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Here’s a secret: GC uses approximations to compute $\displaystyle{\int_a^x r_f(t)dt}$ for any function $r_f$.
When GC integrates a function over a large domain it can be quite inaccurate because it always uses 512 subintervals. So when GC computes $$\int_0^{1024} r_f(t)dt$$ it uses subintervals of size 2, which could produce a highly inaccurate result if $r_f$ varies dramatically within those subintervals.
However, Section 6.2 gave us the insight that we can compute an integral of a function $r_f$ exactly whenever we know an accumulation function f in closed form that has $r_f$ as its rate of change function.
This fact bears repeating.
Whenever we know a closed form representation of a function f that that has $r_f$ as its exact rate of change function, we can use the closed form representation of f to compute $\int_a^x r_f(t)dt$ exactly, namely $$\int_a^x r_f(t)dt=f(x)f(a).$$We do not need an approximate accumulation function when we can represent the exact integral in closed form.
This fact motivates us to derive rate of change functions in closed form from accumulation functions in closed form. As we derive more accumulationrate pairs in closed form, our ability to compute integrals exactly will grow.
A general remark will be helpful in all the remaining sections.
When $0<h<1$ and n is a positive integer, $h^{n+k}$ is smaller than $h^n$ by a factor of $h^k.$ If $h=0.01,$ then $0.01^{15}$ is smaller than $0.01^{10}$ by a factor of $0.01^5,$ or $\left(\frac{1}{100}\right)^5.$ Stated generally, for $0<h<1$, as $h$ becomes small, $h^n$ becomes even smaller, and becomes smaller orders of magnitude more rapidly as the value of n increases.
Here is a sheet of practice problems. Examine it repeatedly during your study of this chapter for problems that you can solve with methods you have learned up to the moment that you are examining the sheet. You will learn a method for checking your work graphically. Use it.
Intuitively, if a quantity has the same measure at every moment of its independent variable, then its value is not varying and its rate of change is 0. More formally, for $h\gt0$
$$\color{red}{\text{(Eq. 6.3.1)}}\qquad \begin{align} f(x)&=c\\[1ex] r(x)&=\frac{f(x+h)f(x)}{h}\\[1ex] &=\frac{cc}{h}\\[1ex] &=0\\[1ex]r_f(x)&=0 \end{align}$$
In regard to recovering f from its rate of change function, we have the seemingly odd result that when $f(x)=c$, $\int_a^x r_f(t)dt=0$, not c.
This oddity is only apparent. If you think of f as an accumulation function and $\int_a^x r_f(t)dt$ as f’s accumulation from a to x, nothing additional has accumulated. Put another way, we know that $f(x) = c$ for some constant c when $r_f(x)=0$ for all x, so by the FTC, $\int_a^x 0dt = f(x)f(a) = cc=0$.
When $g(x)=c \cdot f(x)$, where c is a constant, then
$$\begin{align} r_g(x) & = \frac{g(x+h)g(x)}{h} \\ & = \frac{c \cdot (f(x+h)f(x))}{h} \\ & = c \cdot \frac {f(x+h)f(x)} {h} \\ & = c \cdot r_f(x) \\ \end{align} $$
Therefore, by the FTC, $ \int_a^x c r_f(t)dt = c \int_a^x r_f(t)dt = c (f(x)f(a))$
Let $f(x)=cx^n$, n a positive integer and $c≠0$. We want $r_f$, f's exact rate of change function, in closed form.
We’ll start with $n=3$ and use that to generalize to arbitrary integer values of $n≥0$. We’ll also start with the approximate rate of change function and get the exact rate of change function by making the value of h so small that $h \doteq 0$.
If our derivation is correct, then the graphs of the open form and closed form rate of change functions should coincide, and the graphs of the open form and closed form integrals should coincide.
$$\begin{align} f(x)&=4x^3\\[1ex] r(x)&=\frac{f(x+h)f(x)}{h}\\[1ex] &=\frac{4(x+h)^34x^3}{h}\\[1ex] &=\frac{4\left(x^3+3x^2h+3xh^2+h^2\right)4x^3}{h}\\[1ex] &=\frac{4\cdot x^2h+4\cdot 3xh^2+4h^3}{h}\\[1ex] &=4\cdot x^2+4\cdot 3xh + 4h^2\\[1ex] r_f(x)&\doteq 4\cdot 3x^2 \text{ when }h \doteq 0 \end{align}$$
$f(x)$ 
$=cx^n, c \neq 0, n \ge 0$ 
$r(x)$ 
$=\dfrac{f(x+h)f(x)}{h}$ 
$=\dfrac {c(x+h)^n  cx^n}{h}$ 

$= \dfrac{c(x^n + nx^{n1}h + n(n1)x^{n2}h^2 + ... + nxh^{n1} + h^n)  cx^n}{h}$  
$= \dfrac{cnx^{n1}h + cn(n1)x^{n2}h^2 + ... + cnxh^{n1} + ch^n)}{h}$  
$= cnx^{n1} + cn(n1)x^{n2}h + ... + cnxh^{n2} + ch^{n1}$  
$r_f(x)$  $ \doteq cnx^{n1}$ when $h \doteq 0$ 
Let $f(x)=4x^3$. Use GC to see whether the graph of $y=12x^2$ coincides with the graph of $y=\dfrac{f(x+h)f(x)}{h}$ when $h=0.0001$. What does this tell you?
Use GC to see whether the graphs of $y=\int_a^x 12t^2dt$ and $y=4x^34a^3$ coincide. What does this tell you?
Let $f(x)=cx^n$. Use GC to see whether the graph of $y = cnx^{n1}$ coincides with the graph of $y=\dfrac{f(x+h)f(x)}{h}$ when $h=0.0001$ and n is any arbitrary value.
What does this tell you?
Use GC to see whether the graphs of $\displaystyle{y=\int_a^x cnt^{n1}dt}$ and $y=c(x^na^n)$ coincide for arbitrary values of c and n. What does this tell you? In what way is this result an instance of the FTC?
Accum Fn: $f(x)=$ 
Rate of Change:
$r_f(x)=$ 
Accum Fn from Rate: ${\displaystyle \int_a^x r_f(t)dt}$ 
c 
0 
0 
$cx^n$ 
$cnx^{n1}$ 
$cx^nca^n$ 
Let $g(x)=4$. What is $\int_a^x g(t)dt$?
Rewrite g as $g(x)=4 \cdot 1x^0$. Then $\int_a^x g(t)dt = \int_a^x 4 \cdot 1t^0dt = 4(x^1a^1)$.
Reflection 6.3.1. Think of what $g(x)=4$ means when g is a rate of change function. Give a concrete example that illustrates how it make sense that $\int_a^x 4dt = 4(xa)$.
Let $g(x)=7x^3$. What is $\int_a^x g(t)dt$ in closed form?
Were $g$ defined as $g(x)=4x^3$ we would know by the FTC that $\int_a^x 4t^3 dt = x^4  a^4$. But the coefficient of $x^3$ is 7, not 4. We can adjust for this by noting that $7 = \frac{7}{4} \cdot 4$, so we can write $g(x)=7x^3$ as $g(x)=\frac{7}{4}(4t^3)dt$, so $\int_a^x 7t^3dt = \int_a^x \frac{7}{4}(4t^3)dt=\frac{7}{4}(x^4a^4)$.
Suppose that you walk away from your friend on a straight sidewalk at 6 km/h and your friend walks away from you at 4 km/h.
Every hour that the two of you walk, you will get 6 km further from him and he will get 4 km further from you. So, the distance between the two of you increases at a rate of $(6 + 4)$ km/h.
In general, if $k(x) = f(x)+g(x)$ and if $r_f$ and $r_g$ exist, then $r_k(x) = r_f(x)+r_g(x)$. We can prove the general statement as follows.
Assume that $k(x) = f(x)+g(x)$ and that $r_f$ and $r_g$ exist for all values of x. Then,
$$\color{red}{\text{(Eq. 6.3.2)}}\qquad \begin{align} k(x)&=f(x)+g(x)\\[1ex] r(x)&=\frac{k(x+h)k(x)}{h}\\[1ex] &=\frac{(f(x+h)+g(x+h))(f(x)+g(x))}{h}\\[1ex] &=\frac{(f(x+h)f(x))+(g(x+h)g(x))}{h}\\[1ex] &=\frac{(f(xh)f(x))}{h} + \frac{(g(xh)g(x))}{h}\\[1ex] $r_k(x)&\doteq r_f(x) + r_g(x) \text{ when }h \doteq 0 \end{align}$$
Library of ROC/Integral pairs, Version 2
Accum Fn: $f(x)=$ 
Rate of Change:
$r_f(x)=$ 
Accum Fn from Rate: ${\displaystyle \int_a^x r_f(t)dt}$ 
c 
0 
0 
$cx^n$ 
$cnx^{n1}$ 
$cx^nca^n$ 
$q(x)+p(x)$  $r_q(x)+r_p(x)$  $(q(x)+p(x))(q(a)+p(a))$ 
Let $h(x)=5x^23x+4$. What is $\displaystyle{\int_a^x h(t)dt}$ in closed form?
Rewrite h as $h(x)=f(x)+g(x)+k(x)$, where $f(x)=5x^2$, $g(x)3x$, and $k(x)=4$. We can use the FTC to find $\displaystyle{\int_a^x h(t)dt}$ in closed form:
$$\begin{align} \int_a^x h(t)dt &= \int_a^x f(t)dt + \int_a^x g(t)dt +\int_a^x k(t)dt\\[1ex] &= \int_a^x 5t^2dt + \int_a^x 3t^1dt +\int_a^x 4t^0dt\\[1ex] &= \int_a^x \frac{5}{3} \cdot 3t^2dt + \int_a^x \frac{3}{2} \cdot 2t^1dt +\int_a^x 4t^0dt\\[1ex] &= \frac{5}{3} \int_a^x 3t^2dt  \frac{3}{2}\int_a^x 2t^1dt +4\int_a^x t^0dt\\[1ex] &=\frac{5}{3}(x^3a^3)\frac{3}{2}(x^2a^2)+4(xa)\end{align}$$
Notice in this last example that we changed, for example, $\int_a^x5t^2dt$ to the equivalent statement $\int_a^x \frac{5}{3}\cdot 3t^2dt$.
We then rewrote the second statement as $\frac{5}{3}\int_a^x 3t^2dt$.
This sleight of hand was simply to put the integral into a form that makes it easier to apply the FTC, namely that $\displaystyle{\int_a^x 3t^2dt=x^3a^3}$.
Suppose that k is a composite function such that $k(x)=g(f(x))$ for functions $f$ and $g$. Suppose further that $f$ and $g$ have rate of change functions $r_f$ and $r_g$.
It seems intuitive to think that $r_k$, the exact rate of change function for k, would be $r_k(x)=r_g(f(x))$.
Were this true, then for $k(x)=\left(e^x\right)^2$, $r_k$ would be $r_k(x)=2\left(e^x\right)^1$.
Figure 6.3.1 shows that our intuition is wrong. The graph of r, which we know should be a reasonably good approximation to k’s exact rate of change function, is not even close to the graph of $y=2e^x$. So $r_k(x)≠2e^x$ when $k(x)=(e^x)^2$.
A useful way to think about the composite function k where $k(x)=g(f(x))$ is to remind ourselves that $f(x)$ is the argument to g, not g’s independent variable.
If we were to say $u=f(x)$ and ask about the rate of change of g with respect to u, then $r_k(u)$ would indeed be 2u.
But u is varying at a rate of its own with respect to x, so it is as if variations in x get magnified by f, and then variations in f get magnified by g.
Let’s go further with thinking of rate of change as magnification.
Suppose you have two lenses, f and g, and that f magnifies objects to make them appear 2 times as large, and g magnifies objects to make them appear 3 times as large.
Here is where the idea of rate of change as magnification comes in.
So it is with functions in general: variations in x are magnified by $r_f$, and variations in f are magnified by $r_g$. When $k(x)=g(f(x))$, we have $r_k(x)=r_g(f(x))r_f(x)$. This chain of magnifications is shown in Figure 6.3.2.
Note that Figure 6.3.2 shows three number lines, vertically. The first is the x number line, the second is the f(x) number line, and the third is the k(x) = g(f(x)) number line. In all three, up is positive and down is negative.
Figure 6.3.2 also shows variations in each of x, f(x), and $k(x)=g(f(x))$. The arrows show how the variations are related.
Figure 6.3.2 shows two views of the chain of variations in x, f(x), and g(f(x).
The equation $r_k(x)=r_g(f(x))r_f(x)$, being a chain of magnifications, is called the chain rule.
Reflection 6.3.2 Move the movie slider in Figure 6.3.2 to 9 seconds. Answer each of the following:
Let the functions j, k, and s be defined as $j(u)=u^2+1$, $k(x)=x^3$, and $s(t)=k(j(t))$. According to Version 2 of our Library, $r_j(u)=2u$ and $r_k(x)=3x^2$. Therefore, by the general rule that $r_s(x)=r_k(j(x))r_j(x)$, we have that $r_s(x)=(3(x^2+1)^2)(2x)$.
Let’s check this in GC.
Figure 6.3.3 starts with the closed form accumulation functions $j(u)=u^2+1$, $k(x)=x^3$, and $s(t)=k(j(t))$. The left side also shows GC’s graph of s’s approximate rate of change function, r.
The right graph shows that GC’s graph of our derived function $r_s$ coincides with the graph of r.
Since we trust GC’s graph of r to be a close approximation of s’s actual rate of change function’s graph, Figure 6.3.3’s right image supports our analysis.
(a) 
(b) 
We derived the general rate of change function for a power function whose exponent is a natural number, namely $r_f(x)=nx^{n1}$ when $f(x)=x^n$.
We will use the chain rule together with the fact that $x^{n}=\left(x^{1}\right)^n$ to extend the power rule to the case of any accumulation function that is raised to a negative integer power.
To do this we must first derive the rate of change function for $f(x)=x^{1}$, or $f(x)=\frac{1}{x}$. We do this in Equation 6.3.2.
$$\color{red}{\text{(Eq. 6.3.2)}}\qquad \begin{align}
f(x)&=\frac{1}{x}\\[1ex]
r(x)&=\frac{f(x+h)f(x)}{h}\\[1ex]
&=\frac{\left(\dfrac{1}{x+h}\right)\left(\dfrac{1}{x}\right)}{h}\\[1ex]
&=\frac{\dfrac{x(x+h)}{x(x+h)}}{h}\\[1ex]
&=\frac{xxh}{hx(x+h)}\\[1ex]
&=\frac{h}{hx(x+h)}\\[1ex]
&=\frac{1}{x(x+h)}\\[1ex]
r_f(x)&\doteq \frac{1}{x^2}
\end{align}$$
Now let h be defined as $h(x)=x^{n}$, n a positive integer. We can rewrite h as $h(x)=f(g(x))$, where $f(x)=x^n$ and $g(x)=x^{1}$. By the chain rule, we have
$$\color{red}{\text{(Eq. 6.3.3)}}\qquad \begin{align}In Equation 6.3.3, we used the chain rule to generalize the power rule to $r_h(x)=nx^{n1}$ when $h(x)=x^{n}$ for positive integers $n$. Therefore, $r_h(x)=nx^{n1}$ when $h(x)=x^n$ whether $n$ is negative or positive.
Using the chain rule effectively requires more than remembering it. The primary difficulty in using it effectively is recognizing that a function definition entails a composition of functions. There are two ways of thinking (WOT) that can help.
WOT #1: 
Think of variables in the Library of ROC/Integral Pairs above as arguments instead of as variables. As variables they stand for numbers. As arguments they can stand for numbers, expressions, or functions. 
WOT #2: 
Look for structure in the definition of a function. Look for any expression that could define a function

Another question you will certainly have is, “When should I use the chain rule?”. The answer to this question is, “Always!!”. You can never go wrong by always using the chain rule. The worst that can happen is that your last rate of change function is 1, just like the example in WOT #2, above.
Recall from Chapters 4 and 5 that if a function g has a rate of change at a moment for all values of x, then $dy=r_g(x)dx$ for all moments of x, where $r_g(x)$ is the rate of change of g at the moment x.
We can use this relationship between y and x, together with the chain rule, to integrate rate of change functions that would look too messy to handle normally.
Let $g(x)=x^2+1$, and let $h(x)=3(x^2+1)^2 2x$.
Suppose that $h(x)=r_k(x)$ for some function k. If you look closely, you’ll see that $2x=r_g(x)$, and therefore that $r_k(x)=3(g(x))^2 r_g(x)$.
We know that $\int_a^x r_k(t)dt = k(x)k(a)$. If we let $w=g(x)$, then $dw=r_g(x)dx$, and therefore:
$$\color{red}{\text{(Eq. 6.3.3a)}}\qquad \begin{align} k(x)k(a)&=\int_a^x r_k(t)dt\\[1ex] &=\int_a^x (g(t))^2 r_g(t)dt\\[1ex] &=\int_{g(a)}^{g(x)}3w^2dw\\[1ex] &=(g(x))^3(g(a))^2\\[1ex] &=(x^2+1)^3(a^2+1)^3 \end{align}$$
Therefore, when $r_k(x)=3(x^2+1)^2 2x$, we get $k(x)=(x^2+1)^3$ by recognizing that $3(x^2+1)^2 2x$ is the rate of change function for a composite function and then reversing the chain rule.
While there were several key moves in Equation 6.3.3a, the most critical one was to adjust the limits of the integral upon substituting w for $g(t)$.
How might we understand the decision to make this adjustment? Like this:
In $\int_a^x 3(t^2+1) 2tdt$, t varies from a to x. How does w vary as t varies from a to xwhen we substitute w for $g(t)$ in this integral?
Clearly, w varies from $g(a)$ to $g(x)$ as t varies from a to x. So, $$\int_a^x 3(g(t))^2 r_g(t)dt = \int_{g(a)}^{g(x)} 3w^2 dw.$$
In general, if you recognize that the rate of change function $r_k$ for an unknown function k has the structure $r_k(x)=r_f(g(x))r_g(x)$, and you recognize the functions f and g that have $r_f$ and $r_g$ as their rate of change functions, then:
$$\color{red}{\text{(Eq. 6.3.3b)}}\qquad \begin{align} k(x)k(a)&=\int_a^x r_k(t)dt\\[1ex] &=\int_{g(a)}{g(b)}r_f(w)dw\\[1ex] &=f((g(x)f(g(a)) \end{align}$$
Forthcoming.
We have used the notation $r_f$ (called rate notation) consistently to represent a function f’s exact rate of change function. We did this continually to drive home the point that we are talking about rate of change functions—functions whose values give the rate of change at a moment of another function.
Now that we are deriving rate of change functions, we’ll begin to use other notations: $\frac{d}{dx}$ (called differential notation) and $f'$ (called prime notation). There is a fourth notational system, $D_f$ (called operator notation) that we’ll not use.
Suppose that $f(x)=(x^2+1)^3$. Up to now we represented f’s rate of change function using rate notation as $r_f$. By applying the derivations we have developed we see that $r_f(x)=3(x^2+1)^2(2x)$.
We would write $r_f(x)=3(x^2+1)^2(2x)$ in differential notation as $$\frac{d}{dx}(x^2+1)^3=3(x^2+1)^2(2x).$$Using prime notation, we would write $$f'(x)=3(x^2+1)^2(2x).$$
An advantage of rate notation is that we are reminded that the function we are representing is a rate of change function for another function. Another advantage is that rate notation allows us to represent a rate of change at a moment explicitly, such as $r_f(\sqrt{2})$ to represent $f$'s eact rate of change at the moment $x=2$.
An advantage of differential notation is that you see the function’s original defining formula in close proximity with the derived function’s formula. This makes it easier to remember rules, but it also runs the risk that you cease to think that you are finding a rate of change function.
Differential notation is often used in the context of having defined a function without the use of function notation. If we write $y=(x^2+1)^3$, then we would write $\dfrac{dy}{dx}=3(x^2+1)^2(2x)$ and interpret it as giving us the momentary rate of change of the function defined by $(x^2+1)^3$ at every value of x.
Another advantage to differential notation is that it reminds us that we are talking about differentials in a function’s independent and dependent variables.
A major downside of differential notation is that it gives no way to represent the rate of change of the function at, say, $x=2$ other than to substitute 2 for x in $3(x^2+1)^2(2x)$.
Another disadvantage of differential notation is that it fosters the idea that a function's defining formula is the function itself. We do not have a name for the original function nor for its derived function. Rather, the focus is on having a formula to compute with.
Prime notation is very similar to rate notation in that it reminds you that the derived function is a function and not the symbolic expression that defines it.
A practical drawback for us of prime notation for us is that GC interprets the prime sign ($’$) as a command to draw a graph in a second graphing pane. To GC, prime notation has nothing to do with rate of change functions.
The chain rule, for example, is expressed differently in the three notational systems.
Let f and g be functions and let $h(x)=g(f(x))$.
Keep in mind that all three statements say the same thing about the rate of change function for a composite function.
Do not read these statements literally (e.g., “dee y dee x equals dee y dee u times dee u dee x”). Instead, read them in terms of what they mean.
Recall from Chapter 2 the meaning of θ and y in $y=\sin(θ)$: θ is the an angle measure in radians and y is a fraction of a circle’s radius. The animation in Figure 6.3.4 shows sin(θ) and cos(θ) varying in relation to θ. Focus on sin(θ) in relation to θ. Examine where sin(θ) varies faster and slower in relation to variations in the value of θ. Pause the animation and move the playhead manually to get a sense of how sin(θ) varies for equal increments of θ. Then do the same for cos(θ).
Reflection 6.3.3. Over what intervals of length 0.5 radians does $\sin(\theta)$ vary the most? The least?
We can derive closed form representations for f and g defined as $f(x)=\sin(x)$ and $g(x)=\cos(x)$. In doing so, we will use two identities from trigonometry and two new insights into the sine and cosine functions.
The trigonometric identities are $$\sin(u+v)=\sin(u)\cos(v)+\sin(v)\cos(u)$$which is used in the derivation of sine’s rate of change function in Equation 6.3.4, and $$\cos(u+v)=\cos(u)\cos(v)\sin(u)\sin(v)$$which is used in the derivation of cosine’s rate of change function in Equation 6.3.6.
Let $f(x) = \sin(x)$. We start the derivation of the closed form representation of f’s rate of change function by starting with its approximate rate of change function:
$$\color{red}{\text{(Eq. 6.3.4)}}\qquad\begin{align} r(x)&=\frac{f(x+h)f(x)}{h}\\[1ex] &=\frac{\sin(x+h)sin(x)}{h}\\[1ex] &=\frac{(\sin(x)\cos(h)+\sin(h)\cos(x))\sin(x)}{h}\\[1ex] &=\frac{\sin(x)(\cos(h)1)+\sin(h)cos(x)}{h}\\[1ex] &=\sin(x)\left(\frac{\cos(h)1}{h}\right)+\cos(x)\left(\frac{\sin(h)}{h}\right) \end{align}$$
In Equation 6.3.4 we came to line 5, where we need to understand the behavior of $\dfrac{\cos(h)1}{h}$ and of $\dfrac{\sin(h)}{h}$ for values of h that are indistinguishable from 0 but not 0. We need to say that $h\neq0$ because neither expression is defined for $h=0$.
Examine the two frames in Figure 6.3.5. Figure 6.3.5a shows GC’s graph of $y=\dfrac{\sin(x)}{h}$. Figure 6.3.5b shows GC’s graph of $y=\dfrac{\cos(x)1}{x}$. GC’s graph of $y=\dfrac{\sin(x)}{x}$ suggests that $\dfrac{\sin(x)}{x}$ is essentially 1 for values of x near 0. GC’s graph of $y=\dfrac{\cos(x)1}{x}$ suggests that $\dfrac{\cos(x)1}{x}$ is essentially 0 for values of x near 0.
(a) 
(b) 
If we take as fact that $\dfrac{\sin(x)}{x} \doteq1$ for $h \doteq 0$ and that $\dfrac{cos(x)1}{x} \doteq 0$ for $h \doteq 0$, then we can continue Equation 6.3.4 to conclude that
$$\color{red}{\text{(Eq. 6.3.5)}}\qquad r_f(x)=\cos(x)\text{ when }f(x)=\sin(x)\text{ and that }\int_a^x \cos(t)dt = \sin(x)  \sin(a)$$
From Equation 6.3.5 we can represent the rate of change function for sin(x) in three ways:
We can determine a closed form representation of the rate of change function for $g(x)=\cos(x)$ as before: Start with an approximate rate of change function, derive a form that allows us to omit terms that involve h by letting h become so small that $h \doteq 0$.
$$\color{red}{\text{(Eq. 6.3.6)}}\qquad \begin{align} r(x)&=\frac{g(x+h)g(x)}{h}\\[1ex] &=\frac{\cos(x+h)\cos(x)}{h}\\[1ex] &=\frac{(\cos(x)\cos(h)\sin(x)\sin(h))\cos(x)}{h}\\[1ex] &=\frac{\cos(x)(\cos(h)1)\sin(x)\sin(h)}{h}\\[1ex] &=\cos(x)\left(\frac{\cos(h)1}{h}\right)\sin(x)\left(\frac{\sin(h)}{h}\right)\\[1ex] &\doteq \cos(x)\cdot 0  \sin(x)\cdot 1\text{ when }h\doteq 0\\[1ex] &\doteq \sin(x)\text{ when }h\doteq 0$ \end{align}$$
From the derivation in Equation 6.3.6 we can conclude that
$$\color{red}{\text{(Eq. 6.3.7)}}\qquad r_g(x)=\sin(x)\text{ when }g(x)=\cos(x)\\[1ex] \text{ and that }\int_a^x \sin(t)dt = \cos(x)  \cos(a)$$
From Equation 6.3.6 we can represent the rate of change function for $\cos(x)$ in three ways:
$$\begin{align}r_g(x)&=\sin(x)\text{ when }g(x)=\cos(x)\\[1ex] \frac{d}{dx}\cos(x) &= \sin(x)\\[1ex] g'(x)&=\sin(x)\text{ when }g(x)=\cos(x)\end{align}$$
Because the tangent function is the quotient of sine and cosine functions, we will derive the closed form rate of change function for the tangent function in the next section. It develops the rate of change function for product and quotent functions.
Version 3 of our Library of Rate of Change/Integral pairs is:
Accum Fn: $f(x)=$ 
Rate of Change:
$r_f(x)=$ 
Accum Fn from Rate: ${\displaystyle \int_a^x r_f(t)dt}$ 
c 
0 
0 
$cx^n$ 
$cnx^{n1}$ 
$cx^nca^n$ 
$q(x)+p(x)$  $r_q(x)+r_p(x)$  $(q(x)+p(x))(q(a)+p(a))$ 
$q(p(x))$  $r_q(p(x))r_p(x)$  $q(p(x))q(p(a))$ 
$\sin(x)$  $\cos(x)$  $\sin(x)\sin(a)$ 
$\cos(x)$  $\sin(x)$  $\cos(x)\cos(a)$ 
In this section we will derive the general form of rate of change functions for an accumulation function that is a product of two functions that each have exact rate of change functions.
A quotient function h defined as $$h(x)=\frac{q(x)}{p(x)},\, p(x)\neq 0$$can be expressed as $$h(x)=q(x)(p(x))^{1}.$$where $(p(x))^{1}$ means $\dfrac{1}{p(x)}$.
We will therefore treat a quotient of two functions as a product of two functions.
Let u be a product of functions f and g that each have a rate of change function at every moment. Then $u(x)=g(x)f(x)$.
We know that in finding u’s exact rate of change function we will form an approximate rate of change function that has $g(x+h)f(x+h)g(x)f(x)$ in its numerator. This form, however, is not very helpful for expressing the rate of change function for $u$ in relation to the rate of change functions for $g$ and $f$.
Suppose we were to factor out $g(x+h)$. We could do this if we had $g(x+h)f(x)$ as a term. Well, we can add $g(x+h)f(x)$ and then compensate by subtracting $g(x+h)f(x)$, too, in order to keep the numerator equivalent to $g(x+h)f(x+h)g(x)f(x)$. This is like adding zero when we add and subtract the same term.
The numerator in its new, equivalent form is
which can be simplified to
which is far more useful as a numerator because it represents both a change in g and a change in f.
Equation 6.3.8 employs the trick of adding and subtracting the same term in order to put the approximate rate of change function’s numerator in a more useful form. It ends with the general result that $r_u(x)=g(x)r_f(x)+r_g(x)f(x)$ when u is the product of two functions g and f that each have exact rate of change functions.
Figure 6.3.6 shows a more intuitive derivation of the rate of change function for a product of functions. It uses area as a model of multiplication. The rectangle has side lengths of $g(x)$ and $f(x)$, so the area function A defined as $A(x)=g(x)f(x)$ gives the rectangle’s area for every value of x. The question is at what rate the area increases when we increase x by dx.
Figure 6.3.6 shows the value of x increasing by a tiny amount dx.
We can express the product rule in three ways, using rate, differential, and prime notation:
$$\begin{align} \text{Let }u(x)&=g(x)f(x).\text{ Then:}\\[2ex] r_u(x)&=g(x)r_f(x)+r_g(x)f(x)\\[1.5ex] \frac{d}{dx}u(x)&=g(x)\left(\frac{d}{dx}f(x)\right)+\left(\frac{d}{dx}g(x)\right)f(x)\\[1.5ex] u'(x)&=g(x)f'(x)+g'(x)f(x) \end{align}$$
The tangent function is defined as $$\tan(x)=\frac{\sin(x)}{\cos(x)}, \cos(x)\neq 0,$$ which can be rewritten as a product: $$\tan(x)=(\sin(x))(\cos(x))^{1}.$$ We can derive the tangent function’s exact rate of change function by applying the product rule, the power rule, and the chain rule, and using the definition of the secant function as $\sec(x)=\dfrac{1}{\cos(x)}$.
$$\color{red}{\text{(Eq. 6.3.9)}}\qquad \begin{align} \tan(x)&=\frac{\sin(x)}{\cos(x)},\, \cos(x)\neq 0\\[1ex] &=\sin(x)(cos(x))^{1}\\[2ex] \frac{d}{dx}\tan(x) &=\sin(x)\left(\frac{d}{dx}(\cos(x))^{1}\right)+\left(\frac{d}{dx}\sin(x)\right)(\cos(x))^{1}\\[1ex] &=\left(\sin(x)\left(1(\cos(x))^{2}(\sin(x))\right)\right)+(\cos(x))(\cos(x))^{1}\\[1ex] &=\frac{(\sin(x))^2}{(\cos(x))^2}+\frac{\cos(x)}{\cos(x)}\\[1ex] &=\frac{(\sin(x))^2+(\cos(x))^2}{(\cos(x))^2}\\[1ex] &=\frac{1}{(\cos(x))^2}\\[1ex] &=\left(\frac{1}{\cos(x)}\right)^2\\[1ex] &=(\sec(x))^2 \end{align}$$
Therefore, $r_{\tan}(x)=(\sec x)^2$.
We can express the derived rate of change function for $\tan(x)$ in three ways using our three notation schemes:
Accum Fn: $f(x)=$ 
Rate of Change:
$r_f(x)=$ 
Accum Fn from Rate: ${\displaystyle \int_a^x r_f(t)dt}$ 
c 
0 
0 
$cx^n$ 
$cnx^{n1}$ 
$cx^nca^n$ 
$q(x)+p(x)$  $r_q(x)+r_p(x)$  $(q(x)+p(x))(q(a)+p(a))$ 
$q(p(x))$  $r_q(p(x))r_p(x)$  $q(p(x))q(p(a))$ 
$\sin(x)$  $\cos(x)$  $\sin(x)\sin(a)$ 
$\cos(x)$  $\sin(x)$  $\cos(x)\cos(a)$ 
$\tan(x)$  $(\sec x)^2$  $\tan(x)\tan(a)$ 
$p(x)q(x)$  $p(x)r_q(x)+r_p(x)q(x)$  $p(x)q(x)p(a)q(a)$ 
In Chapter 3, Section 3.10, we distinguished between functions defined explicitly and functions defined implicitly.
In the equation $x^2xy+y^2=3$ we can envision x being a function of y or y being a function of x over suitably restricted intervals of x and y.
Suppose that we envision x as a function of y. This supposition has the same effect as rewriting $$x^2xy+y^2=3\text{ as}$$ $$(g(y))^2g(y)y+y^2=3,$$ where we now interpret "3" as the constant function $f(y)=3$.
We do not know the actual definition of g, nor do we know immediately how y must be restricted so that g is a function. But we can proceed nevertheless with the assumption that g is defined as a function of y and that 3 is a function of y without the definition of g and without knowing ahead of time how y must be restricted.
Assume that g is an accumulation function and that y is its independent variable.
We can then apply what we know about derived rate of change functions to the equation $(g(y))^2g(y)y+y^2=3$.
Since the left side and right side are equal for all values in the domain of y, the left side will change at the same rate with respect to y as will the right side.
We can therefore perform the derivation given in Equation 6.17.
Print Equation 6.3.10 so that you can write on it as you read the following moves that we made in this derivation.
We:
It might seem strange that we end with a rate function for $g(y)$ that involves both x and y. However, what our definition of $r_g$ says is that values of $r_g(y)$ depend on values of both x and y.
While it might seem more accurate to express $r_g$ as a function of both x and y, this would not be appropriate. To define $r_g$ as a function of x and y would imply that both x and y are independent variables for $r_g$. But this is not the case. Values of x are dependent upon values of y and values of y are dependent upon values of x. Neither x nor y determines values of the other uniquely.
We can test the derivation in Equation 6.3.10 in this way:
(a) 
(b)

The method exemplified in Equation 6.3.10, deriving the rate of change function for a function defined implicitly, is often called implicit differentiation. It is important to realize that implicit differentiation (deriving the rate of change function for a function defined implicitly) is a method. It is not a rule.
As an immediate application of the method of implicitly derived rate functions we will address the issue that, to this point, the power rule applies only to functions that are to an integer power.
Suppose that $y=x^\frac{p}{q}$, where p and q are integers and $q \neq 0$. If $y=x^\frac{p}{q}$, then it is also true that $y^p=(x^\frac{p}{q})^q$, or $y^q=x^p$. Using the method of implicitly defined rate functions, we have
$$\color{red}{\text{(Eq. 6.3.11)}}\qquad \begin{align} y&=x^{p/q},\, p\text{ and }q \text{ integers, }q\neq 0\\[2ex] y^p&=x^q\\[1ex] \frac{d}{dx}y^p&=\frac{d}{dx}x^p\\[1ex] qy^{q1}\frac{dy}{dx}&=px^{p1}\frac{dx}{dx}\\[1ex] \frac{dy}{dx}&=\frac{px^{p1}}{qy^{q1}}\\[1ex] &=\left(\frac{p}{q}\right)\frac{x^{p1}}{y^{q1}}\\[1ex] &=\left(\frac{p}{q}\right)x^{p1}y^{1q}\\[1ex] &=\left(\frac{p}{q}\right)x^{p1}\left(x^{p/q}\right)^{1q}\\[1ex] &=\left(\frac{p}{q}\right)x^{p1}x^{(p/q)(1q)}\\[1ex] &=\left(\frac{p}{q}\right)x^{p1}x^{p/qp}\\[1ex] &=\left(\frac{p}{q}\right)x^{p1p/qp}\\[1ex] &=\left(\frac{p}{q}\right)x^{p/q1}\\[2ex] \frac{dy}{dx}&=\left(\frac{p}{q}\right)x^{p/q1}\text{ when }y=x^{p/q} \end{align}$$
Therefore, $$\frac{dy}{dx}=(\dfrac{p}{q})x^{p/q1} \text{ when } y=x^{p/q}.$$
We could also write this as $$\dfrac{d}{dx}x^{p/q}=\dfrac{p}{q}x^{\frac{p}{q}1},$$
or as$$r_f(x)=x^{p/q1} \text{ when } f(x)=x^{p/q}.$$
All of these statements assume that p and q are integers, $q\neq0$.
In other words, the power rule works for rational exponents as well as integers. Indeed, the power rule works for all real numbers, but the argument for the power rule working for all real numbers demands an understanding of real numbers that is beyond the scope of an introductory calculus course.
In this development of rate of change functions for inverse trigonometric functions, we will denote inverse sine of x as “asin(x)”, which is short for “arcsine of x”, instead of the oftenused notation “$\sin^{1}(x)$”.
This is for two reasons:
(1) students often confuse the negative exponent to mean $\dfrac{1}{\sin(x)}$, and
(2) GC uses the notation asin(x) to mean the arc sine functionthe arc which produces $\sin(x)$.
We will similarly use “acos(x)” in place of “$\cos^{1}(x)$” and “atan(x)” in place of “$\tan^{1}(x)"$.
We will use several facts and relationships in the derivation of the rate of change function for asin(x), which we will denote in differential notation as $\frac{d}{dx}\asin(x)$. These are:
The deriviation of $\dfrac{d}{dx}\asin(x)$ is given in Equation 6.3.12. We derive the rate of change function for $\asin(x)$ implicitly by deriving the rate of change function for sin(y), where $y=\asin(x)$.
$$\color{red}{\text{(Eq. 6.3.12)}}\qquad \begin{align} y&=\asin(x),\, 1\lt x \lt 1\\[1ex]\sin(y)&=x&\text{Def. of inverse}\\[1ex] \frac{d}{dx}\sin(y) &=\frac{d}{dx}x\\[1ex] \cos(y)\frac{dy}{dx} &=1 & \text{Chain rule}\\[1ex] \frac{d}{dx}y&=\frac{1}{\cos(y)}&\cos(y)\ge 0\text{ since }\pi/2\le y \le \pi/2\\[1ex] &=\frac{1}{\sqrt{1\left(\sin(y)\right)^2}}&\cos^2(y)+\sin^2(y)=1\\[1ex] &=\frac{1}{{\sqrt{1x^2}}}&x=\sin(y)\\[1ex] \text{Therefore, }\frac{d}{dx}\asin(x)&=\frac{1}{\sqrt{1x^2}},\,1\lt x\lt 1 \end{align}$$
Or, in rate notation, $r_f(x)=\dfrac{1}{\sqrt{1x^2}}$ when $f(x)=\asin(x), 1<x<1$.
It is left as exercises for you to use the same method as above to derive the rate of change functions for acos(x) and atan(x).
Confirm Equation 6.3.10 by clicking on the graph of $x^2xy+y^2=3$ at several points, recording their coordinates, and modifying the GC statements in Figure 6.3.7 accordingly.
In Equation 6.3.10 we assumed that x is a function of y. Repeat Equation 6.3.10 with the assumption that y is a function of x, then test your derivation in the same way as in Figure 6.3.7.
Repeat the derivation in Equation 6.3.9 twice, first using rate notation then using prime notation.
Assume that $y=g(x)$ in each of ac. Find the exact rate of change function $r_g$ for each. Use any notational system that is convenient. Check your derivation by graphing it along with the graph of g’s approximate rate function.
Find the exact rate of change function, $r_f(x)$, for each function $f(x)$ shown below. Check your derivation by graphing it along with the graph of f’s approximate rate function.
For each of 5c and 5d, use your work to find $\displaystyle{\int_a^x r_f(t)dt}$ in closed form. What do you notice?
A 10foot ladder leans against a wall. A mischievous monkey kicks the ladder so that its bottom slides along the floor (see animation below). The bottom of the ladder is initially 0 feet from the wall. The bottom moves away from the wall at a rate of 1.3 feet per second.
A man having a height of 1.75 meters stood under a street light, then walked away it at a rate of 1.7 meters per second. The streetlight is 6 meters above the street. The streetlight casts a shadow in front of him.
Use the method of deriving a rate of change function implicitly to show that $r_f(x)=\dfrac{1}{1+x^2}$ when $f(x)=\atan(x)$. Use these facts:
Restrict the value of x appropriately in $r_f(x)$.
Check your derivation by graphing it along with the graph of $y=\dfrac{f(x+0.001)f(x)}{0.001}$.
Use the method of deriving a rate of change function implicitly to show that $r_f(x)=\dfrac{1}{1x^2}$ when $f(x)=\acos(x)$. Use these facts:
Restrict the value of x appropriately in $r_f(x)$.
Check your derivation by graphing it along with the graph of $y=\dfrac{f(x+0.001)f(x)}{0.001}$.
The diagram below shows a kite 100 ft above the ground moves horizontally away from you at a rate of 8 feet per second. The kite’s height does not change, but the angle made by the ground and the kite string varies as the kite moves away from you.
The animation below shows the hands of a large clock rotating rapidly. The hour hand is 1.5 meters long; the minute hand is 4 meters long. Define a function whose values give the rate at which the distance between the tips of the hour and minute hands varies with respect to the measure of the angle between them.
In this section we will first review the ideas of exponential and logarithmic functions and then derive rate of change functions for them in closed form.
Paal Payasam is an Indian dish made of milk and rice. The Legend of Paal Payasam illustrates the immensity of exponential growth.
The legend is that Lord Krishna played a game of chess against King Rhada for this bet: If Krishna won, the King must give him 1 grain of rice on the board’s first square, 2 grains on the second, 4 grains on the third, and $2^{n1}$ grains on the $n^{th}$ square, up to $n=64$.
Krishna won the game, and King Rhada quickly understood the foolishness of his bet. He foresaw having to put 4.29 billion grains of rice on the $33^{rd}$ square and 9.22 billion billion grains on the $64^{th}$ squarea number of grains of rice that would cover all of India.
The expressions $x^b$ and $b^x$ have similar forms, but they have entirely different meanings and, as illustrated by the Legend of Pall Paysam, entirely different behaviors.
When we say $y=x^b$, we are saying that for any value of x, the value of y contains x as a factor b times. If $b=2$, then y contains 2 factors of x, or $y=x \cdot x$. The value of x varies, but the number of factors does not vary.
$y=x^b$ 
$y=b^x$ 
y has x as a factor
b times; x varies, b is constant 
y has b as a factor
x times; b is constant, x varies 
Functions of the form $f(x)=x^b$ are called power functions, as you already know. In a power function, the exponent of x is constant. It does not vary.
Functions of the form $f(x)=b^x$ are called exponential functions. In an exponential function, it is the exponent that varies.
It would be convenient if exponential functions obeyed the power rule. However, they do not.
Figure 6.3.8 shows the power rule applied to $f(x)=2^x$ (see GC’s graph of $r_1$) compared to f’s approximate rate of change function (see GC’s graph of $r_2$).
We trust the approximate rate function with $h=0.001$ to be reasonably close to the exact rate of change function for f.
GC’s graph of $r_1$, which uses the power rule on f, is not even close to GC’s graph of $r_2$, f’s approximate rate function. The power rule does not apply to exponential functions.
To understand the derivation of the general form of an exponential function’s rate of change function you must recall properties of exponents. As a reminder:
Two special cases are when $a=0$ or $n = 0$ in $a^n$.
The derivation of the rate of change function for $f(x)=b^x$ is given in Equation 6.3.13.
The derivation starts by defining the approximate rate function r, and deriving the statement $r(x)=b^x(\frac{b^h1}{h})$.
That is, the rate of change function for $f(x)=b^x$ is a product of $b^x$ and the expression $\dfrac{b^h1}{h}$. We cannot let $h=0$ because $\frac{b^01}{0}=\frac{0}{0}$, which is undefined.
The question we must answer, then, is whether $\frac{b^h1}{h}$ represents a number when $h$ is indistinguishable from zero but not equal to zero.
$$\color{red}{\text{(Eq. 6.3.13)}}\qquad \begin{align} f(x)&=b^x,\,b\gt 0\\[2ex] r(x)&=\frac{f(x+h)f(x)}{h},\, h\neq 0\\[1ex] &=\frac{b^{x+h}b^x}{h}\\[1ex] &=\frac{b^xb^hb^x}{h}\\[1ex] &=\frac{b^x\left(b^h1\right)}{h}\\[1ex] &=b^x\left(\frac{b^h1}{h}\right) \end{align}$$
Equation 6.3.13 ends with $b^x$ multiplied by $(b^h1)/h$. At this moment we don't know what to make of $(b^h1)/h$.
Figure 6.3.9 investigates the behavior of $\dfrac{b^x1}{x}$ around $x=0$ for various values of b from 0 to 500.
When you play the animation in Figure 6.3.9 you should see:
For $b=0$, $\dfrac{b^x1}{x}$ has $x=0$ as a vertical asymptote.
For values of $b>0$, it appears that $\dfrac{b^x1}{x}$ is always essentially equal to some number for $x \doteq 0$, although it is a different number for each value of b.
The main point of Figure 6.3.9 is that $(b^h1)/h$ is always equal to some number when $h \neq 0$ and $b>0$. But the value of $b^h1)/h$ depends on the value of b.
For the meantime, we’ll represent $(b^h1)/h$ as $c_b$ to remind us that $(b^h1)/h$ is a number for $h \neq 0$. We use “sub b” in $c_b$ to remind us that this number depends on the value of b.
When $f(x)=b^x, b>0$, we have that $r_f(x)=c_b b^x$, where the value of $c_b$ depends on the value of b.
For some value of b, $c_b$ is essentially equal to 1 when $h \doteq 0$.
In honor of Leonard Euler, we give the special symbol e to the value of b that makes $\dfrac {b^h1}{h}\doteq 1$ when $h\doteq 0$. Therefore $c_e=1$ by definition of e.
The importance of the value of b that makes $c_b=1$ is that the rate function for this exponential function is $r_f(x)=f(x)$ when $f(x)=e^x$.
In differential notation, $\dfrac{d}{dx}e^x=e^x$.
The value of e is irrational. Its value is approximately 2.71828 (to 5 decimal places). The history of attempts to approximate the value of e is fascinating. See Eli Maor’s excellent account of this history.
A logarithmic function is the inverse of an exponential function, and thus its definition depends on the base of the exponential function for which it is inverse. In other words,
For $b>0$, $\log_b(x)=y$ such that $b^y=x$
So $\log_3(81)=y$ such that $3^y=81$. Therefore $\log_3(81)=4$, because $3^4=81$. Also, $\log_{\frac{1}{3}}(81)=4$ because $\left(\dfrac{1}{3}\right)^{4}=81$.
We can use the inverse relationship between $\log_b$ and $b^x$ and the method of implicitly derived rate functions to find the derived rate of change function for f defined as $f(x)=\log_b(x), b>0$.
Equation 6.3.14 starts by stating the inverse relationship between exponential and logarithmic functions and then uses the method of implictly derived rate functions and the chain rule to end with the derived rate function for $y=\log_b(x)$.
$$\color{red}{\text{(Eq. 6.3.14)}}\qquad \begin{align} \log_b(x)&=y\text{ means that $b^y=x$} & \text{Def. of Inverse Function}\\[1ex] \frac{d}{dx}\log_b(x)&=\frac{d}{dx}y\text{ means that }\frac{d}{dx}b^y=\frac{d}{dx}x\\[1ex] \text{Derivation:}\\[1ex] \frac{d}{dx}b^y&=\frac{d}{dx}x\\[1ex] c_b \frac{d}{dx}y&=1 & \text{Chain rule}\\[1ex] \frac{d}{dx}y&=\frac{1}{c_b b^y}\\[1ex] \frac{d}{dx}y&=\frac{1}{c_b x} & \text{Substitute $x$ for }b^y \end{align}$$
In the case where $b=e$, $c_e=1$, so we have
$$\frac{d}{dx}\log_e(x)=\frac{1}{c_e x}\text{, or}$$ $$\frac{d}{dx}\log_e(x)=\frac{1}{x}.$$
Reflection 6.3.4. Why do we include the restrictions $b>0$ and $x>0$ in the definition of a logarithm function? Hint: Suppose that we allowed $b≤0$ or $b=1$? What would go wrong? What about values of $x≤0$?
Because $e^x$ has the special property that $\dfrac{d}{dx} e^x=e^x$, and because $\log_e(x)$ has the special property that $\dfrac{d}{dx}\log_e(x)=\dfrac{1}{x}$, e is called the natural base of the exponential and log functions.
The natural log is denoted “ln” (from Latin, logarithmus naturalis).
In the future we will write $\log_e(x)$ as $\ln(x)$, and we will call $\ln(x)$ the natural log of x.
Equation 6.3.15 gives us the opportunity to determine the value of this ubiquitous constant $c_b=\dfrac{b^h1}{h}$ for $h \doteq 0$.
In determining $c_b$ we will use the simplicity of the derived rate function for $f(x)=e^x$ together with the fact that $b^{\log_b(x)}=x$ for any value of $b>0$ and for any value of $x>0$.
Reflection 6.3.5. Check for yourself that $b^{\log_b(x)}=x$. Use the fact that log and exponential functions are inverse functions:
Now we use the fact that $b^x=e^{\ln b^x}$ to determine the value of $c_b$ in $\dfrac{d}{dx}b^x=c_b b^x$.
$$\color{red}{\text{(Eq. 6.3.15)}}\qquad \begin{align} b^x&=e^{\ln b^x}\\[1ex] &=e^{x\ln b}\\[1ex] \frac{d}{dx}b^x&=\frac{d}{dx}e^{x\ln b}\\[1ex] &=e^{x\ln b}\frac{d}{dx}x\ln b & \text{Chain rule}\\[1ex] &=e^{x\ln b}\ln b\\[1ex] &=(\ln b)b^x & \text{Substitute }b^x\text{ for }e^{x \ln b}\\[2ex] \text{Therefore, }c_b&=\ln b\\[1ex] \frac{b^h1}{h}&= \ln b\text{ for }h\doteq 0 \end{align}$$
In summary, and in differential notation:
The case of $\dfrac{d}{dx}\ln(x)=\dfrac{1}{x}$ deserves more consideration.
$\ln(x)$ is defined only when $x>0$. However, $\dfrac{1}{x}$ is defined when $x>0$ and when $x<0$. The question therefore becomes,
Figure 6.3.10 addresses this question. It shows that $\int_a^x \frac{1}{t}dt$ is defined for $a<0$ and $x<0$ as well as for $a>0$ and $x>0$.
Figure 6.3.10 also demonstrates that the graphs of $\displaystyle{\int_a^x \frac{1}{t}dt}$ coincides with the graph of $y=\ln(x)\ln(a)$ when a and x are both negative.
We can therefore say that $\int_a^x \frac{1}{t}dt = \ln(x)\ln(a)$ when a and x are both negative or both positive.
We cannot integrate $\frac{1}{t}$ over an interval that includes 0 because $\frac{1}{t}$ is undefined for $t=0$, and, as we will see later, $\int_0^c \frac{1}{t}dt$ is infinite for any value of c.
Make an updated library of ROC/Integral pairs.
Show that $\log_3(x)$ can be expressed as $\dfrac{\ln(x)}{\ln(3)}$.
Show that $\log_b(x)$ can be expressed as $\dfrac{\ln(x)}{\ln(b)}$. Use this fact to graph $y=\log_7(x)$.
In this section we will derive rate of change functions for functions of the form $h(x)=f(x)^{g(x)}$. Functions in this form are not power functions, since the exponent is not constant, and they are not exponential functions, since the base is not constant.
Briggs, Cochran, Gillet, and Schultz (2011) propose to call functions of the form $h(x)=f(x)^{g(x)}$ tower functions. We will adopt Briggs et al.’s proposal.
Suppose that h is defined as $h(x)=g(x)^{f(x)}$ and that f and g have rate of change functions $r_f$ and $r_g$. We do not have a method for deriving h’s rate of change function.
Just as in much of mathematics, our question is not so much about how to do something, but how to represent what we want to act upon so that we can use the methods that we have. We ask, “How can we represent the function $h(x)=g(x)^{f(x)}$ in an equivalent form so that we can use existing methods?”
We can rewrite the definition of h using the identities $g(x)=e^{\ln(g(x))}$ and $\ln(x^y)=y\ln x$ to put the definition of h in a form where we do have methods to derive its rate of change function.
Before we do this, however, we must point out that $g(x)$ and $e^{\ln(g(x))}$ are not always equivalent. Because $\ln(u)$ is defined only for values of $u>0$, the expression $e^{\ln(g(x))}$ is defined only for values of x such that $g(x)>0$.
Figure 6.3.11 illustrates the constraint that $g(x)$ must be greater than 0. The first pane in Figure 6.3.11 shows GC’s graphs of $y=x^2$ and $y=e^{\ln(x^2)}$ being identical. The second pane shows that GC’s graphs of $y=\sin(3x)$ and $y=e^{\ln(\sin(3x))}$ are identical only for values of x such that $\sin(3x)>0$.
(a) 
(b) 
Suppose that $h(x)=g(x)^{f(x)}$, $g(x)>0$, and f and g have rate of change functions $r_f$ and $g_f$. We will use two identities in the derivation of $r_h$, namely:
We also will use the chain rule several times, and the derived functions for products and logs. The derivation of $r_h$ where $h(x)=g(x)^{f(x)}$, $g(x)>0$ is given in Equation 6.3.17.
$$\color{red}{\text{(Eq. 6.3.17)}}\qquad \begin{align} h(x)&=g(x)^{f(x)}\\[1ex] &=e^{f(x)\ln(g(x))},\, g(x)\gt 0\\[1ex] r_h(x)&=\frac{d}{dx}e^{f(x)\ln(g(x))}\\[1ex] &=e^{f(x)\ln(g(x))}\frac{d}{dx}\left(f(x)(\ln(g(x))\right) & \text{Chain rule}\\[1ex] \frac{d}{dx}f(x)\ln(g(x))&=f(x)\frac{d}{dx}\ln(g(x))+\ln(g(x))\frac{d}{dx}f(x) & \text{Product rule}\\[1ex] &=f(x)\left(\frac{1}{g(x)}\right)\frac{d}{dx}g(x)+\ln(g(x))\frac{d}{dx}f(x) & \text{ln and chain rules}\\[1ex] &=\frac{f(x)}{g(x)}r_g(x)+\ln\left(g(x)\right)r_f(x)\\[1ex] r_h(x)&=g(x)^{f(x)}\left(\frac{f(x)}{g(x)}r_g(x)+\ln\left(g(x)\right)r_f(x)\right),\, g(x)\gt 0 \end{align}$$
Reflection 6.3.6. Download this file. Then do the following.
Reflection 6.3.7. What is $r_f(x)$ when $f(x)=x^x$? Over what domain is $r_f$ defined?
Reflection 6.3.8. Examine GC’s graph of $y=x^x$. Why is it reasonable to adopt the convention that $0^0=1$ even though, logically, $0^0$ is undefined?
Examine GC’s graph of $f(x)=(\cos(x))^x$. (Zoom out so that you can see at least $20<x<20$.)