## Section4.2The Derivative Function

In Section 4.1, we have seen how to create, or derive, a new function $f'(x)$ from a function $f(x)\text{,}$ and that this new function carries important information. In one example we saw that $f'(x)$ indicates how steep the graph of $f(x)$ is; in another we saw that $f'(x)$ tells us the velocity of an object if $f(x)$ represents the position of the object at time $x\text{.}$ As we said earlier, this same mathematical idea is useful whenever $f(x)$ represents some changing quantity and we want to know something about how it changes, or roughly, the “rate” at which it changes. Most functions encountered in practice are built up from a small collection of “primitive” functions in a few simple ways, for example, by adding or multiplying functions together to get new, more complicated functions. To make good use of the information provided by $f'(x)$ we need to be able to compute it for a variety of such functions.

We begin by formally defining the derivative.

###### Definition4.6. Definition of Derivative.

The derivative of a function $y=f(x)$ with respect to $x$ is the function $f'$ defined by

\begin{equation*} f'(x)=\lim_{\Delta x\to0} \dfrac{f(x + \Delta x) - f(x)}{\Delta x}\text{,} \end{equation*}

provided this limit exists.

Some textbooks use $h$ in place of $\Delta x$ in the definition of derivative:

\begin{equation*} f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\text{,} \end{equation*}

provided this limit exists.

The process of finding the derivative is called differentiation. There are several different notations in use for the derivative.

###### Derivative Notations.

Given a function $y = f(x)\text{,}$ its derivative function

\begin{equation*} f'(x)=\lim_{\Delta x\to0} \dfrac{f(x + \Delta x) - f(x)}{\Delta x}\text{,} \end{equation*}

provided it exists, can be written in the following ways:

The symbol $d/dx$ is called a differential operator which means to take the derivative of the function $f(x)$ with respect to the variable $x\text{.}$

The notation, $dy/dx\text{,}$ and its derivations remind us that the derivative is related to an actual slope between two points. This notation is called Leibniz notation, after Gottfried Leibniz, who developed the fundamentals of calculus independently at about the same time that Isaac Newton did. This notation has the added benefit that it indicates what we are differentiating with respect to, which is important in applications such as related rates (see Section 5.2), or multi-variable calculus (see Chapter 7).

Let us come back to our familiar function $y = \sqrt{625-x^{2}}$ to provide an example of how Leibniz notation is used. If the function $f(x)$ is written out in full we often write the derivative as something like this

\begin{equation*} f'(x)={d\over dx}\sqrt{625-x^2} \end{equation*}

with the function written to the side, instead of trying to fit it into the numerator.

###### Example4.7. Derivative of $y=t^2$ Using $\Delta t$ Notation.

Find the derivative of $\ds y=f(t)=t^2\text{.}$

Solution

We compute

\begin{equation*} \begin{array}{rll} y' \amp =\amp \ds\lim_{\Delta t\to0}\ds\frac{\Delta y}{\Delta t} \\ \\ \amp =\amp \ds\lim_{\Delta t\to0}\ds\frac{(t+\Delta t)^2-t^2}{\Delta t} \\ \\ \amp =\amp \ds\lim_{\Delta t\to0}\ds\frac{t^2+2t\Delta t+\Delta t^2-t^2}{\Delta t} \\ \\ \amp =\amp \ds\lim_{\Delta t\to0}\ds\frac{2t\Delta t+\Delta t^2}{\Delta t} \\ \\ \amp =\amp \ds\lim_{\Delta t\to0} \left(2t+\Delta t\right) = 2t. \end{array} \end{equation*}

Remember that $\Delta t$ is a single quantity, not a ''$\Delta$'' times a “$t$”, and so $\ds \Delta t^2$ is $\ds (\Delta t)^2$ not $\ds \Delta (t^2)\text{.}$ Doing the same example using the second formula for the derivative with $h$ in place of $\Delta t$ gives the following. Note that we compute $f(t+h)$ by substituting $t+h$ in place of $t$ everywhere we see $t$ in the expression $f(t)\text{,}$ while making no other changes (at least initially). For example, if

\begin{equation*} f(t)=t+\sqrt{(t+3)^2-t} \end{equation*}

then

\begin{equation*} f(t+h)=(t+h)+\sqrt{((t+h)+3)^2-(t+h)}=t+h+\sqrt{(t+h+3)^2-t-h}\text{.} \end{equation*}
###### Example4.8. Derivative of $y=t^2$ Using $h$ Notation.

Find the derivative of $\ds y=f(t)=t^2\text{.}$

Solution

We compute

\begin{equation*} \begin{split} f'(t) \amp = \ds\lim_{h\to0}\ds\frac{f(t+h)-f(t)}{h} \\ \amp =\ds\lim_{h\to0}\ds\frac{(t+h)^2-t^2}{h} \\ \amp = \ds\lim_{h\to0}\ds\frac{t^2+2th+h^2-t^2}{h} \\ \amp = \ds\lim_{h\to0}\ds\frac{2th+h^2}{h} \\ \amp = \ds\lim_{h\to0} \left(2t+h\right) = 2t. \end{split} \end{equation*}
###### Example4.9. Derivative.

Find the derivative of $y=f(x)=1/x\text{.}$

Solution

We compute

\begin{equation*} \begin{split} y' \amp = \lim_{\Delta x\to0}{\Delta y\over\Delta x}\\ \amp = \lim_{\Delta x\to0}{ {1\over x+\Delta x} - {1\over x}\over \Delta x}\\ \amp = \lim_{\Delta x\to0}{ {x\over x(x+\Delta x)} - {x+\Delta x\over x(x+\Delta x)}\over \Delta x}\\ \amp = \lim_{\Delta x\to0}{ {x-(x+\Delta x)\over x(x+\Delta x)}\over \Delta x}\\ \amp = \lim_{\Delta x\to0} {x-x-\Delta x\over x(x+\Delta x)\Delta x}\\ \amp =\ \lim_{\Delta x\to0} {-\Delta x\over x(x+\Delta x)\Delta x}\\ \amp = \lim_{\Delta x\to0} {-1\over x(x+\Delta x)}={-1\over x^2} \end{split} \end{equation*}
Note: If you happen to know some “derivative formulas” from an earlier course, for the time being you should pretend that you do not know them. In examples like the ones above and the exercises below, you are required to know how to find the derivative function using the definition of the derivative, i.e from basic principles. We will later develop some formulas so that we do not always need to do such computations, but we will continue to need to know how to do the more involved computations.

To recap, given any function $f$ and any number $x$ in the domain of $f\text{,}$ we define $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ wherever this limit exists, and we call the number $f'(x)$ the derivative of $f$ at $x\text{.}$ Geometrically, $f'(x)$ is the slope of the tangent line to the graph of $f$ at the point $(x,f(x))\text{.}$ The following symbols also represent the derivative:

\begin{equation*} f'(x)=y'=\frac{dy}{dx}=\frac{df}{dx}=\frac{d}{dx}f(x)\text{.} \end{equation*}

As above, and as you might expect, for different values of $x$ we generally get different values of the derivative $f'(x)\text{.}$ Could it be that the derivative always has the same value? This would mean that the slope of $f\text{,}$ or the slope of its tangent line, is the same everywhere. One curve that always has the same slope is a line; it seems odd to talk about the tangent line to a line, but if it makes sense at all the tangent line must be the line itself. It is not hard to see that the derivative of $f(x)=mx+b$ is $f'(x)=m\text{.}$ In the next example we emphasize the geometrical interpretation of derivative.

###### Example4.10. Geometrical Interpretation of Derivative.

Consider the function $f(x)$ given by the graph below. Verify that the graph of $f'(x)$ is indeed the derivative of $f(x)$ by analyzing slopes of tangent lines to the graph at different points.

Solution

We must think about the tangent lines to the graph of $f\text{,}$ because the slopes of these lines are the values of $f'(x)\text{.}$

We start by checking the graph of $f$ for horizontal tangent lines, since horizontal lines have a slope of 0. We find that the tangent line is horizontal at the points where $x$ has the values -1.9 and 1.8 (approximately). At each of these values of $x\text{,}$ we must have $f'(x)=0\text{,}$ which means that the graph of $f'$ has an $x$-intercept (a point where the graph intersects the $x$-axis).

Note that horizontal tangent lines have a slope of zero and these occur approximately at the points $(-1.9,-3.2)$ and $(1.8,3.2)$ of the graph. Therefore $f'(x)$ will cross the $x$-axis when $x=-1.9$ and $x=1.8\text{.}$

Analyzing the slope of the tangent line of $f(x)$ at $x=0$ gives approximately $3.0\text{,}$ thus, $f'(0)=3.0\text{.}$ Similarly, analyzing the slope of the tangent lines of $f(x)$ at $x=1$ and $x=-1$ give approximately $2.0$ for both, thus, $f'(1)=f'(-1)=2.0\text{.}$

In the next example we verify that the slope of a straight line is $m\text{.}$

###### Example4.11. Derivative of a Linear Function.

Let $m,b$ be any two real numbers. Determine $f'(x)$ if $f(x)=mx+b\text{.}$

Solution

By the definition of derivative (using $h$ in place of $\Delta x$) we have,

\begin{equation*} \begin{split} f'(x)\amp =\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{(m(x+h)+b)-(mx+b)}{h}\\ \amp =\lim_{h\to 0}\frac{mh}{h}=\lim_{h\to 0}m=m. \end{split} \end{equation*}

This is not surprising. We know that $f'(x)$ always represents the slope of a tangent line to the graph of $f\text{.}$ In this example, since the graph of $f$ is a straight line $y=mx+b$ already, every tangent line is the same line $y=mx+b\text{.}$ Since this line has a slope of $m\text{,}$ we must have $f'(x)=m\text{.}$

### Subsection4.2.1Differentiable

Now that we have introduced the derivative of a function at a point, we can begin to use the adjective differentiable.

###### Definition4.12. Differentiable at a Point.

A function $f$ is differentiable at point $a$ if $f'(a)$ exists.

###### Definition4.13. Differentiable on an Interval.

A function $f$ is differentiable on an open interval $\left(a,b\right)$ if it is differentiable at every point in the interval.

Sometimes one encounters a point in the domain of a function $y=f(x)$ where there is no derivative, because there is no tangent line. In order for the notion of the tangent line at a point to make sense, the curve must be “smooth” at that point. This means that if you imagine a particle traveling at some steady speed along the curve, then the particle does not experience an abrupt change of direction. There are four types of situations you should be aware of—discontinuities, vertical tangent lines, corners and cusps—where the slope takes on two equal values at the same $x$-value, or the slope is undefined at $x\text{,}$ and hence no derivative can exist at $x\text{.}$

###### Non-Existence of Derivative.

Given a function $y = f(x)\text{,}$ the derivative does not xists at $x$ if there is a discontinuity $(x_{1},x_{2},x_{3})\text{,}$ vertical tangent $(x_{4})\text{,}$ cusp $(x_{5}$) or corner $(x_{6})$ as shown. We say that $f$ is not differentiable at $x\text{.}$

###### Example4.14. Derivative of the Absolute Value.

Discuss the derivative of the absolute value function $y=f(x)=|x|\text{.}$

Solution

If $x$ is positive, then this is the function $y=x\text{,}$ whose derivative is the constant 1. (Recall that when $y=f(x)=mx+b\text{,}$ the derivative is the slope $m\text{.}$) If $x$ is negative, then we're dealing with the function $y=-x\text{,}$ whose derivative is the constant $-1\text{.}$ If $x=0\text{,}$ then the function has a corner, i.e., there is no tangent line. A tangent line would have to point in the direction of the curve—but there are two directions of the curve that come together at the origin.

We can summarize this as

\begin{equation*} y'=\left\{\begin{array}{rl} 1, \amp \mbox{if $x>0$,} \\ -1, \amp \mbox{if $x\lt 0$,} \\ \hbox{undefined,} \amp \mbox{if $x=0$.} \end{array} \right. \end{equation*}

In particular, the absolute value function $f(x)=|x|$ is not differentiable at $x=0\text{.}$

We note that the following theorem can be proved using limits.

Suppose that $f$ is differentiable at $a\text{.}$ That is,

\begin{equation*} f^{\prime }\left( a\right) =\lim_{h\rightarrow 0}\frac{f\left( a+h\right) -f\left( a\right) }{h} \end{equation*}

exists. At this stage, we find it convenient to write this limit in an alternative form so that its connection with continuity can become more easily seen. If we let $x=a+h\text{,}$ then $h=x-a\text{.}$ Furthermore, $h\rightarrow 0$ is equivalent to $x\rightarrow a\text{.}$ So,

\begin{equation*} f^{\prime }\left( a\right) =\lim_{x\rightarrow a}\frac{f\left( x\right) -f\left( a\right) }{x-a}\text{.} \end{equation*}

(This alternative formulation of the derivative is also standard. We will use it whenever we find it convenient to do so. You should get familiar with it.) Continuity at $a$ can now be proved as follows:

\begin{equation*} \begin{split} \lim_{x\rightarrow a}f\left( x\right) \amp = \lim_{x\rightarrow a}\left( \frac{ f\left( x\right) -f\left( a\right) }{x-a}\cdot \left( x-a\right) +f\left( a\right) \right)\\ =\amp \lim_{x\rightarrow a}\frac{f\left( x\right) -f\left( a\right) }{x-a}\cdot \lim_{x\rightarrow a}\left( x-a\right) +\lim_{x\rightarrow a}f\left( a\right)\\ =\amp f^{\prime }\left( a\right) \cdot \left( a-a\right) +f\left( a\right)\\ =\amp f\left( a\right) \end{split} \end{equation*}

Note: If $f$ is continuous at $a$ it is not necessarily true that $f$ is differentiable at $a\text{.}$ For example, it was shown that $f(x)=|x|$ is not differentiable at $x=0$ in the previous example, however, one can observe that $f(x)=|x|$ is continuous everywhere.

###### Example4.16.

Discuss the derivative of the function $\ds y=x^{2/3}\text{,}$ shown in Figure 4.2.

Solution

We will later see how to compute this derivative; for now we use the fact that $\ds y'=(2/3)x^{-1/3}\text{.}$ Visually this looks much like the absolute value function, but it technically has a cusp, not a corner. The absolute value function has no tangent line at 0 because there are (at least) two obvious contenders—the tangent line of the left side of the curve and the tangent line of the right side. The function $\ds y=x^{2/3}$ does not have a tangent line at 0, but unlike the absolute value function it can be said to have a single direction: as we approach 0 from either side the tangent line becomes closer and closer to a vertical line; the curve is vertical at 0. But as before, if you imagine traveling along the curve, an abrupt change in direction is required at 0: a full 180 degree turn.

In practice we won't worry much about the distinction between these examples; in both cases the function has a “sharp point” where there is no tangent line and no derivative.

### Subsection4.2.2Tangent Line Equation

Recall from Section 1.2 that the point-slope form of a straight line passing through the point $\left(x_{1},y_{1}\right)$ with slope $m$ is given by $y - y_{1} = m\left(x-x_{1}\right)\text{.}$ We are now in a position to provide the equation of a tangent line to a curve described by $y=f(x)$ at $x = a\text{,}$ provided the derivative exists.

###### Definition4.17. Tangent Line Equation.

Given a function $y=f(x)\text{,}$ the tangent line equation at $x=a$ is given by

\begin{equation*} y-f(a) = f'(a)\left(x-a\right)\text{,} \end{equation*}

provided $f'(a)$ exists.

###### Example4.18. Horizontal Tangent Line.

Let $f(x) = x^{2} - 4x\text{.}$

1. Compute $f'(x)\text{.}$

2. Find the point on the graph of $f$ where the tangent line to the curve is horizontal.

3. Sketch the graph of $f$ and the tangent line to the curve at the point found above.

4. What is the rate of change of $f$ at this point?

Solution
1. To find $f'(x)\text{,}$ we compute

\begin{equation*} \begin{split} f(x+h) - f(x) \amp = (x+h)^{2} - 4(x+h) \\[1ex] \amp = x^{2} + 2xh + h^{2} - 4x -4h - (x^{2} - 4x)\\[1ex] \amp = 2xh + h^{2} - 4h \\[1ex] \amp = h\left(2x + h -4 \right) \\[1ex] \dfrac{f(x+h) - f(x)}{h} \amp = 2x +h - 4 \end{split} \end{equation*}

Therefore,

\begin{equation*} f'(x) = \lim\limits_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = \lim\limits_{h \to 0} \left(2x +h - 4\right) = 2x - 4\text{.} \end{equation*}
2. At a point on the graph of $f$ where the tangent line to the curve is horizontal and hence has zero slope, the derivative $f'$ of $f$ is zero. Accordingly, to find such points, we set $f'(x) = 0\text{,}$ which gives $2x-4 = 0\text{,}$ or $x = 2\text{.}$ The corresponding value of $y$ is given by $y = f(2) = -4\text{,}$ and so the desired point is $\left(2,-4\right)\text{.}$

3. The graph of $f$ and the tangent line are shown below:

4. The rate of change of $f$ at $x = 2$ is zero.

### Subsection4.2.3Second and Higher Derivatives

If $f$ is a differentiable function then its derivative $f'$ is also a function and so we can take the derivative of $f'\text{.}$ The new function, denoted by $f''\text{,}$ read “$f$ double prime”, is called the second derivative of $f\text{,}$ since it is the derivative of the derivative of $f\text{.}$

The following symbols represent the second derivative:

\begin{equation*} f''(x)=y''=\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\text{.} \end{equation*}

We can continue this process to get the third derivative of $f\text{.}$ In general, the n-th derivative of $f$ is denoted by $f^{(n)}$ and is obtained from $f$ by differentiating $n$ times. If $y=f(x)\text{,}$ then we write:

\begin{equation*} y^{(n)}=f^{(n)}(x)=\frac{d^ny}{dx^n}\text{.} \end{equation*}

We have establised that the derivative of a function $f$ at a point $x$ measures the rate of change of the function $f$ at that point, and so the second derivative of $f$ (the derivative of $f'$) measures the rate of change of the derivative $f'$ of the function $f\text{.}$ Similarly, the third derivative of the function $f\text{,}$ $f'''\text{,}$ measures the rate of change of $f''\text{,}$ and so on.

In Chapter 5 we will discuss applications such as curve sketching involving the geometric interpretation of the second derivative of a function. The following examples provide an interpretation of both the first and the second derivative in familiar roles.

### Subsection4.2.4Applications

In economics, the derivative $y=E'(q)$ of a certain function $y=E(q)$ representing either cost, average cost, revenue, or profit is referred to by yet another name, namely marginal cost function , marginal average cost function, marginal revenue function, marginal profit function respectively. The study of the rate of change of economic functions is referred to as marginal analysis. As before, the marginal function $y=E'(q)$ indicates the rate of change of the function $y=E(q)$ with respect to the number $q$ of units produced. For example: We are given a revenue function $R(q)=pq= qf(q)\text{,}$ where $p$ is the unit selling price of the commodity, $q$ is the quantity of the commodity demanded, and $f$ is the demand function. Then the marginal revenue function $R'(q)$ yields the actual revenue realized from the sale of an additional unit of the commodity given that sales are already at a certain level.

###### Example4.19. Marginal Cost Function.

A certain manufacturer produces tactical flashlights with a daily total manufacturing cost in dollars of

\begin{equation*} C(q) = 0.0001q^{3}-0.08q^{2}+40q+5000 \end{equation*}

where $q$ stands for the number of tactical flashlights produced.

1. Determine the marginal cost function.

2. Calculate the marginal cost when $q = 200,300,400\text{,}$ and $600\text{.}$

3. What can you deduce from your results?

Solution
1. The marginal cost $C'$ is given by the derivative of the total cost function $C\text{.}$ Thus,

\begin{equation*} \begin{split} C'(q) \amp = \lim_{h\to 0} \dfrac{C(q+h)-C(q)}{h}\\ \amp = \lim_{h\to 0} \dfrac{0.0001(q+h)^{3} -0.08(q+h)^{2}+40(q+h)+5000-0.001q^{3}+0.08q^{2}-40q-5000}{h} \end{split} \end{equation*}

We expand, combine like-terms, and simplify by $h$ to find

\begin{equation*} C'(q) = 0.003q^{2}-0.16q+40\text{.} \end{equation*}
2. The marginal cost when $q = 200,300,400\text{,}$ and $600$ is given by

\begin{equation*} \begin{split} \amp C'(200)=0.003(200)^{2}-0.16(200)+40=20\\ \amp C'(300)=0.003(300)^{2}-0.16(300)+40=19\\ \amp C'(400)=0.003(400)^{2}-0.16(400)+40=24\\ \amp C'(600)=0.003(600)^{2}-0.16(600)+40=52 \end{split} \end{equation*}

or $$20\text{,}$$$19\text{,}$ $$24$ and$$52\text{,}$ respectively.

###### Example4.21. Marginal Profit Function.

Refer to Example 4.20. Suppose the cost of producing $q$ units of this product is

\begin{equation*} C(q) = 100q + 200,000 \end{equation*}

dollars.

1. Determine the profit function $P\text{.}$

2. Determine the marginal profit function $P'\text{.}$

3. Compute $P'(2000)\text{.}$ What can you deduce from your result?

4. Sketch the graph of the profit function $P\text{.}$

Solution
1. From part (a) of example 4.20, we have

\begin{equation*} R(q)= -0.02q^{2}+400q\text{.} \end{equation*}

Hence, we determine the profit function $P$ to be

\begin{equation*} \begin{split} P(q) \amp = R(q) - C(q) \\ \amp = \left(-0.02q^{2}+400q\right) - \left(100q+200,000\right)\\ \amp =-0.02q^{2} + 300q - 200,000 \end{split} \end{equation*}
2. The marginal profit function $P'$ is given by

\begin{equation*} \begin{split} P'(q) \amp = \lim_{h\to 0} \dfrac{P(q+h)-P(q)}{h} \\ \amp = \lim_{h\to 0} \dfrac{-0.02(q+h)^{2}+300(q+h)-200,000+0.02q^{2}-300q+200,000}{h} \end{split} \end{equation*}

As above, we expand and simplify to find

\begin{equation*} P'(q) = -0.04q + 300\text{.} \end{equation*}
3. $P'(2000)=-0.04(2000)+300=220\text{.}$ Therefore, the actual profit from selling the 2001st product is approximated by 220 4. The graph of the profit function $P$ in thousands of dollars is shown below. The Consumer Price Index (CPI) provides a broad measure of some countries living cost by periodically measuring changes in the price level of market basket of consumer goods and services purchased by households. Let $I(t)$ with $a\leq t \leq b$ describe the CPI of an economy between the years $a$ and $b$ as shown in Figure 4.3. Then we can determine the rate of change of $I$ for some value $c$ with $a\lt c\lt b$ by calculating the derivative $I'(c)\text{.}$ The relative rate of change of $I(t)$ with respect to $t$ at $t=c$ is given by the quantity \begin{equation*} \dfrac{I'(c)}{I(c)} \end{equation*} and measures the inflation rate of the economy at $t=c\text{.}$ We can also determine the rate of change of $I'(c)$ for some value $c$ with $a\lt c\lt b$ by calculating the second derivative $I''(c)\text{.}$ When $I'(t)$ is positive and $I''(t)$ is negative at $t=c$ as determined in Example 4.21, then we deduce that the economy is experiencing inflation (the CPI is increasing) at $t=c\text{,}$ but the rate at which inflation is growing is in fact slowing down. This is often used by a politician to claim that because inflation is slowing, the prices of goods and services are about to drop. A claim that is often false! ###### Example4.22. Working with CPI. Suppose the function \begin{equation*} I(t) = -0.2t^{3} + 3t^{2} + 100 \ \ \ (0 \leq t \leq 9) \end{equation*} gives the CPI of an economy, where $t = 0$ corresponds to the beginning of $1995\text{.}$ 1. Determine $I'(t)$ using the definition of the derivative. 2. Determine the inflation rate at the beginning of $2001$ ($t=6$). 3. Determine $I''(t)$ from $I'(t)$ using the definition of the derivative. 4. Calculate $I''(6)\text{.}$ What you can deduce from your result? Solution 1. From the definition of the derivative, we have \begin{equation*} \begin{split} I'(t) \amp= \lim_{h \to 0} \dfrac{I(t+h)-I(t)}{h}\\ \amp= \lim_{h \to 0} \dfrac{-0.2(t+h)^{3}+3(t+h)^{2}+100 +0.2t^{3}-3t^{2}-100}{h}\end{split}\text{.} \end{equation*} We expand, group like-terms, and simplify by $h$ to find \begin{equation*} I'(t) = \lim_{h \to 0} \left(-0.6t^{2}-0.2h^{2}-0.6ht + 6t +3h\right)= -0.6t^{2}+6t\text{.} \end{equation*} 2. We compute \begin{equation*} \ I'(6) = -0.6(6)^{2} + 6)6) = 14.4, \ \ \text{ and } \end{equation*} \begin{equation*} I(6) = -0.2(6)^{3}+3(6)^{2} + 100 = 164.8 \end{equation*} From which we see that the inflation rate is \begin{equation*} \dfrac{I'(6)}{I(6)} = \dfrac{14.4}{164.8} \simeq 0.0874 \end{equation*} or approximately $8.7$ percent. 3. We must now compute \begin{equation*} I''(t) = \lim_{h \to 0} \dfrac{I'(t+h)-I'(t)}{h}\text{.} \end{equation*} Using the above result, we get \begin{equation*} I''(t) = \lim_{h \to 0} \dfrac{(0.6(t+h)^{2}+6(t+h))-(-0.6t^{2} + 6t)}{h}\text{.} \end{equation*} We again expand, simplify and take the limit to find \begin{equation*} I''(t) = -1.2t+6\text{.} \end{equation*} 4. Since \begin{equation*} I''(6) = -1.2(6) + 6 = -1.2\text{,} \end{equation*} we see that $I'$ is indeed decreasing at $t = 6$ and conclude that inflation was levelling off at the time (see figures below). Suppose $f(t)$ is a position function of an object, representing the displacement of the object from the origin at time $t\text{.}$ In terms of derivatives, the velocity of an object is: \begin{equation*} v(t)=f'(t)\text{.} \end{equation*} The change of velocity with respect to time is called the acceleration and can be found as follows: \begin{equation*} a(t)=v'(t)=f''(t)\text{.} \end{equation*} Acceleration is the derivative of the velocity function and the second derivative of the position function. ###### Example4.23. Position, Velocity and Acceleration. Suppose the position function of an object is $f(t)=t^2$ metres at $t$ seconds. Find the velocity and acceleration of the object at time $t=1s\text{.}$ Solution By the definition of velocity and acceleration we need to compute $f'(t)$ and $f''(t)\text{.}$ Using the definition of derivative, we have, \begin{equation*} f'(t)=\lim_{h\to 0}\frac{(t+h)^2-t^2}{h}=\lim_{h\to 0}\frac{2th+h^2}{h}=\lim_{h\to 0}(2t+h)=2t\text{.} \end{equation*} Therefore, $v(t)=f'(t)=2t\text{.}$ Thus, the velocity at time $t=1$ is $v(1)=2~m/s\text{.}$ We now have that the acceleration at time $t$ is: \begin{equation*} a(t)=f''(t)=\lim_{h\to 0}\frac{2(t+h)-2t}{h}=\lim_{h\to 0}\frac{2h}{h}=2\text{.} \end{equation*} Therefore, $a(t)=2\text{.}$ Substituting $t=1$ into the function $a(t)$ gives $a(1)=2~m/s^2\text{.}$ ##### Exercises for Section 4.2. Find the derivatives of the following functions. 1. $\ds f(x)=\sqrt{169-x^2}$ Answer $\ds -x/\sqrt{169-x^2}$ Solution We find the derivative $f'(x)$ of $f(x) = \sqrt{169-x^{2}}$ using the definition of the derivative. \begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \ \dfrac{\sqrt{169-(x+h)^{2}}-\sqrt{169-x^{2}}}{h} \\ \amp = \lim_{h\to 0} \ \frac{1}{h} \cdot \dfrac{\left(\sqrt{169-(x+h)^{2}}-\sqrt{169-x^{2}}\right)\left(\sqrt{169-(x+h)^{2}}+\sqrt{169-x^{2}}\right)}{\sqrt{169-(x+h)^{2}}+\sqrt{169-x^{2}}} .\end{split} \end{equation*} Simplifying the above expression gives \begin{equation*} \begin{split} f'(x) \amp = \lim_{h \to 0} \ \frac{1}{h} \cdot \dfrac{-h(h+2x)}{\sqrt{169-(x+h)^{2}}+\sqrt{169-x^{2}}} \\ \amp = \lim_{h \to 0} \ \dfrac{-(h+2x)}{\sqrt{169-(x+h)^{2}} + \sqrt{169-x^{2}}} \\ \amp = \dfrac{-x}{\sqrt{169-x^{2}}} \cdot \end{split} \end{equation*} 2. $\ds h(t)=80-4.9t^2$ Answer $\ds -9.8t$ Solution \begin{aligned}h'(t) \amp = \lim\limits_{k\to 0} \dfrac{h(t+k)-h(t)}{k}\\ \amp = \lim\limits_{k\to 0} \dfrac{80-4.9(t+k)^2-80+4.9t^2}{k}\\ \amp = \lim\limits_{k\to 0} \dfrac{-4.9(t^2+2tk+k^2) + 4.9t^2}{k}\\ \amp = \lim\limits_{k\to 0} \dfrac{-4.9(2tk+k^2)}{k} \\ \amp = \lim\limits_{k\to 0} -4.9(2t+k) = -9.8t \end{aligned} 3. $\ds g(x)=x^2-(1/x)$ Answer $\ds 2x+1/x^2$ Solution \begin{aligned}g'(x) \amp = \lim\limits_{h \to 0} \dfrac{g(x+h)-g(x)}{h}\\ \amp =\lim\limits_{h \to 0} \dfrac{\left((x+h)^2-\frac{1}{x+h}\right) - \left(x^2-\frac{1}{x}\right)}{h} \\ \amp =\lim\limits_{h \to 0} \frac{ x^2 + 2xh + h^2 - x^2 - \frac{1}{x+h} + \frac{1}{x} } {h}\\ \amp =\lim\limits_{h \to 0} \frac{2xh + h^2 + \frac{-x + x + h}{x(x+h)}}{h}\\ \amp =\lim\limits_{h \to 0} \left(2x + h + \frac{1}{x(x+h)}\right)\\ \amp = 2x + \frac{1}{x^2} \end{aligned} 4. $\ds f(x)=ax^2+bx+c\text{,}$ where $a\text{,}$ $b\text{,}$ and $c$ are constants. Answer $\ds 2ax+b$ Solution Let $f(x) = ax^{2}+bx+c$ for any constants $a,b,c\text{.}$ We take the derivative with respect to $x\text{,}$ \begin{equation*} \diff{f}{x}= f'(x) = \lim_{h\to 0} \dfrac{a(x+h)^{2}+b(x+h)+c-ax^{2}-bx-c}{h} \cdot \end{equation*} Expand $a(x+h)^{2}$ and $b(x+h)$ and notice the terms which cancel. We find \begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \ \dfrac{2axh+ah^{2}+bh}{h} \amp = \lim_{h\to 0} \ \dfrac{2axh+ah^{2}+bh}{h} \amp = \lim_{h\to 0} \ \left(2ax+ah+b\right) \amp = 2ax+b. \end{split} \end{equation*} And so we find that $\diff{}{x} \left(ax^{2}+bx+c\right) = 2ax + b\text{.}$ 5. $\ds h(x)=x^3$ Answer $\ds 3x^2$ Solution \begin{aligned}h'(x) \amp = \lim\limits_{k \to 0} \frac{h(x+k)-h(x)}{k} \\ \amp = \lim\limits_{h \to 0} \frac{(x+k)^3 - x^3}{h}\\ \amp =\lim\limits_{h \to 0} \frac{ x^3 +3x^2k + 3xk^2 + k^3 - x^3}{k}\\ \amp =\lim\limits_{h \to 0} \frac{3x^2k + 2xk^2}{k}\\ \amp =\lim\limits_{h \to 0} \left(3x^2+2xk\right)\\ \amp = 3x^2 \end{aligned} 6. $\ds f(t)=2/\sqrt{2t+1}$ Answer $\ds -2/(2t+1)^{3/2}$ Solution \begin{aligned}f'(t) \amp = \lim\limits_{h\to 0} \dfrac{f(t+h)-f(t)}{h} \\ \amp = \lim\limits_{h\to 0} \dfrac{\frac{2}{\sqrt{2(t+h)+1}}-\frac{2}{\sqrt{2t+1}}}{h}\\ \amp = 2\lim\limits_{h\to 0} \dfrac{1}{h} \cdot \left(\dfrac{1}{\sqrt{2(t+h)+1}}-\dfrac{1}{\sqrt{2t+1}}\right)\\ \amp = 2\lim\limits_{h\to 0} \dfrac{1}{h} \cdot \left(\dfrac{1}{\sqrt{2(t+h)+1}}-\dfrac{1}{\sqrt{2t+1}}\right) \cdot \left(\dfrac{\frac{1}{\sqrt{2(t+h)+1}}+\frac{1}{\sqrt{2t+1}}}{\frac{1}{\sqrt{2(t+h)+1}}+\frac{1}{\sqrt{2t+1}}}\right)\\ \amp = 2\lim\limits_{h\to 0} \dfrac{1}{h} \cdot \left(\frac{-2h}{(2t+1)(2h+2t+1)}\right) \cdot \left(\dfrac{1}{\frac{1}{\sqrt{2(t+h)+1}}+\frac{1}{\sqrt{2t+1}}}\right)\\ \amp = 2\lim\limits_{h\to 0} \left(\frac{-2}{(2t+1)(2h+2t+1)}\right) \cdot \left(\dfrac{1}{\frac{1}{\sqrt{2(t+h)+1}}+\frac{1}{\sqrt{2t+1}}}\right)\\ \amp = 2 \cdot \left(\frac{-2}{(2t+1)(2t+1)}\right) \cdot \left(\dfrac{1}{\frac{1}{\sqrt{2t+1}}+\frac{1}{\sqrt{2t+1}}}\right)\\ \amp =2 \cdot \left(\frac{-2}{(2t+1)^2 \cdot 2(2t+1)^{-1/2}}\right) = \dfrac{-2}{(2t+1)^{3/2}}. \end{aligned} 7. $\ds g(t)=(2t-1)/(t+2)$ Answer $\ds 5/(t+2)^2$ Solution Now, we consider the function $g(t)=\dfrac{2t-1}{t+2}$ and take the derivative with respect to $t\text{.}$ \begin{equation*} \begin{split} \diff{g}{t} = g'(t) \amp = \lim_{h\to 0} \ \frac{1}{h} \cdot \left(\dfrac{2(t+h)-1}{(t+h)+2}-\dfrac{2t-1}{t+2}\right) \\ \amp = \lim_{h\to 0} \ \frac{1}{h} \cdot \dfrac{(2t+2h-1)(t+2) - (2t-1)(t+2+h)}{(t+2)(t+h+2)} \\ \amp = \lim_{h\to 0} \ \frac{1}{h} \cdot \dfrac{5h}{(t+2)(t+h+2)} \\ \amp = \lim_{h\to 0} \ \dfrac{5}{(t+2)(t+h+2)} \\ \amp = \dfrac{5}{(t+2)^{2}} \cdot \end{split} \end{equation*} Shown is the graph of a function $f(x)\text{.}$ Sketch the graph of $f'(x)$ by estimating the derivative at a number of points in the interval: estimate the derivative at regular intervals from one end of the interval to the other, and also at “special” points, as when the derivative is zero. Make sure you indicate any places where the derivative does not exist. Answer Solution First, we notice that the slope of $f$ is symmetric about the line $x=2\text{.}$ This means that the function $f'(x)$ will be symmetric about $x=2\text{.}$ We next indicate on the graph of $f$ the points at which $f'(x) = 0\text{,}$ and add these points to our graph of $f'(x)\text{:}$ Next, we estimate the slope of the tangent line to $f$ at the points $x=0.5$ and $x=1.5$ using the points $(0,2)$ and $(1,3)$ and $(2,2)\text{.}$ As shown the graph below to the left, the slope of these lines are good approximations to the slopes of the desired tangent lines. Therefore, \begin{equation*} f'(0.5) \approx \frac{2-3}{0-1} = 1, \text{ and } f'(1.5) \approx \frac{3-2}{1-2} = -1\text{.} \end{equation*} We add the points $(0.5,1)$ and $(1.5,-1)$ to our graph of $f'(x)$ and then reflect across our axis of symmetry: We can connect the points smoothly to obtain a rough graph of $f'(x)\text{.}$ The exact result is shown below. Shown is the graph of a function $f(x)\text{.}$ Sketch the graph of $f'(x)$ by estimating the derivative at a number of points in the interval: estimate the derivative at regular intervals from one end of the interval to the other, and also at “special” points, as when the derivative is zero. Make sure you indicate any places where the derivative does not exist. Solution First, we consider the derivative $f'(x)$ on the interval $\left(0,2\right)\text{.}$ Here, $f(x)$ is a line, and so the slope of the tangent line to $f$ is constant, and therefore its derivative is constant (and equal to the slope of the line, here $1$). At $x = 2\text{,}$ there is a corner, thus the derivative does not exist. On the interval $\left(2,3\right)\text{,}$ the slope of the tangent lines are negative, and so $f'(x)$ will take on negative values. Now, look at the slope of the tangent line between, say , $x_{1}=2.1$ and $x_{2}=2.2\text{.}$ The slope is steeper (more negative) at $x_{1}=2.1\text{,}$ by an amount that appears proportional to $\Delta x = 0.1$ — we verify with a careful drawing that $\dfrac{m_{1}-m_{2}}{\Delta x} = 2\text{.}$ This behaviour is the same on the entire interval, and so we see that, here, $f'(x)$ is an increasing line with slope $2$ which takes on negative values. At $x=3\text{,}$ the tangent line is horizontal and so $f'(x)=0\text{.}$ Finally, on the interval $\left(3,\infty\right)\text{,}$ we notice that the tangent lines have positive slope, but otherwise exhibit the same behaviour as on the interval $\left(2,3\right)\text{.}$ Putting this all together, we get the following sketch for $f'(x)\text{:}$ Find an equation for the tangent line to the graph of $\ds f(x)=5-x-3x^2$ at the point $x=2$ Answer $y=-13x+17$ Solution We first find the point of tangency: \begin{equation*} (2,f(2)) = (2, 5-2-3(4)) = (2,-9) \end{equation*} Next, we calculate the derivative of $f$ using the definition of the derivative: \begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}\\ \amp = \lim_{h \to 0} \frac{(5-(x+h)-3(x+h)^2) - (5-x-3x^2)}{h} \\ \amp = \lim_{h\to 0} \frac{5 - x - h - 3(x^2+2xh+h^2) - 5 + x +3x^2}{h}\\ \amp = \lim_{h\to 0} \frac{-h-6xh-3h^2}{h} \\ \amp = \lim_{h\to 0} \frac{-h(1+6x+3h)}{h}\\ \amp = \lim_{h\to 0} -(1+6x+3h)\\ \amp = -(1+6x) \end{split} \end{equation*} Therefore, the slope of the tangent line is \begin{equation*} f'(2) = -(1+6(2)) = -13\text{.} \end{equation*} All together, the equation of the tangent line (in point-slope form) is \begin{equation*} y + 9 = -13(x-2)\text{.} \end{equation*} Find a value for $a$ so that the graph of $\ds f(x)=x^2+ax-3$ has a horizontal tangent line at $x=4\text{.}$ Answer $-8$ Solution We wish to find a value of $a$ for which the function $f(x)=x^{2}+ax-4$ has a horizontal tangent line at $x=4\text{.}$ We first compute $f'(x)\text{:}$ \begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ \amp = \lim_{h\to 0} \frac{((x+h)^2+a(x+h)-4)-(x^2+ax-4)}{h}\\ \amp = \lim_{h\to 0} \frac{(x^2+2xh+h^2+ax+ah-4)-(x^2+ax-4)}{h}\\ \amp = \lim_{h\to 0} \frac{2xh+h^2+ah}{h}\\ \amp = \lim_{h\to 0} \left(2x+h+a\right) = 2x+a \end{split} \end{equation*} Therefore, $f'(4)= 8+a\text{.}$ The function $f$ will have a horizontal tangent line here if $f'(4)=0 \implies a=-8\text{.}$ We verify with a sketch: Find the slope of the tangent line to the graph of the given function at any point. 1. $f(x)=7$ Answer $0$ Solution Since $f(x)$ is a constant function, the slope of the tangent line will be zero for all $x\text{.}$ 2. $f(x)=x+5$ Answer $1$ Solution The function $f(x) = x+5$ is a line, and therefore has a constant slope, $m=1\text{.}$ Thus, for all $x\text{,}$ the slope of the tangent line will be 1. 3. $f(x)=2x^{2}$ Answer $4x$ Solution The function $f(x)=2x^2$ is a parabola, and so has a non-constant slope. Therefore, we compute the derivative: \begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ \amp = \lim_{h\to 0} \frac{2(x+h)^2 - 2x^2}{h}\\ \amp = \lim_{h\to 0} \frac{2(x^2+2xh+h^2) - 2x^2}{h}\\ \amp = \lim_{h\to 0} \frac{2x^2 + 4xh + 2h^2 - 2x^2}{h}\\ \amp = \lim_{h\to 0} \frac{h(4x+2h)}{h}\\ \amp = \lim_{h\to 0} (4x+2h)\\ \amp = 4x \end{split} \end{equation*} Therefore, at any point $x=a$ is $4a\text{.}$ 4. $f(x)=\sqrt{3x-1}$ Answer $\dfrac{3}{2\sqrt{3x-1}}$ Solution The square root function has a non-constant slope, and so we need to compute $f'(x)\text{:}$ \begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ \amp = \lim_{h\to 0} \frac{\sqrt{3(x+h)-1}-\sqrt{3x-1}}{h}\\ \amp = \lim_{h\to 0} \frac{\sqrt{3(x+h)-1}-\sqrt{3x-1}}{h} \cdot \frac{\sqrt{3(x+h)-1}+\sqrt{3x-1}}{\sqrt{3(x+h)-1}+\sqrt{3x-1}}\\ \amp = \lim_{h\to 0} \frac{3h}{h\sqrt{3(x+h)-1}+\sqrt{3x-1}}\\ \amp = \lim_{h\to 0} \frac{3}{\sqrt{3(x+h)-1}+\sqrt{3x-1}}\\ \amp = \frac{3}{2\sqrt{3x-1}} \end{split} \end{equation*} Hence, the slope of the tangent line at any point $x=a$ is given by $\dfrac{3}{2\sqrt{3a-1}}\text{.}$ Find the slope of the tangent line to the graph of each function at the given point and determine an equation of the tangent line. 1. $f(x)=2x+7$ at $(1,1)$ Answer $2\text{;}$ $y=2x-1$ Solution We first compute the derivative of $f(x)$ for any $x$ using the definition of the derivative: \begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ \amp = \lim_{h\to 0} \frac{(2(x+h)+7) - (2x+7)}{h}\\ \amp = \lim_{h\to 0} \frac{2x+2h+7 - 2x - 7}{h}\\ \amp = \lim_{h\to 0} \frac{2h}{h} \\ \amp = 2 \end{split} \end{equation*} Therefore, at the point $(1,1)$ the slope of the tangent line is 2. The equation of the tangent line is thus \begin{equation*} y - 1 = 2(x-1)\text{.} \end{equation*} 2. $f(x)=-\dfrac{1}{x}$ at $(2,-1)$ Answer $\frac{1}{4}\text{;}$ $y=\frac{1}{4}x-\frac{3}{2}$ Solution The derivative of $f(x)$ is found to be: \begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ \amp = \lim_{h\to 0} \frac{\frac{-1}{x+h} - \frac{-1}{x}}{h}\\ \amp = \lim_{h\to 0} \frac{\frac{-x+(x+h)}{x(x+h)}}{h}\\ \amp = \lim_{h\to 0} \frac{\frac{h}{x(x+h)}}{h}\\ \amp = \lim_{h\to 0} \frac{1}{x(x+h)}\\ \amp = \frac{1}{x^2} \end{split} \end{equation*} Hence, at the point $(2,-1)$ the slope of the tangent line is $\dfrac{1}{(2)^2} = \dfrac{1}{4}\text{.}$ Therefore, the equation of the tangent line is \begin{equation*} y + 1 = \frac{1}{4} (x-2)\text{.} \end{equation*} Let $f(x) = x^{2}-2x+1\text{.}$ 1. Determine the derivative $f'$ of $f\text{.}$ Answer $f'(x)=2x-2$ Solution Using the definition of the derivative, we compute $f'(x)\text{:}$ \begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \frac{\frac{h}{x(x+h)}}{h}\\ \amp = \lim_{h\to 0} \frac{\left((x+h)^2-2(x+h)+1\right)-\left(x^2-2x+1\right)}{h}\\ \amp = \lim_{h\to 0} \frac{\left(x^2+2xh+h^2 - 2x-2h +1\right) -\left(x^2-2x+1\right)}{h}\\ \amp = \lim_{h\to 0} \frac{2xh+h^2-2h}{h}\\ \amp = \lim_{h\to 0} \frac{h(2x+h-2)}{h}\\ \amp = \lim_{h\to 0} (2x+h-2)\\ \amp = 2x -2 \end{split} \end{equation*} 2. Determine the point on the graph of $f$ where the tangent line to the curve is horizontal. Answer $(1,0)$ Solution The tangent line to $f$ is horizontal when $f'(x)=0\text{.}$ Therefore, we set \begin{equation*} \begin{split} f'(x) \amp = 0 \\ 2x -2 \amp = 0\\ x \amp = 1. \end{split} \end{equation*} Therefore, the tangent line is horizontal at the point \begin{equation*} \left(1,f(1)\right) = \left(1, 1-2+1\right) = \left(1,0\right)\text{.} \end{equation*} 3. Sketch the graph of $f$ and the tangent line to the curve at the point found in part (b). Answer Solution Notice that the function \begin{equation*} f(x) = x^2-2x+1 \end{equation*} is a parabola (opening upwards) with vertex $(1,0)\text{,}$ which is the point identified above. We make the following sketch: 4. Determine the rate of change of $f$ at the point found in part(b). Answer $0$ Solution From our work above, we determine that the rate of change at $x=1$ is zero. Let $y=h(t)=t^{2}+t\text{.}$ 1. Find the average rate of change of $y$ with respect to $t$ in the interval from $t=2$ to $t=3\text{,}$ from $t=2$ to $t=2.5\text{,}$ and from $t=2$ to $t=2.1\text{.}$ Answer $6\text{;}$$5.5\text{;}$$5.1$ Solution The average rate of change of the function $y=h(t) = t^2+t$ from $t=2$ to $t=2+\Delta t$ is given by \begin{equation*} \begin{split} \frac{\Delta y}{\Delta t} \amp = \frac{(2+\Delta t)^2 + (2+ \Delta t) - (2^2+2)}{\Delta t} \\ \amp = \frac{4+4\Delta t + \Delta t^2 + 2 + \Delta t - 4 - 2}{\Delta t}\\ \amp = \frac{4\Delta t + \Delta t^2 + \Delta t}{\Delta t}\\ \amp = 5 + \Delta t \end{split} \end{equation*} Therefore, between $t=2$ and $t=3$ ($\Delta t=1$), the average rate of change must be $5 + 1 = 6\text{.}$ Between $t=2$ and $t=2.5$ ($\Delta t=0.5$), the average rate of change is $5 + 0.25 = 5.25\text{.}$ Between $t=2$ and $t=2.1$ ($\Delta t =0.1$), the average rate of change is $5+0.1 = 5.1\text{.}$ 2. Find the (instantaneous) rate of change of $y$ at $t=2\text{.}$ Answer $5$ Solution The instantaneous rate of change at $t=2$ is therefore \begin{equation*} \lim_{\Delta t \to 0} 5 + \Delta t = 5\text{.} \end{equation*} 3. Compare the results obtained in part (a) with that of part (b). Answer We notice that as we refine $\Delta t$ in part (b), our answers get closer and closer to our answer in part (c), as expected. The total cost $C(q)$ (in dollars) incurred by a certain manufacturer in producing $q$ units a day is given by \begin{equation*} C(q)=-10q^{2}+300q+130 \ \ \ (0\leq q \leq 15) \end{equation*} 1. Find $C'(q)\text{.}$ Answer $C'(q)=-20q+300$ Solution We differentiate with respect to $q\text{:}$ \begin{equation*} \begin{split} C'(q) \amp = \lim_{h\to 0} \frac{C(q+h)-C(q)}{h}\\ \amp = \lim_{h\to 0} \frac{\left(-10(q+h)^2+300(q+h)+130\right) - \left(-10q^2+300q+130\right)}{h}\\ \amp =\lim_{h\to 0} \frac{-10(q^2+2qh+h^2) + 300q + 300h + 130 + 10q^2 - 300q - 130}{h}\\ \amp = \lim_{h\to 0} \frac{ -200qh -200h^2 + 300h}{h}\\ \amp = \lim_{h\to 0} \frac{h(-200q-200h+300)}{h}\\ \amp = \lim_{h\to 0} \left(-200q- 200h + 300\right)\\ \amp = -200q+300 \end{split} \end{equation*} 2. What is the rate of change of the total cost when the level of production is ten units? Answer$100$ per unit
Solution

If production level is set to 10 units, then the rate of change of the cost is

\begin{equation*} C'(10) = -200(1) + 300 = 100\text{.} \end{equation*}

This means that the cost is increasing at a rate of $100. 3. What is the average cost the manufacturer incurs when the level of production is ten units? Answer$$213$ per unit
Solution

The average cost when $q=10$ is

\begin{equation*} \overline{C} = \frac{C(10)}{10} = \frac{-10(100)+300(10)+130}{10} = -100+300+13 = 213\text{.} \end{equation*}

Therefore, when the production level is set to 10 units, the average cost incurred is $213 per unit. A certain manufacturer produces beach umbrellas with a daily cost $C(q)$ of \begin{equation*} C(q)=0.000002q^{3}+5q+400\text{.} \end{equation*} Calculate \begin{equation*} \dfrac{C(100+h)-C(100)}{h} \end{equation*} for $h=1,0.1,0.001\text{,}$ and $0.0001\text{.}$ Use your results to estimate the rate of change of the total cost function when the level of production is $100$ units/day. Answer$$5.06060\text{,}$ $$5.06006\text{,}$$$5.060006\text{,}$$$5.0600006\text{.}$ Estimate$$5.06\text{.}$
Solution

When $q=100$ and $h=1\text{,}$ we compute

\begin{equation*} \frac{C(100+1) - C(100)}{1} = 5.06060\text{.} \end{equation*}

When $h=0.1\text{,}$ we compute

\begin{equation*} \frac{C(100+0.1) - C(100)}{0.1} = 5.06006\text{.} \end{equation*}

Next, when $h=0.001\text{,}$ we have

\begin{equation*} \frac{C(100+0.001)-C(100)}{0.001} = 5.060006\text{.} \end{equation*}

Finally, when $h=0.0001\text{,}$ we have

\begin{equation*} \frac{C(100+0.0001)-C(100)}{0.0001} = 5.0600006\text{.} \end{equation*}

Therefore, we estimate that the instantaneous rate of change when $q=100$ is $5.06\text{.}$

The gross domestic product (GDP) of a certain country over 8 years is approximated by

\begin{equation*} G(t)=-0.2t^{3}+2.4t^{2}+60 \ \ \ (0 \leq t \leq 8)\text{,} \end{equation*}

billions of dollars, where $t=0$ corresponds to $1992\text{.}$ The derivative of $G(t)$ is given by

\begin{equation*} G'(t)=-0.6t^{2}+4.8t \end{equation*}
1. Compute $G'(0)\text{,}$$G'(1)\text{,...,}$$G'(8)\text{.}$

$0\text{;}$$4.2\text{;}$$7.2\text{;}$$9\text{;}$$9.6\text{;}$$9\text{;}$$7.2\text{;}$$4.2\text{;}$$0$
Solution

We are given that $G'(t)=-0.6t^2+4.8t\text{.}$ Therefore, we compute:

\begin{equation*} G'(0) = -0.6(0)+4.8(0) = 0 \end{equation*}
\begin{equation*} G'(1) = -0.6+4.8 = 4.2 \end{equation*}
\begin{equation*} G'(2) = -0.6(4)+4.8(2) = 7.2 \end{equation*}
\begin{equation*} G'(3) = -0.6(9) + 4.8(3) = 9 \end{equation*}
\begin{equation*} G'(4) = -0.6(16)-4.8(4) = 9.6 \end{equation*}
\begin{equation*} G'(5) = -0.6(25)+4.8(5) = 9 \end{equation*}
\begin{equation*} G'(6) = -0.6(36)+4.8(6) = 7.2 \end{equation*}
\begin{equation*} G'(7) = -0.6(49)+4.8(7) = 4.2 \end{equation*}
\begin{equation*} G'(8) = -0.6(64)+4.8(8) = 0 \end{equation*}
2. Compute $G''(0)\text{,}$$G''(1)\text{,...,}$$G''(8)\text{.}$

$4.8\text{;}$$3.6\text{;}$$2.4\text{;}$$1.2\text{;}$$0\text{;}$$-1.2\text{;}$$-2.4\text{;}$$-3.6\text{;}$$-4.8$
Solution

We first compute the derivative of $G'(t)\text{:}$

\begin{equation*} \begin{split} G''(t) \amp = \lim_{h\to 0} \frac{G'(t+h)-G'(t)}{h}\\ \amp =\lim_{h\to 0} \frac{(-0.6(t+h)^2+4.8(t+h)) - (-0.6t^2+4.8t)}{h}\\ \amp = \lim_{h\to 0} \frac{-0.6(t^2+2th+h^2) +4.8t + 4.8 h -0.6t^2 - 4.8 t }{h}\\ \amp = \lim_{h\to 0} \frac{-1.2th-0.6h^2+4.8h}{h}\\ \amp = \lim_{h\to 0} (-1.2t -0.6h + 4.8)\\ \amp = -1.2t +4.8 \end{split} \end{equation*}

Now compute:

\begin{equation*} G''(0) = -1.2(0) + 4.8 = 4.8 \end{equation*}
\begin{equation*} G''(1) = -1.2(1) + 4.8 = 3.6 \end{equation*}
\begin{equation*} G''(2) = -1.2(2)+4.8 = 2.4 \end{equation*}
\begin{equation*} G''(3) = -1.2(3)+4.8 = 1.2 \end{equation*}
\begin{equation*} G''(4) = -1.2(4)+4.8 = 0 \end{equation*}
\begin{equation*} G''(5) = -1.2(5)+4.8 = -1.2 \end{equation*}
\begin{equation*} G''(6) = -1.2(6)+4.8 = -2.4 \end{equation*}
\begin{equation*} G''(7) = -1.2(7) + 4.8 = -3.6 \end{equation*}
\begin{equation*} G''(8) = -1.2(8) + 4.8 = -4.8 \end{equation*}
3. What can you deduce from your results?

We conclude that after a spectacular growth rate in the early years, the growth of the GDP cooled off. This cool off starts to occur around year 5.

It is estimated that the number $N(t)$ of individuals infected with a certain contagious disease is

\begin{equation*} N(t)=-0.1t^{3}+1.5t^{2}+100 \ \ \ (0 \leq t \leq 7) \end{equation*}

where $t$ is in months and $t=0$ corresponds to the initial outbreak. The derivative of $N(t)$ is given by

\begin{equation*} N'(t)=-0.3t^{2}+3t\text{.} \end{equation*}

After 4 months, a drug which reduces the infectiousness of the disease is developped.

1. Verify that the number of infected individuals was increasing for 7 months. (Hint: Compute $N'(0)\text{,}$$N'(1)\text{,...,}$$N'(7)$).

0;2.7;4.8;6.3;7.2;7.5;7.2;6.3
Solution

We are given that $N'(t) = -0.3t^2+3t\text{.}$ Therefore, we compute:

\begin{equation*} N'(0) = -0.3(0) + 3(0) = 0 \end{equation*}
\begin{equation*} N'(1) = -0.3(1) + 3(1) = 2.7 \end{equation*}
\begin{equation*} N'(2) = -0.3(2) + 3(2) =4.8 \end{equation*}
\begin{equation*} N'(3) = -0.3(3) + 3(3) =6.3 \end{equation*}
\begin{equation*} N'(4) = -0.3(4) + 3(4) = 7.2 \end{equation*}
\begin{equation*} N'(5) = -0.3(5) + 3(5) = 7.5 \end{equation*}
\begin{equation*} N'(6) = -0.3(6) + 3(6) = 7.2 \end{equation*}
\begin{equation*} N'(7) = -0.3(7) + 3(7) = 6.3 \end{equation*}

Since $N'(t) > 0$ for all $t\in (0,7]\text{,}$ the number of infected individuals was increasing during this time.

2. Show that the drug was working by computing $N''(4)\text{,}$$N''(5)\text{,}$$N''(6)$ and $N''(7)\text{.}$

0.6;0;-0.6;-1.2
Solution

In order to show the drug which was developped was working, we need to show that the rate at which the number of infected individuals was increasing starts to decrease after $t=4$ months. To do so, we first compute the second derivative:

\begin{equation*} \begin{split} N''(t) \amp = \lim_{h\to 0} \frac{N(t+h)-N(t)}{h}\\ \amp = \lim_{h\to 0} \frac{\left(-0.3(t+h)^2 +3(t+h)\right) - \left(0.3t^2 + 3t\right)}{h}\\ \amp = \lim_{h\to 0} \frac{-0.3(t^2+2th+h^2) + 3t + 3h - 0.2t^2 - 3t}{h}\\ \amp = \lim_{h\to 0} \frac{-0.6th-0.3h^2+3h}{h}\\ \amp = \lim_{h\to 0} \frac{h(-0.6t-0.3h+3)}{h}\\ \amp = \lim_{h\to 0} (-0.6t-0.3h+3)\\ \amp = -0.6t + 3 \end{split} \end{equation*}

We now compute:

\begin{equation*} N''(4) = -0.6(4) + 3 = 0.6 \end{equation*}
\begin{equation*} N''(5) = -0.6(5) + 3 = 0 \end{equation*}
\begin{equation*} N''(6) = -0.6(6)+3 = -0.6 \end{equation*}
\begin{equation*} N''(7) = -0.6(7)+3 = -1.2 \end{equation*}

This means that after 6 months—2 months after the introduction of the drug—the rate at which number of infected individuals was increasing starts to decrease. This indicates that the drug had started to work.

A certain manufacturer determines that the amount of defective products coming out of their new Ontario plant $t$ days after it opened can be estimated by

\begin{equation*} A(t) = -0.00006t^{5} + 0.00468t^{4}-0.1316t^{3} + 1.915t^2 - 17.63t + 100 \end{equation*}

percent of the total number of units produced. The first and second derivatives of $A(t)$ are given:

\begin{equation*} A'(t) = -0.0003t^{4}+0.01872t^{3}-0.3948t^{2}+3.83t-17.63 \end{equation*}
\begin{equation*} A''(t)=-0.0012t^{3}+0.05616t^{2}-0.7896t+3.83\text{.} \end{equation*}

Compute $A'(10)$ and $A''(10)$ and interpret your results.

Solution

From the formulas given for $A'(t)$ and $A''(t)\text{,}$ we compute

\begin{equation*} A'(10) = -3.09, \text{ and } A''(10) = 0.35\text{.} \end{equation*}

This means that 10 days after the plant had opened, the amount of defective products was decreasing at a rate of approximately 3% per day. From the second derivative, we see that the rate at which the amount of defective products was decreasing was increasing at the rate of 0.35% per day per day.

Between the years $1980$ and $2001\text{,}$ it is estimated that the percentage of young families who own their own home can be approximated by the function

\begin{equation*} P(t)=33.55(t+5)^{0.205} \ \ \ (0 \leq t \leq 21) \end{equation*}

where $t$ is measured in years, with $t=0$ corresponding to the beginning of $1980\text{.}$ The second derivative of $P(t)$ is given by

\begin{equation*} -5.46781 (t+5)^{-1.795}\text{.} \end{equation*}

Compute $P''(20)$ and interpret your results.

Solution

We are given that

\begin{equation*} P''(t) = -5.46781(t+5)^{-1.795}\text{.} \end{equation*}

Therefore, when $t=20\text{,}$ we have

\begin{equation*} P''(20) = -5.46781(25)^{-1.795}= -0.0169243\text{.} \end{equation*}

This means that in the year 2000, The rate of change of young families owning their own homes is decreasing at a rate of approximately 0.02%/yr$^2\text{.}$

A certain manufacturer estimates that the total weekly cost in producing $q$ units is

\begin{equation*} C(q) = 2000+2q-0.0001q^{2} \ \ \ 0 \leq q \leq 6000\text{,} \end{equation*}

dollars.

1. What is the actual cost incurred in producing the $1001$st and the $2001$st unit?

$$1.80\text{;}$$$1.60\text{.}$
Solution

The actual cost incurred in producing the 1001st unit is

\begin{equation*} C(1001)-C(1000) = 3901.8 - 3900 = 1.80 \end{equation*}

dollars. Similarly, the actual cost incurred in producing the 2001st unit is

\begin{equation*} C(2001)-C(2000) = 5601.6 - 5600 = 1.60 \end{equation*}

dollars.

2. What is the marginal cost when $q=1000$ and $2000\text{?}$

$$1.80\text{;}$$$1.60\text{.}$
Solution

We first compute the derivative:

\begin{equation*} \begin{split} C'(q) \amp = \lim_{h\to 0} \frac{C(q+h)-C(q)}{h}\\ \amp = \lim_{h\to 0} \frac{(2000+2(q+h)-0.0001(q+h)^2) - (2000 + 2q - 0.0001q^2)}{h}\\ \amp = \lim_{h\to 0} \frac{2h - 0.0002qh - 0.0001h^2}{h} \\ \amp = \lim_{h\to 0} (2-0.0002q-0.0001h)\\ \amp = 2-0.0002q \end{split} \end{equation*}

The marginal cost when $q=1000$ is

\begin{equation*} C'(1000) = 2-0.0002(1000) = 1.80\text{.} \end{equation*}

The marginal cost when $q=2000$ is

\begin{equation*} C'(2000) = 2-0.0002(2000) = 1.60\text{.} \end{equation*}

A movie theatre determines that their monthly revenue can be estimated by

\begin{equation*} R(q) = 8000q-100q^{2} \end{equation*}

dollars when the price per movie ticket is $q$ dollars.

1. Find the marginal revenue $R'\text{.}$

$8000-200q$
Solution

We differentiate:

\begin{equation*} \begin{split} R'(q) \amp = \lim_{h\to 0} \frac{R(q+h)-R(q)}{h}\\ \amp = \lim_{h\to 0} \frac{(8000(q+h)-100(q+h)^2) - (8000q - 100q^2)}{h}\\ \amp = \lim_{h\to 0} \frac{8000q + 8000h -100(q^2+2qh+h^2) - 8000q + 100q^2}{h}\\ \amp = \lim_{h\to 0} \frac{8000h - 200qh-100h^2}{h}\\ \amp = \lim_{h\to 0} (8000 - 200q - 100h)\\ \amp = 8000 - 200q \end{split} \end{equation*}
2. Compute $R'(39)\text{,}$ $R'(40)\text{,}$ and $R'(41)\text{.}$

$200\text{,}$$0\text{,}$$-200\text{.}$
Solution

We compute:

\begin{equation*} R'(39) = 8000-200(39) = 200 \end{equation*}
\begin{equation*} R'(40) = 8000-200(40)= 0 \end{equation*}
\begin{equation*} R'(41) = 8000-200(41) = -200 \end{equation*}
3. Based on the results above, what price should the theatre charge in order to maximize their revenue?

We conclude that the movie theatre should charge $40 per ticket. The demand function for a certain product is given by \begin{equation*} p = -0.04q+800 \ \ \ 0\leq q \leq 20,000 \end{equation*} where $p$ denotes the unit price in dollars and $q$ denotes the quantity demanded. 1. Determine the revenue function $R\text{.}$ Solution $R(q)= pq = (-0.04q + 800)q = -0.04q^{2}+800q\text{.}$ 2. Determine the marginal revenue function $R'\text{.}$ Answer We can again use the result from 4.2.1(d) to find $R'(q)=-0.08q+800\text{.}$ 3. Compute $R'(5000)\text{.}$ What can you deduce from your results? Answer $R'(5000)=400\text{.}$ We see that, when the level of production is set at $5000$ units, the production of the $5001$st unit will bring an additional revenue of approximately$$400\text{.}$
4. If the total cost in producing $q$ units is given by

\begin{equation*} C(q)=200q+300,000 \end{equation*}

determine the profit function $P(q)\text{.}$

$P(q)= R(q) - C(q) = -0.04q^{2}+600q-300,000\text{.}$
5. Find the marginal profit function $P'\text{.}$

$P'(q)=-2(0.04)q+600 = -0.08q + 600\text{.}$
6. Compute $P'(5000)$ and $P'(8000)\text{.}$
$P'(5000)=200\text{,}$ $P'(8000)=-40$
The profit increases as production increases, peaking at $7500$ units. Beyond this level, profit falls. $P(q)$ shown in millions of dollars.