $f(x)=a{x}^2+bx+c$ where $a\ne 0$.
  2. Solve $f^{\prime }(x)=0$ for $x$. Does this look familiar? Explain.

In Exercises -, find the limit of the following difference quotients:

• $\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(-1+\mathrm{\Delta }x)-f(-1)}{\mathrm{\Delta }x}$
• $\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}$
• $f(x)=\frac{2}{x}$
• $f(x)=\frac{3}{1-x}$
• $f(x)=\frac{1}{x^2}$
• $f(x)=\frac{2}{x+5}$

In Exercises -, find the limit of the following:

• $$f^{\prime }(-1)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(-1+\mathrm{\Delta }x)-f(-1)}{\mathrm{\Delta }x}$$
• The equation of the tangent line at $(-1,f(-1))$.
• $$f^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}$$
• The equation of the tangent line at $(0,f(0))$.
• $f(x)=\frac{1}{4x-3}$
• $f(x)=\frac{3x}{x+2}$
• $f(x)=\frac{x}{x-9}$
• $f(x)=\frac{x^2}{2x+1}$

In Exercises -, find the limit of the following difference quotients:

• $\underset{\mathrm{\Delta }t\to 0}{lim}\frac{g(\mathrm{\Delta }t)-g(0)}{\mathrm{\Delta }t}$
• $\underset{\mathrm{\Delta }t\to 0}{lim}\frac{g(t+\mathrm{\Delta }t)-g(t)}{\mathrm{\Delta }t}$
• $g(t)=\sqrt{9-t}$
• $g(t)=\sqrt{2t+1}$

In Exercises -, find the following:

• $$g^{\prime }(0)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{g(\mathrm{\Delta }t)-g(0)}{\mathrm{\Delta }t}$$
• The equation of the tangent line at $(0,g(0))$.
• $$g^{\prime }(t)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{g(t+\mathrm{\Delta }t)-g(t)}{\mathrm{\Delta }t}$$
• The equation of the tangent line at $(1,g(1))$.
• $g(t)=\sqrt{-4t+5}$
• $g(t)=\sqrt{4-t}$

• For $g(t)=t\sqrt{t}$:

  1. Explain why $$g^{\prime }(0)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{g(\mathrm{\Delta }t)-g(0)}{\mathrm{\Delta }t}$$ does not exist.
  2. Find the derivative from the right at $t=0$: $$g_{+}^{\prime }(0)=\underset{\mathrm{\Delta }t\to 0^{+}}{lim}\frac{g(\mathrm{\Delta }t)-g(0)}{\mathrm{\Delta }t}$$
  3. Find $$y=g_{+}^{\prime }(0)(x-0)+g(0)$$ and interpret.
  4. Find $$g^{\prime }(t)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{g(t+\mathrm{\Delta }t)-g(t)}{\mathrm{\Delta }t}$$. Assume $t>0$.
• 1. Find $$g^{\prime }(t)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{g(t+\mathrm{\Delta }t)-g(t)}{\mathrm{\Delta }t}$$ for $g(t)=\sqrt{at+b}$, $a\ne 0$.
  2. What restrictions do you place on $t$ so your formula is valid?

• Let $f(x)=|x|$.

  1. Explain why $f$ is continuous at $x=0$.
  2. Show $f^{\prime }(0)$ does not exist by showing $$\underset{h\to 0^{-}}{lim}\frac{f(h)-f(0)}{h}=-1$$ but $$\underset{h\to 0^{+}}{lim}\frac{f(h)-f(0)}{h}=1$$.
  3. Graph $y=f(x)$ near $(0,0)$. Interpret your answer to number graphically.

  4. Find and simplify $$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{|x+h|-|x|}{h}$$ assuming $x\ne 0$.

     HINT: Consider the two cases $x>0$ and $x<0$ …

• Let $g(t)=\sqrt[3]{t}$.

  1. Explain why $g$ is continuous at $t=0$.
  2. Show $g^{\prime }(0)$ does not exist by showing $$\underset{\mathrm{\Delta }t\to 0}{lim}\frac{g(\mathrm{\Delta }t)-g(0)}{\mathrm{\Delta }t}=\mathrm{\infty }$$.
  3. Graph $y=g(t)$ near $(0,0)$. Interpret your answer to number graphically.

  4. Find and simplify $$g^{\prime }(t)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{g(t+\mathrm{\Delta }t)-g(t)}{\mathrm{\Delta }t}$$ assuming $t\ne 0$.

     HINT: $$(a-b)\left(a^2+ab+b^2\right)=a^3-b^3$$

• Let $h(x)=x^{\frac{2}{3}}$.

  1. Explain why $h$ is continuous at $x=0$.
  2. Show $h^{\prime }(0)$ does not exist by showing $$\underset{\mathrm{\Delta }x\to 0^{-}}{lim}\frac{h(\mathrm{\Delta }x)-h(0)}{\mathrm{\Delta }x}=-\mathrm{\infty }$$ and $$\underset{\mathrm{\Delta }x\to 0^{+}}{lim}\frac{h(\mathrm{\Delta }x)-h(0)}{\mathrm{\Delta }x}=\mathrm{\infty }$$
  3. Graph $y=h(x)$ near $(0,0)$. Interpret your answer to number graphically.

  4. Find and simplify $$h^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{h(x+\mathrm{\Delta }x)-h(x)}{\mathrm{\Delta }x}$$ assuming $x\ne 0$.

     HINT: $$(a-b)\left(a^2+ab+b^2\right)=a^3-b^3$$ and $$a^2-b^2=(a+b)(a-b)$$.

• Recall from Exercise in Section 1.4 that Carl's friend Jason participates in the Highland Games. In one event, the hammer throw, the height $h(t)$ in feet of the hammer above the ground $t$ seconds after Jason lets it go is modeled by the function $$h(t)=-16t^2+22.08t+6$$.

  1. Find and simplify a formula for the velocity of the hammer, $$v(t)=h^{\prime }(t)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{h(t+\mathrm{\Delta }t)-h(t)}{\mathrm{\Delta }t}$$.
  2. Find and interpret $v(0)$.
  3. Solve $v(t)=0$ and interpret.
  4. Find the velocity of the hammer when it hits the ground, rounded to three decimal places.

• In Exercise in Section 1.4, the average fuel economy $F(t)$ in miles per gallon (mpg) for passenger cars in the US $t$ years after 1980 is modeled by $$F(t)=-0.0076t^2+0.45t+16$$, $0\le t\le 28$.

  1. Find and simplify a formula for $$F^{\prime }(t)=\underset{h\to 0}{lim}\frac{F(t+h)-F(t)}{h}$$.
  2. Find and interpret $F^{\prime }(0)$, $F^{\prime }(5)$ and $F^{\prime }(10)$.
  3. Interpret the trend you observe in your answers to part.

• Let us return to Example 5.2.4 where $$C(x)=.03x^3-4.5x^2+225x+250$$ denotes the cost, in dollars, of producing $x$ PortaBoy game systems.

  1. Find and interpret $$C^{\prime }(75)=\underset{h\to 0}{lim}\frac{C(75+h)-C(75)}{h}$$.
  2. Recall in Exercise in Section 5.2, we found the marginal cost, $MC(75)=58.53$, which means it will cost an additional $\$58.53$ to produce the $76$th item. Compare $C^{\prime }(75)$ and $MC(75)$.

In Exercises -, match the graph of the function with a plausible graph of its derivative.

1. $y=f(x)$:

Graph A:

1. $y=g(x)$:

Graph B:

1. $y=h(x)$:

Graph C:

In Exercises -, match the graph of the function with a plausible graph of its derivative.

1. $y=f(x)$:

Graph A:

1. $y=g(x)$:

Graph B:

1. $y=h(x)$:

Graph C:

Answers

1. $\underset{h\to 0}{lim}2=2$, $\underset{h\to 0}{lim}2=2$
2. $\underset{h\to 0}{lim}-3=-3$, $\underset{h\to 0}{lim}-3=-3$,
3. $\underset{h\to 0}{lim}0=0$, $\underset{h\to 0}{lim}0=0$
4. $\underset{h\to 0}{lim}(3h+11)=11$, $$\underset{h\to 0}{lim}(6x+3h-1)=6x-1$$
5. $\underset{h\to 0}{lim}(-h-2)=-2$, $$\underset{h\to 0}{lim}(-2x-h+2)=-2x+2$$
6. $\underset{h\to 0}{lim}(4h+16)=16$, $\underset{h\to 0}{lim}(8x+4h)=8x$
7. • $$f^{\prime }(2)=\underset{h\to 0}{lim}(-h-3)=-3$$
   • $$y=f^{\prime }(2)(x-2)+f(2)=(-3)(x-2)+(-2)$$ so $y=-3x+4$.
   • $$f^{\prime }(x)=\underset{h\to 0}{lim}(-2x-h+1)=-2x+1$$
   • $$y=f^{\prime }(0)(x-0)+f(0)=(1)(x-0)+0$$ so $y=x$.
8. • $$f^{\prime }(2)=\underset{h\to 0}{lim}\left(h^2+6h+12\right)=12$$
   • $$y=f^{\prime }(2)(x-2)+f(2)=12(x-2)+9$$ so $y=12x-15$.
   • $$f^{\prime }(x)=\underset{h\to 0}{lim}\left(3x^2+3xh+h^2\right)=3x^2$$
   • $$y=f^{\prime }(0)(x-0)+f(0)=0(x-0)+1$$ so $y=1$.
9. $f^{\prime }(x)=\underset{h\to 0}{lim}m=m$.
10. • $$f^{\prime }(x)=\underset{h\to 0}{lim}(2ax+ah+b)=2ax+b$$.
    • Solving $f^{\prime }(x)=2ax+b=0$ for $x$ gives $x=-\frac{b}{2a}$ which is the formula for the $x$-coordinate of the vertex of the parabola $y=f(x)$. If we zoom in near the the vertex of a parabola, the graph becomes locally flat so it makes sense the slope of the tangent line, $f^{\prime }(x)=0$ there.
11. $\underset{\mathrm{\Delta }x\to 0}{lim}\frac{2}{\mathrm{\Delta }x-1}=-2$, $\underset{\mathrm{\Delta }x\to 0}{lim}\frac{-2}{x(x+\mathrm{\Delta }x)}=-\frac{2}{x^2}$
12. $\underset{\mathrm{\Delta }x\to 0}{lim}\frac{-3}{2(\mathrm{\Delta }x-2)}=\frac{3}{4}$, $$\underset{\mathrm{\Delta }x\to 0}{lim}\frac{3}{(x+\mathrm{\Delta }x-1)(x-1)}=\frac{3}{{(x-1)}^2}$$
13. $\underset{\mathrm{\Delta }x\to 0}{lim}\frac{2-\mathrm{\Delta }x}{{(\mathrm{\Delta }x-1)}^2}=2$, $\underset{\mathrm{\Delta }x\to 0}{lim}\frac{-(2x+\mathrm{\Delta }x)}{x^2{(x+\mathrm{\Delta }x)}^2}=-\frac{2}{x^3}$
14. $\underset{\mathrm{\Delta }x\to 0}{lim}\frac{-1}{2(\mathrm{\Delta }x+4)}=-\frac{1}{8}$, $$\underset{\mathrm{\Delta }x\to 0}{lim}\frac{-2}{(x+5)(x+\mathrm{\Delta }x+5)}=-\frac{2}{{(x+5)}^2}$$
15. • $$f^{\prime }(-1)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{4}{7(4\mathrm{\Delta }x-7)}=-\frac{4}{49}$$
    • $$y=f^{\prime }(-1)(x-(-1))+f(-1)=-\frac{4}{49}(x+1)+\left(-\frac{1}{7}\right)$$ so $y=-\frac{4}{49}x-\frac{11}{49}$.
    • $$f^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{-4}{(4x-3)(4x+4\mathrm{\Delta }x-3)}=-\frac{4}{{(4x-3)}^2}$$
    • $$y=f^{\prime }(0)(x-0)+f(0)=-\frac{4}{9}(x-0)+\left(-\frac{1}{3}\right)$$ so $y=-\frac{4}{9}x-\frac{1}{3}$.
16. • $$f^{\prime }(-1)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{6}{\mathrm{\Delta }x+1}=6$$
    • $$y=f^{\prime }(-1)(x-(-1))+f(-1)=6(x+1)+(-3)$$ so $y=6x+3$.
    • $$f^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{6}{(x+2)(x+\mathrm{\Delta }x+2)}=\frac{6}{{(x+2)}^2}$$
    • $$y=f^{\prime }(0)(x-0)+f(0)=\frac{3}{2}(x-0)+0$$ so $y=\frac{3}{2}x$.
17. • $$f^{\prime }(-1)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{9}{10(\mathrm{\Delta }x-10)}=-\frac{9}{100}$$
    • $$y=f^{\prime }(-1)(x-(-1))+f(-1)=-\frac{9}{100}(x+1)+\frac{1}{10}$$ so $y=-\frac{9}{100}x+\frac{1}{100}$.
    • $$f^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{-9}{(x-9)(x+\mathrm{\Delta }x-9)}=-\frac{9}{{(x-9)}^2}$$
    • $$y=f^{\prime }(0)(x-0)+f(0)=-\frac{1}{9}(x-0)+0$$ so $y=-\frac{1}{9}x$.
18. • $$f^{\prime }(-1)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }x}{2\mathrm{\Delta }x-1}=0$$
    • $$y=f^{\prime }(-1)(x-(-1))+f(-1)=(0)(x+1)+(-1)$$ so $y=-1$.
    • $$f^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{2x^2+2x\mathrm{\Delta }x+2x+\mathrm{\Delta }x}{(2x+1)(2x+2\mathrm{\Delta }x+1)}=\frac{2x^2+2x}{{(2x+1)}^2}$$
    • $$y=f^{\prime }(0)(x-0)+f(0)=(0)(x-0)+0$$ so $y=0$.
19. $\underset{\mathrm{\Delta }t\to 0}{lim}\frac{-1}{\sqrt{9-\mathrm{\Delta }t}+3}=-\frac{1}{6}$, $$\underset{\mathrm{\Delta }t\to 0}{lim}\frac{-1}{\sqrt{9-t-\mathrm{\Delta }t}+\sqrt{9-t}}=-\frac{1}{2\sqrt{9-t}}$$
20. $\underset{\mathrm{\Delta }t\to 0}{lim}\frac{2}{\sqrt{2\mathrm{\Delta }t+1}+1}=1$, $$\underset{\mathrm{\Delta }t\to 0}{lim}\frac{2}{\sqrt{2t+2\mathrm{\Delta }t+1}+\sqrt{2t+1}}=\frac{2}{2\sqrt{2t+1}}$$
21. • $$g^{\prime }(0)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{-4}{\sqrt{5-4\mathrm{\Delta }t}+\sqrt{5}}=-\frac{2}{\sqrt{5}}$$
    • $$y=g^{\prime }(0)(x-0)+g(0)=-\frac{2}{\sqrt{5}}(x-0)+\sqrt{5}$$ so $y=-\frac{2}{\sqrt{5}}x+\sqrt{5}$.
    • $$g^{\prime }(t)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{-4}{\sqrt{-4t-4\mathrm{\Delta }t+5}+\sqrt{-4t+5}}=-\frac{2}{\sqrt{-4t+5}}$$
    • $$y=g^{\prime }(1)(x-1)+g(1)=(-2)(x-1)+1$$ so $y=-2x+3$.
22. • $$g^{\prime }(0)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{-1}{\sqrt{4-\mathrm{\Delta }t}+2}=-\frac{1}{4}$$
    • $$y=g^{\prime }(0)(x-0)+g(0)=-\frac{1}{4}(x-0)+2$$ so $y=-\frac{1}{4}x+2$.
    • $$g^{\prime }(t)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{-1}{\sqrt{4-t-\mathrm{\Delta }t}+\sqrt{4-t}}=-\frac{1}{2\sqrt{4-t}}$$
    • $$y=g^{\prime }(1)(x-1)+g(1)=-\frac{1}{2\sqrt{3}}(x-1)+\sqrt{3}$$ so $y=-\frac{1}{2\sqrt{3}}x+\frac{7\sqrt{3}}{6}$.
23. 1. If $\mathrm{\Delta }t<0$, then $\sqrt{\mathrm{\Delta }t}$ is not a real number. Hence, $$g^{\prime }(0)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{g(\mathrm{\Delta }t)-g(0)}{\mathrm{\Delta }t}$$ does not exist.
    2. $$g_{+}^{\prime }(0)=\underset{\mathrm{\Delta }t\to 0^{+}}{lim}{(\mathrm{\Delta }t)}^{\frac{1}{2}}=0$$

    3. $$y=g_{+}^{\prime }(0)(x-0)+g(0)=(0)(x-0)+0$$ so $y=0$.

       This line is a tangent line to the graph of $y=t\sqrt{t}$ at $(0,0)$ for $t\ge 0$.

    4. $$g^{\prime }(t)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{3t^2+3t\mathrm{\Delta }t+{(\mathrm{\Delta }t)}^2}{{(t+\mathrm{\Delta }t)}^{3/2}+t^{3/2}}=\frac{3t^2}{2t^{3/2}}=\frac{3}{2}{t}^{1/2}$$ provided $t>0$.
24. 1. $$g^{\prime }(t)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{a}{\sqrt{at+a\mathrm{\Delta }t+b}+\sqrt{at+b}}=\frac{a}{2\sqrt{at+b}}$$.
    2. We are told $a\ne 0$. In order for the square roots to be happy, we assume $at+b>0$. Hence, $t>-\frac{b}{a}$ if $a>0$ and $t<-\frac{b}{a}$ if $a<0$.
25. 1. $f$ is continuous at $x=0$ since $$\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}|x|=0=|0|=f(0)$$.
    2. $$\underset{h\to 0^{-}}{lim}\frac{f(h)-f(0)}{h}=\underset{h\to 0^{-}}{lim}\frac{-h}{h}=-1$$, $$\underset{h\to 0^{+}}{lim}\frac{f(h)-f(0)}{h}=\underset{h\to 0^{+}}{lim}\frac{h}{h}=1$$
    3. Near $x=0$, the graph of $y=f(x)$ looks like a `$\vee$' shape¹² consisting of a line of slope $-1$ to the left of $x=0$ and a line with a slope of $+1$ to the right of $x=0$.
    4. $$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{|x+h|-|x|}{h}=-1$$ if $x<0$ and $$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{|x+h|-|x|}{h}=1$$ of $x>0$.
26. 1. $g$ is continuous at $t=0$ since $$\underset{t\to 0}{lim}g(t)=\underset{t\to 0}{lim}\sqrt[3]{t}=0=\sqrt[3]{0}=g(0)$$.
    2. $$\underset{\mathrm{\Delta }t\to 0}{lim}\frac{g(\mathrm{\Delta }t)-g(0)}{\mathrm{\Delta }t}=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{1}{{(\mathrm{\Delta }t)}^{\frac{2}{3}}}=\mathrm{\infty }$$
    3. The graph of $y=g(t)$ near $(0,0)$ is a vertical line.¹³ Since $g$ is increasing through $(0,0)$, the `slope' of this vertical line could be seen as $+\mathrm{\infty }$.
    4. $$g^{\prime }(t)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{1}{{(t+\mathrm{\Delta }t)}^{\frac{2}{3}}+{(t+\mathrm{\Delta }t)}^{\frac{1}{3}}{t}^{\frac{1}{3}}+t^{\frac{2}{3}}}=\frac{1}{3t^{\frac{2}{3}}}$$, $t\ne 0$.
27. 1. $$\underset{x\to 0}{lim}h(x)=\underset{x\to 0}{lim}{x}^{\frac{2}{3}}=0=0^{\frac{2}{3}}=h(0)$$.
    2. $$\underset{\mathrm{\Delta }x\to 0^{-}}{lim}\frac{h(\mathrm{\Delta }x)-h(0)}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0^{-}}{lim}\frac{1}{{(\mathrm{\Delta }x)}^{\frac{1}{3}}}=-\mathrm{\infty }$$ and $$\underset{\mathrm{\Delta }x\to 0^{+}}{lim}\frac{h(\mathrm{\Delta }x)-h(0)}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0^{+}}{lim}\frac{1}{{(\mathrm{\Delta }x)}^{\frac{1}{3}}}=\mathrm{\infty }$$
    3. The graph $y=h(x)$ near $(0,0)$ shows a steeply decreasing graph as we approach $(0,0)$ from the left followed by a steeply increasing graph as we approach $(0,0)$ from the right.¹⁴
    4. $$h^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{h(x+\mathrm{\Delta }x)-h(x)}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{2x+\mathrm{\Delta }x}{{(x+\mathrm{\Delta }x)}^{\frac{4}{3}}+{(x+\mathrm{\Delta }x)}^{\frac{2}{3}}{x}^{\frac{2}{3}}+x^{\frac{4}{3}}}=\frac{2}{3x^{\frac{1}{3}}}$$, $x\ne 0$.
28. 1. $$v(t)=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{h(t+\mathrm{\Delta }t)-h(t)}{\mathrm{\Delta }t}=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{-16{(\mathrm{\Delta }t)}^2-32t\mathrm{\Delta }t+22.08\mathrm{\Delta }t}{\mathrm{\Delta }t}=-32t+22.08$$.
    2. $$v(0)=-32(0)+22.08=22.08$$. This means initially (when Jason lets go of the hammer), the hammer is traveling upwards at $22.08$ feet per second.
    3. $v(t)=0$ when $t=\frac{22.08}{32}=0.69$. This means the (vertical) velocity zeros out $0.69$ seconds after Jason lets go of the hammer. In this scenario, this corresponds to when the hammer reaches its peak height
    4. We first find when the hammer hits the ground by solving $h(t)=0$. The positive answer here is $t\approx 1.612$ seconds. The velocity of the hammer is: $$v(1.612)=-32(1.612)+22.08=-29.504$$. The hammer hits the ground going (approximately) $29.504$ feet per second.¹⁵
29. 1. $$F^{\prime }(t)=\underset{h\to 0}{lim}\frac{F(t+h)-F(t)}{h}=\underset{h\to 0}{lim}\frac{-0.0076h^2-0.0152th+0.45h}{h}=-0.0152t+0.45$$.

    2. $F^{\prime }(0)=0.45$, so fuel economy was increasing at a rate of $0.45$ mpg per year in 1980.¹⁶

       $F^{\prime }(5)=0.374$, so fuel economy was increasing at a rate of $0.374$ mpg per year in 1985.

       $F^{\prime }(10)=0.298$, so fuel economy was increasing at a rate of $0.298$ mpg per year in 1990.

    3. Based on the model, we have that during the years 1980 - 1990, fuel economy was increasing, but less so as the decade wore on. Technical and cost limitations could be at work here.
30. 1. $$C^{\prime }(75)=\underset{h\to 0}{lim}\frac{C(75+h)-C(75)}{h}=\underset{h\to 0}{lim}\frac{0.03h^3+2.25h^2+56.25h}{h}=56.25$$. This means when producing 75 systems, the cost is increasing at a rate of $\$56.25$ per system.

    2. We see that $C^{\prime }(75)$ is numerically close to $MC(75)$ but the former is a rate of change (measured in dollars per system) where the latter is a change (measured in dollars). Note that if we set $h=1$ in the difference quotient:

       $$\frac{C(75+h)-C(75)}{h}=\frac{C(75+1)-C(75)}{1}=C(76)-C(75)$$

       we see these two quantities can be used to approximate each other.

31. $f^{\prime }(x)$ is Graph B
32. $g^{\prime }(x)$ is Graph C
33. $h^{\prime }(x)$ is Graph A
34. $f^{\prime }(x)$ is Graph B
35. $g^{\prime }(x)$ is Graph A
36. $h^{\prime }(x)$ is Graph C