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Calculus Concepts and Modelling: An introduction to the concepts of calculus

Jeremy Sylvestre

Chapter 12 The derivative function

Section 12.1 Definition

In Section 6.3, we turned the process of calculating an accumulation for a rate model over a specific time interval into a function by replacing the specific end time with a variable one. The resulting function accepted inputs that represented that variable end time, and returned outputs that represented accumulation values over time domains of different lengths, but always with the same start time.
The process of computing derivative values can be turned into a function in a much simpler way. In Section 11.5, we defined a Derivative value at a specific input time for a given quantity function. To turn this process into a function, we merely have to replace “at a specific time” with “at a variable time”, so that the new derivative function takes a time as input and, if the quantity function is differentiable at that input time, returns the associated derivative value.

Definition 12.1.1. Derivative function.

Given quantity function \(q(t)\text{,}\) the associated derivative function is the function \(q'(t)\) so that
\begin{equation*} q'(t) = \dqdt \end{equation*}
if \(q\) is differentiable at \(t\text{,}\) and \(q'(t)\) is undefined if \(q\) is not differentiable at \(t\text{.}\)

Notation.

We use \(q'(t)\) when we want to refer to the derivative function for \(q\) in function notation, while we use
\begin{equation*} \dqdt \end{equation*}
to represent an input-output formula in \(t\) for \(q'\text{.}\) Of course, both of these forms are relative to the initial choice of letter \(q\) — if we were dealing with a position function \(s(t)\) as our quantity function, then we would use \(s'(t)\) and
\begin{equation*} \ddt{s} \end{equation*}
to refer to the derivative. In some contexts, particularly physics, you might see \(\dot{q}\) used to representative the derivative function for \(q(t)\text{.}\)
It is also common practice in mathematics and other science disciplines to use a “bare” \(\opddt\) as an operator that applies the differentiation process to a function or formula. For example, notation in the form
\begin{equation*} \ddt{e^t} = \text{(some formula)} \end{equation*}
or
\begin{equation*} \opddt e^t = \text{(some formula)} \text{,} \end{equation*}
is meant to be read as “a formula for the derivative function associated to the exponential function is (some formula).” Or, more simply, “the derivative of the exponential function is (some formula)” — that is, the symbol \(\opddt\) represents the phrase “the derivative of.”

Principle 12.1.2. Derivative function provides a rate model.

If \(q(t)\) is a differentiable quantity model, then \(q'(t)\) is a rate model for that quantity.
We won’t be able to verify Principle 12.1.2 until Chapter 13, where we will see that if we start with a rate model \(r(t)\text{,}\) use an accumulation function to create an associated quantity model
\begin{equation*} q(t) = q(t_0) + \ccmint{t_0}{t}{r(u)}{u} \text{,} \end{equation*}
then this quantity model will be differentiable and its derivative function will be exactly the rate function we began with:
\begin{equation*} q'(t) = r(t) \text{.} \end{equation*}
Intuitively it seems simple that the above description is a circular process, so of course we should return back to the rate function we began with. But mathematically, it is far from obvious that the processes behind the two concepts integral and derivative are reverse processes.

Section 12.2 Computing algebraically

In Section 11.5, we calculated individual derivative values by looking for patterns in average-rate approximations to rate at a specific time as we made \(\change{t}\) smaller. By replacing the specific time \(t^\ast\) with a variable time \(t\) and working conceptually with a time domain of “infinitesimal” length \(dt\) instead of a time domain of substantial length \(\change{t}\text{,}\) in many cases we can obtain a formula for all output values of \(q'(t)\text{.}\) We think of \(dt\) as being so small as to be negligible, but always consider \(dt \neq 0\) (or else the denominator of \(dt\) represents division by zero).
For this, we will exclusively use the forward approximation formula, so that the approximation at a specific point in time
\begin{equation*} \funceval{\dqdt}{t = t^\ast} \approx \slope{q}{t} = \frac{ q(t^\ast + \change{t}) - q(t^\ast) }{ \change{t} } \end{equation*}
becomes the general, exact formula
\begin{equation*} \dqdt = \frac{ q(t + dt) - q(t) }{ dt } \text{.} \end{equation*}
The key to the process is to not consider the “infinitesimal” length \(dt\) negligible too early, since if we take \(dt \approx 0\) right away, we will always end up with
\begin{align*} \dqdt \amp = \frac{ q(t + 0) - q(t) }{ 0 } \\ \amp = \frac{ 0 }{ 0 } \text{,} \end{align*}
which is undefined. Instead, we simplify as much as possible until we can definitively see the effect of \(dt \to 0\text{.}\)
Here is a basic example of the process.

Example 12.2.1. Velocity for parabolic motion.

Let’s return to the example situation we made us of repeatedly in Chapter 7 of an object launched from the surface of an airless planet whose motion is then determined by the gravitational pull of the planet. (See Example 7.2.1, Example 7.3.1, and Example 7.3.4.) In those examples, we began with a constant acceleration model and traced it back to a parabolic position model
\begin{equation*} s(t) = -\frac{g}{2} t^2 + v_0 t \text{.} \end{equation*}
For simplicity in this example, let’s use actual numbers for \(g\) and \(v_0\) — say, \(g = 4\) and \(v_0 = 6\text{.}\) Our positional model is then
\begin{equation*} s(t) = -2 t^2 + 6 t \text{,} \end{equation*}
and we have
\begin{align*} s(t + dt) \amp = -2 (t + dt)^2 + 6 (t + dt) \\ \amp = -2 (t^2 + 2 t dt + dt^2) + 6 (t + dt) \\ \amp = -2 t^2 - 4 t dt - 2 dt^2 + 6 t + 6 dt \text{,} \end{align*}
so that
\begin{align*} s(t + dt) - s(t) \amp = (\bcancel{-2 t^2} - 4 t dt - 2 dt^2 + \bcancel{6 t} + 6 dt) - (\bcancel{-2 t^2} + \bcancel{6 t}) \\ \amp = - 4 t dt - 2 dt^2 + 6 dt \\ \amp = (- 4 t + 6 - 2 dt) dt \text{.} \end{align*}
Therefore,
\begin{align*} \ddt{s} \amp = \frac{ s(t + dt) - s(t) }{ dt } \\ \amp = \frac{ (- 4 t + 6 - 2 dt) \bcancel{dt} }{ \bcancel{dt} } \\ \amp = - 4 t + 6 - 2 dt \text{.} \end{align*}
Now treating \(dt\) as negligible, we can clearly see that
\begin{equation*} \ddt{s} = -4 t + 6 \text{.} \end{equation*}
In Example 7.3.4, we obtained the position model
\begin{equation*} s(t) = -\frac{g}{2} t^2 + v_0 t \end{equation*}
from the velocity model
\begin{equation*} v(t) = -g t + v_0 \text{.} \end{equation*}
Substituting in the values for \(g\) and \(v_0\) chosen in this example, we have
\begin{equation*} v(t) = -2 t + 6 \text{,} \end{equation*}
which matches exactly with our \(ds/dt\) formula calculated above. So, taking \(ds/dt\) as a formula for the derivative function \(s'(t)\text{,}\) we have
\begin{equation*} s'(t) = v(t) \end{equation*}
as expected.

Section 12.3 Fundamental differentiation formulas and principles

Subsection 12.3.1 Two fundamental formulas

Function formulas are built out of the independent variable and constants. So we’ll start by analysing the two simplest possible quantity models built on those two components.
Justification.
Consider quantity function \(q(t) = c\text{.}\) Then also \(q(t + dt) = c\text{,}\) so
\begin{equation*} q(t + dt) - q(t) = 0 \text{.} \end{equation*}
Therefore,
\begin{equation*} \dqdt = \frac{ q(t + dt) - q(t) }{ dt } = \frac{0}{dt} \text{.} \end{equation*}
In an actual average-rate approximation (that is, with \(dt \neq 0\)), this would always compute to \(0\text{.}\) So, in fact, these approximations would not be approximations but instead they are already the exact value, and it is correct to write
\begin{equation*} \dqdt = \frac{0}{dt} = 0 \text{.} \end{equation*}
We are not making the mistake of taking \(0/0 = 0\) to be true, because we think of our infinitesimal time domain length \(dt\) as being so small as to be negligible, but we always consider \(dt \neq 0\) to be true.
Justification.
Consider quantity function \(q(t) = t\text{.}\) Then \(q(t + dt) = t + dt\text{,}\) so
\begin{equation*} q(t + dt) - q(t) = dt \text{.} \end{equation*}
Therefore,
\begin{align*} \dqdt \amp = \frac{ q(t + dt) - q(t) }{ dt } \\ \amp = \frac{ dt }{ dt } \\ \amp = 1 \text{.} \end{align*}
Again, there is no \(0/0\) error here, as we think of \(dt\) as negligible but nonzero.

Subsection 12.3.2 Scaling and adding quantities does the same to the derivative

Now we’ll work a little more generally to develop some algebraic rules for derivatives.
Justification.
Consider quantity function \(p(t) = c q(t)\text{.}\) Then
\begin{equation*} p(t + dt) = c q(t + dt) \text{,} \end{equation*}
so
\begin{align*} p(t + dt) - p(t) \amp = c q(t + dt) - c q(t) \\ \amp = c \bbrac{ q(t + dt) - q(t) } \text{.} \end{align*}
Therefore,
\begin{align*} \ddt{p} \amp = \frac{ p(t + dt) - p(t) }{ dt } \\ \amp = \frac{c \bbrac{ q(t + dt) - q(t) } }{ dt } \\ \amp = c \, \frac{\bbrac{ q(t + dt) - q(t) } }{ dt } \\ \amp = c \, \dqdt \text{,} \end{align*}
as claimed.
A tangent to a graph and a scaled tangent to a scaled graph.
Figure 12.3.4. Scaling a tangent line scales its slope.
Pattern 12.3.14 applies equally well when the vertical scale factor is negative. Thinking in terms of tangent lines, if we reflect vertically then the rise becomes negative while the run stays the same.
A tangent to a graph and a reflected tangent to a reflected graph.
Figure 12.3.5. Reflecting a tangent line makes its slope negative.
Example 12.3.6.
In Example 12.2.1, for
\begin{equation*} s(t) = -2 t^2 + 6 t \end{equation*}
we computed
\begin{equation*} \ddt{s} = -4 t + 6 \text{.} \end{equation*}
So if we had
\begin{align*} q(t) \amp = t^2 - 3 t \amp \amp\implies \amp q(t) \amp = \left(- \frac{1}{2}\right) \, s(t)\text{,} \end{align*}
then
\begin{align*} \dqdt \amp = \left(- \frac{1}{2}\right) \, \ddt{s} \\ \amp = \left(- \frac{1}{2}\right) \, (-4 t + 6) \\ \amp = 2 t - 3 \text{.} \end{align*}
Justification.
Consider quantity function
\begin{equation*} p(t) = q_1(t) + q_2(t)\text{.} \end{equation*}
Then
\begin{equation*} p(t + dt) = q_1(t + dt) + q_2(t + dt) \text{,} \end{equation*}
so
\begin{align*} p(t + dt) - p(t) \amp = \bbrac{ q_1(t + dt) + q_2(t + dt) } - \bbrac{ q_1(t) + q_2(t) } \\ \amp = \bbrac{ q_1(t + dt) - q_1(t) } + \bbrac{ q_2(t + dt) - q_2(t)} \text{.} \end{align*}
Therefore,
\begin{align*} \ddt{p} \amp = \frac{ p(t + dt) - p(t) }{ dt } \\ \amp = \frac{ \bbrac{ q_1(t + dt) - q_1(t) } + \bbrac{ q_2(t + dt) - q_2(t)} }{ dt } \\ \amp = \frac{ \bbrac{ q_1(t + dt) - q_1(t) } }{ dt } + \frac{ \bbrac{ q_2(t + dt) - q_2(t)} }{ dt } \\ \amp = \ddt{q_1} + \ddt{q_2} \text{,} \end{align*}
as claimed.
Example 12.3.8. Derivative of a linear quantity function.
For
\begin{equation*} q(t) = r t + q_0 \text{,} \end{equation*}
we may apply a combination of our rules and basic formulas so far:
\begin{align*} \dqdt \amp = \opddt (r t + q_0) \\ \amp = \opddt (r t) + \ddt{q_0} \amp \amp\text{(i)} \\ \amp = \opddt (r t) + 0 \amp \amp\text{(ii)} \\ \amp = r \ddt{t} \amp \amp\text{(iii)} \\ \amp = r \cdot 1 \amp \amp\text{(iv)} \\ \amp = r \text{,} \end{align*}
with step justifications
If we accept Principle 12.1.2, then in Example 12.3.8 we have successfully reversed Pattern 7.2.2 — a linear quantity model represents a quantity that varies at a constant rate, and a quantity that varies at a constant rate can be modelled by a linear function.
In fact, a function is forced to be linear by the property of exhibiting a uniformly constant rate. Graphically, this corresponds to exhibiting a constant slope, so that an average rate calculation
\begin{equation*} \slope{q}{t} \end{equation*}
between any pair of points on the graph leads to the same result. (See Example 11.2.8, and Figure 11.2.9 in particular.)
By definition, a tangent line to a quantity’s graph meets that graph at a particular point with slope equal to the quantity’s derivative value there. Example 12.3.8 shows that the slope of that tangent line will be the same as the slope of the quantity’s graph, so in fact they will be the same line.
Now consider the special case of Pattern 12.3.7 where the second function is a constant function: if
\begin{equation*} q(t) = q_1(t) + c \text{,} \end{equation*}
then
\begin{align*} \dqdt \amp = \opddt (q_1 + c) \\ \amp = \ddt{q_1} + \ddt{c} \\ \amp = \ddt{q_1} + 0 \\ \amp = \ddt{q_1} \text{.} \end{align*}
A tangent to a graph and a shifted tangent to a shifted graph.
Figure 12.3.11. Shifting a tangent line does not change its slope.
Finally, let’s consider the effect of horizontal transformations on derivative.
Justification idea.
Instead of carrying out a detailed calculation of \(dq / dt\text{,}\) let’s just argue in terms of the concept of the derivative. If the value of \(dq / dt\) at a particular time \(t = t_0\) represents the rate of variation of \(q\) at that time, but
\begin{equation*} q(t) = p(t - c) \text{,} \end{equation*}
then we are really just measuring the rate of variation of \(p\) at the “earlier” time \(t = t_0 -c\text{.}\)
A tangent to a graph and a shifted tangent to a shifted graph.
Figure 12.3.13. Shifting a tangent line does not change its slope.
Justification.
First let’s examine the pattern of average rate calculations for the scaled quantity function:
\begin{align*} \slope{q}{t} \amp = \frac{q(t_2) - q(t_1)}{t_2 - t_1} \\ \amp = \frac{p(a t_2) - p(a t_1)}{t_2 - t_1} \text{.} \end{align*}
This almost looks like an average rate calculation for \(p\text{,}\) but the two times in the denominator don’t match the times used in the numerator. Let’s fix that:
\begin{align*} \slope{q}{t} \amp = \frac{p(a t_2) - p(a t_1)}{t_2 - t_1} \cdot \frac{a}{a} \\ \amp = a \cdot \frac{p(a t_2) - p(a t_1)}{a t_2 - a t_1} \text{.} \end{align*}
So an average rate calculation for \(q\) over domain \(t_1 \le t \le t_2\) is a scaling of an average rate calculation for \(p\) over domain \(a t_1 \le t \le a t_2\text{.}\)
A secant line representing an average rate on a compressed graph.
(a) A secant line representing an average rate on a compressed graph.
A stretched secant line representing an average rate on the original graph.
(b) A stretched secant line representing an average rate on the original graph.
Figure 12.3.15. An average rate calculation on one graph stretched into an average rate calculation on another. Note that in the case that \(a \gt 1\) (which is what is pictured here), then the graph of \(q\) is a compression of the graph of \(p\text{.}\) And notice how this makes the slope of the secant line on the graph of \(q\) steeper than the slope of the secant line on the graph of \(p\text{.}\)
If we are considering a specific time \(t = t^\ast\text{,}\) our calculations above say that every average rate calculation for \(q\) on a domain containing \(t^\ast\) can be converted into a scaled average rate calculation for \(p\) on a domain containing \(a t^\ast\text{,}\) since
\begin{equation*} t_1 \le t^\ast \le t_2 \qquad \implies a t_1 \le a t^\ast \le a t_2 \text{.} \end{equation*}
Referring back to Definition 11.5.1, since \(p\) is assumed to be differentiable,
\begin{equation*} p'(a t^\ast) \approx \frac{p(a t_2) - p(a t_1)}{a t_2 - a t_1} \text{,} \end{equation*}
where the approximation can be made arbitrarily close by taking
\begin{equation*} a \change{t} = a t_2 - a t_1 \end{equation*}
sufficiently small. If we take \(\change{t}\) small enough so that \(a \change{t}\) can be described as “sufficiently small” in this sense, then
\begin{align*} p'(a t^\ast) \amp \approx \frac{p(a t_2) - p(a t_1)}{a t_2 - a t_1} \\ a p'(a t^\ast) \amp \approx a \cdot \frac{p(a t_2) - p(a t_1)}{a t_2 - a t_1} \\ a p'(a t^\ast) \amp \approx \slope{q}{t} \text{,} \end{align*}
satisfying Definition 11.5.1 for \(q\) at \(t = t^\ast\text{.}\) That is,
\begin{equation*} q'(t^\ast) = a p'(a t^\ast) \text{,} \end{equation*}
as desired.
Pattern 12.3.14 applies equally well when the horizontal scale factor is negative. Thinking in terms of tangent lines, if we reflect horizontally then the rise becomes negative because the second point has become the first, and vice versa. Meanwhile, the amount of run stays the same.
A tangent to a graph and a reflected tangent to a reflected graph.
Figure 12.3.16. Reflecting a tangent line makes its slope negative.
We need some basic example derivatives to work with before we can consider examples of the previous two patterns, which is what the next section is about. Once you have studied the derivatives of some familiar functions, see Example 12.5.5 for an example of applying Pattern 12.3.12 and Pattern 12.3.14.

Section 12.4 Derivatives of polynomials

Polynomials are built out of constants and powers of the independent variable. We already know the derivatives for constants and the independent variable itself, so let’s tackle higher powers. We’ve already carried out one example involving \(t^2\) (Example 12.2.1), so let’s try a cubic.

Example 12.4.1. Derivative of the basic cubic function.

For \(q(t) = t^3\text{,}\) we have
\begin{equation*} q(t + dt) = (t + dt)^3 \text{.} \end{equation*}
This sum-inside-a-power formula is an example of a binomial expression, and there is a formula for expanding it:
\begin{equation*} (t + dt)^3 = t^3 + 3 t^2 dt + 3 t dt^2 + dt^3 \text{.} \end{equation*}
So we have
\begin{align*} q(t + dt) - q(t) \amp = (\bcancel{t^3} + 3 t^2 dt + 3 t dt^2 + dt^3) - \bcancel{t^3} \\ \amp = 3 t^2 dt + 3 t dt^2 + dt^3 \\ \amp = (3 t^2 + 3 t dt + dt^2) dt \text{,} \end{align*}
leading to
\begin{align*} \dqdt \amp = \frac{ q(t + dt) - q(t) }{ dt } \\ \amp = \frac{ (3 t^2 + 3 t dt + dt^2) \bcancel{dt} }{ \bcancel{dt} } \\ \amp = 3 t^2 + 3 t dt + dt^2 \text{.} \end{align*}
Treating \(dt \approx 0\) as negligible at this point gives us
\begin{align*} \dqdt \amp = 3 t^2 + \cancelto{0}{3 t dt} + \cancelto{0}{dt^2} \\ \amp = 3 t^2 \text{.} \end{align*}
Let’s investigate what happens if we go back-and-forth through our two processes.

Example 12.4.2. Derivative of the quantity model for a quadratic rate model.

Suppose we start with \(r(t) = t^2\text{.}\) Formula 1 of Pattern 7.4.2 says that the associated quantity model will be
\begin{equation*} q(t) = \frac{t^3}{3} + q_0 \text{,} \end{equation*}
where \(q_0 = q(0)\) is the initial quantity. The constant denominator is actually a scale factor:
\begin{equation*} q(t) = \frac{1}{3} \, t^3 + q_0 \text{.} \end{equation*}
Apply Pattern 12.3.3 and Pattern 12.3.7, as well as our result from Example 12.4.1:
\begin{align*} \dqdt \amp = \opddt \left(\frac{1}{3} \, t^3 + q_0\right) \\ \amp = \frac{1}{3} \, \ddt{(t^3)} + \ddt{q_0} \\ \amp = \frac{1}{3} \, (3 t^2) + 0 \\ \amp = t^2 \\ \amp = r(t) \text{.} \end{align*}
So the derivative process has successfully returned our original rate function.
Formula 1 of Pattern 7.4.2 could be described in words as follows: to obtain the accumulation function for a power of \(t\text{,}\) increase the exponent and then divide by that new exponent. The differentiation pattern should be the reverse of this pattern. To reverse a multi-step pattern, we need to both reverse the steps but also reverse the order of the steps. With this in mind, and also using the result of Example 12.4.1 as inspiration, we might guess that the pattern is: to obtain the derivative function for a power of \(t\text{,}\) multiply by the exponent and then decrease the exponent.

Justification for positive integer \(n\).

We follow the template of Example 12.4.1. For quantity function
\begin{equation*} q(t) = t^n\text{,} \end{equation*}
we have
\begin{align*} q(t + dt) \amp = (t + dt)^n \\ \amp = t^n + b_1 t^{n - 1} dt + b_2 t^{n - 2} dt^2 + \dotsb + b_{n - 1} t dt^{n - 1} + dt^n \text{,} \end{align*}
where \(b_1,b_2,\dotsc,b_{n-1}\) are the so-called binomial coefficients for the expansion of a binomial expression of the form \((x+y)^n\text{.}\) Therefore,
\begin{align*} q(t + dt) - q(t) \amp = (\bcancel{t^n} + b_1 t^{n - 1} dt + b_2 t^{n - 2} dt^2 + \dotsb + b_{n - 1} t dt^{n - 1} + dt^n) - \bcancel{t^n} \\ \amp = (b_1 t^{n - 1} + b_2 t^{n - 2} dt + \dotsb + b_{n - 1} t dt^{n - 2} + dt^{n - 1}) dt \text{,} \end{align*}
and so
\begin{align*} \dqdt \amp = \frac{ q(t + dt) - q(t) }{ dt } \\ \amp = \frac{ (b_1 t^{n - 1} + b_2 t^{n - 2} dt + \dotsb + b_{n - 1} t dt^{n - 2} + dt^{n - 1}) \bcancel{dt} }{ \bcancel{dt} } \\ \amp = b_1 t^{n - 1} + b_2 t^{n - 2} dt + \dotsb + b_{n - 1} t dt^{n - 2} + dt^{n - 1} \text{.} \end{align*}
Treating \(dt \approx 0\) as negligible at this point gives us
\begin{align*} \dqdt \amp = b_1 t^{n - 1} + \cancelto{0}{b_2 t^{n - 2} dt} + \dotsb + \cancelto{0}{b_{n - 1} t dt^{n - 2}} + \cancelto{0}{dt^{n - 1}} \\ \amp = b_1 t^{n - 1} \text{.} \end{align*}
Finally, it turns out that the first binomial coefficient \(b_1\) for the expansion of a binomial expression of the form \((x+y)^n\) is \(b_1 = n\) (refer to the Binomial Theorem), so that
\begin{equation*} \ddt{(t^n)} = n t^{n - 1} \text{,} \end{equation*}
as claimed.

Remark 12.4.4. Non-positive and/or non-integer exponents.

While our justification is only valid for positive, integer exponents \(n\text{,}\) Pattern 12.4.3 itself is valid for all exponents, and we will justify this in later chapters after we have a few more techniques and patterns available to us. In the meantime, recall that for negative exponents we cannot attach a derivative value to the quantity function at the singularity at \(t = 0\) (see Subsection 11.7.3), which is the reason for including the phrase “everywhere that it is defined” in the statement of the pattern. We could also consider the pattern to be true for \(n = 0\text{,}\) since in that case the quantity function satisfies
\begin{align*} q(t) \amp = t^0 = 1 \amp \text{for } t \amp \gt 0 \text{,} \end{align*}
while Pattern 12.4.3 gives
\begin{equation*} \ddt{(t^0)} = 0 t^{-1} = 0 \end{equation*}
(again assuming \(t \gt 0\)). So in some sense Pattern 12.4.3 “contains” Pattern 12.3.1. (And, for that matter, it also “contains” Pattern 12.3.2 when we take \(n = 1\text{.}\))
Here are a couple examples of using Pattern 12.4.3.

Example 12.4.5. Derivative of a power with a negative exponent.

\begin{align*} q(t) \amp = t^{-3} \amp \amp\implies \amp \dqdt \amp = -3 t^{-4} \end{align*}

Note 12.4.6.

While we might describe part of Pattern 12.4.3 as decrease the exponent, this does not mean decrease the magnitude of the exponent. For example, a common mistake in applying Pattern 12.4.3 is to use \(t^{-2}\) in the derivative formula for \(t^{-3}\text{,}\) instead of \(t^{-4}\) as in Example 12.4.5.

Example 12.4.7. Derivative of a power with an irrational exponent.

\begin{align*} q(t) \amp = t^\pi \amp \amp\implies \amp \dqdt \amp = \pi t^{\pi - 1} \end{align*}
Finally, we can combine Pattern 12.4.3 with Pattern 12.3.3 and Pattern 12.3.7 to obtain a general formula for the derivative function of any polynomial. We will leave the justification up to you, the reader.

Example 12.4.9. Derivative of a polynomial.

\begin{gather*} q(t) = 3 t^5 - 4 t^3 + t^2 + 7 t + 6 \\ \implies \qquad q'(t) = 15 t^4 - 12 t^2 + 2 t + 7 \end{gather*}

Checkpoint 12.4.10.

Verify for yourself that beginning with a polynomial rate function
\begin{equation*} r(t) = r_0 + r_1 t + r_2 t^2 + \dotsb + r_m t^m \text{,} \end{equation*}
applying Pattern 7.4.4 to obtain an associated quantity model (don’t forget to include the initial value), and then applying Pattern 12.4.8 to obtain the derivative function of that quantity function will return you back to the original rate function.

Section 12.5 Derivatives of the natural logarithm and exponential functions

Subsection 12.5.1 The natural logarithm

If we accept Principle 12.1.2, then Definition 8.1.1 tells us how to differentiate the natural logarithm function.
It is common to extend the domain of the natural logarithm by composing with the absolute value function: whereas \(\ln (t)\) is defined only for \(t \gt 0\text{,}\) \(\ln \abs{t}\) is defined for all \(t\) except at the singularity at \(t = 0\text{.}\) For this function, a negative input has the same output value as the corresponding positive input of the same magnitude, so the graph is symmetric about the vertical axis.
Graph of the natural logarithm composed with the absolute value.
Figure 12.5.2. Graph of the quantity function \(q(t) = \ln \abs{t}\text{.}\)
Consider a point on the graph in Figure 12.5.2 at \(t = a\) with \(a \lt 0\text{.}\) There is a corresponding point at the same height at \(t = \abs{a}\text{.}\) As \(\abs{a} > 0\text{,}\) this second point is on the graph of \(y = \ln (t)\text{,}\) and so we know its derivative value as
\begin{equation*} \funceval{\ddt{\bbrac{\ln (t)}}}{t = \abs{a}} = \frac{1}{\abs{a}} \text{.} \end{equation*}
This derivative value describes the slope of the tangent line at this point. The derivative value
\begin{equation*} \funceval{\ddt{(\ln \abs{t})}}{t = a} \end{equation*}
describes the slope of the tangent line at the original point. But since these two points are reflections of each other in the vertical axis, we can see that the two slopes will be equal in magnitude but have opposite signs.
Two tangent lines at reflected points on the graph of the natural logarithm composed with the absolute value.
Figure 12.5.3. Two tangent lines at reflected points on the graph of the quantity function \(q(t) = \ln \abs{t}\text{.}\)
So by symmetry we have
\begin{equation*} \funceval{\ddt{(\ln \abs{t})}}{t = a} = - \frac{1}{\abs{a}} = \frac{1}{- \abs{a}} \text{.} \end{equation*}
However, as \(a\) was negative to begin with,
\begin{equation*} - \abs{a} = a \text{,} \end{equation*}
and we end up with
\begin{equation*} \funceval{\ddt{(\ln \abs{t})}}{t = a} = - \frac{1}{\abs{a}} = \frac{1}{a} \text{,} \end{equation*}
which is the same “one-over” formula as for the derivative of the natural logarithm.
Example 12.5.5. Derivative of a horizontally transformed logarithm.
To calculate
\begin{equation*} \opddt \ln\abs{3 t - 5} \text{,} \end{equation*}
start with \(p_1(t) = \ln \abs{t}\text{.}\) Then Pattern 12.5.4 says that
\begin{equation*} p'(t) = \frac{1}{t} \text{.} \end{equation*}
Now consider
\begin{equation*} p_2(t) = p_1(t - 5) = \ln \abs{t - 5} \text{.} \end{equation*}
Pattern 12.3.12 says that
\begin{equation*} p_2'(t) = p_1'(t - 5) = \frac{1}{t - 5} \text{.} \end{equation*}
Finally, consider
\begin{equation*} q(t) = p_2(3 t) = \ln \abs{3 t - 5} \text{,} \end{equation*}
the function whose derivative we would like to calculate. Pattern 12.3.14 tells us
\begin{equation*} q'(t) = 3 p_2'(3 t) = \frac{3}{3 t - 5} \text{.} \end{equation*}
So we have determined that
\begin{equation*} \opddt \ln\abs{3 t - 5} = q'(t) = \frac{3}{3 t - 5} \text{.} \end{equation*}
Recall that the natural logarithm is defined to be the accumulation function \(A_1(t)\) for rate function \(r(t) = 1/t\text{.}\) That is, values of \(\ln(t)\) are defined to be results of definite integral calculations. In fact, we can interpret values of \(\ln\abs{t}\) the same way.
Two corresponding areas under the graph of the reciprocal function.
Figure 12.5.6. Two corresponding areas under the graph of \(r(u) = 1/u\text{.}\)
By symmetry, the areas in Figure 12.5.6 have the same magnitude, but are oppositely oriented. We can counteract the negative orientation of the area on the left by integrating backwards in time:
\begin{equation*} \ccmint{-1}{t}{\frac{1}{u}}{u} = - \ccmint{t}{-1}{\frac{1}{u}}{u} = \ccmint{1}{\abs{t}}{\frac{1}{u}}{u}\text{.} \end{equation*}
Thus we can say that
\begin{equation*} \ln\abs{t} = \begin{cases} \displaystyle \ccmint{1}{t}{\frac{1}{u}}{u} \amp t \gt 0 \\[3pt] \displaystyle \ccmint{-1}{t}{\frac{1}{u}}{u} \amp t \lt 0 \end{cases}\text{.} \end{equation*}

Subsection 12.5.2 The natural exponential

Again, if we accept Principle 12.1.2, then Pattern 9.6.1 tells us how to differentiate the natural exponential function. In particular, Pattern 9.6.1 says that rate function \(r(t) = e^t\) leads to the accumulation function \(A_0(t) = e^t - 1\text{.}\) If we assume that \(q(0) = 0\text{,}\) then this accumulation function is also the quantity function:
\begin{equation*} q(t) = e^t - 1 \text{,} \end{equation*}
and Principle 12.1.2 tells us to expect that
\begin{equation*} \opddt (e^t - 1) = r(t) = e^t \text{.} \end{equation*}
Now, the quantity functions
\begin{align*} q(t) \amp = e^t - 1 \amp q(t) \amp = e^t \end{align*}
are vertical shifts of one another, so Pattern 12.3.10 tells us that they have the same derivative.
This is a remarkable fact! It says that a quantity that experiences exponential growth will also have a rate function that grows exponentially. This is why exponential growth can be both exciting (when considering compounding of gains in your investment account, for example) and terrifying (when considering unchecked spread of a virus in a population, for example). — the more there is, the sooner there will be a lot more than that.
In this direction, we can now combine Pattern 12.5.7 with Pattern 12.3.14 to verify exponential solutions to rate equations where the rate is proportional to the quantity, as in our Radioactive decay example.
Example 12.5.8. Verifying an exponential solution to a proportional rate equation.
Suppose \(q(t)\) is a quantity function whose rate of variation is proportional to the quantity, with proportionality constant \(k\text{.}\) Let’s take our proportionality constant to be \(k = 3\text{,}\) so that
\begin{gather} r(t) = 3 q(t) \text{.}\tag{✶} \end{gather}
We expect
\begin{equation*} r(t) = q'(t) \text{,} \end{equation*}
so we are looking for a function whose derivative is a multiple of that original function. The pattern
\begin{equation*} \ddt{e^t} = e^t \end{equation*}
is almost correct, but the “multiple” is only \(1\text{.}\) However, we know from Pattern 12.3.14 that introducing a horizontal scale factor into the quantity function also introduces a vertical scale factor into the derivative function. In this case, if we take
\begin{equation*} q(t) = \exp(3 t) \text{,} \end{equation*}
then Pattern 12.3.14 says that
\begin{align*} q'(t) \amp = 3 \exp'(3 t) \\ \amp = 3 \exp(3 t) \\ \amp = 3 q(t) \text{.} \end{align*}
This shows that \(q(t) = e^{3 t}\) is a solution to the rate equation in (✶).
This is only one particular solution to (✶). Can we guess at the general solution by introducing some parameter? Our particular solution above has initial value
\begin{equation*} q(0) = e^{3 \cdot 0} = 1 \text{.} \end{equation*}
But exponential growth or decay does not turn into some other behaviour if there is a different initial value — otherwise a horizontal shift (so that the point in time that is considered \(t = 0\) is shifted) would change the type of graph, which is not possible. For example, see our direction field for \(k = -1/2\) (Figure 4.3.5) — the example solution curves all look like exponential decay no matter the initial value.
Recall from Definition 9.1.5 that introducing a constant multiple in an exponential function changes the initial amount. So let’s try
\begin{equation*} q(t) = A e^{3 t} \text{,} \end{equation*}
where \(A\) represents the initial amount \(A = q(0)\text{.}\) Now using Pattern 12.3.3 for our first step, we have
\begin{align*} q'(t) \amp = \opddt \bbrac{A \exp(3 t)} \\ \amp = A \opddt \bbrac{\exp(3 t)} \\ \amp = A \bbrac{3 \exp'(3 t)} \\ \amp = 3 A \exp(3 t) \\ \amp = 3 q(t) \text{.} \end{align*}
This verifies that for every possible initial value \(A\text{,}\) the exponential function \(q(t) = A e^{3 t}\) is a solution to the rate equation in (✶).

Section 12.6 Derivatives of the sine and cosine functions

Looking at Pattern 10.5.1, Principle 12.1.2 tells us to expect that
\begin{equation*} \opddt (1 - \cos(t)) = \sin(t) \text{.} \end{equation*}
If this is true, then Pattern 12.3.10 tells us that
\begin{equation*} \opddt (- \cos(t)) = \sin(t) \end{equation*}
as well. Finally, applying Pattern 12.3.3 would lead us to
\begin{equation*} \opddt \cos(t) = - \sin(t) \text{.} \end{equation*}
We will verify this fact using the definition of derivative, but first we need a special property of the sine and cosine functions, each.
Comparing the graph of the cosine function to the horizontal line at 1.
(a) Comparing cosine to \(1\text{.}\)
Comparing the graph of the sine function to the line with slope 1.
(b) Comparing sine to \(t\text{.}\)
Figure 12.6.1. Comparing cosine to \(1\) and sine to \(t\text{.}\)
Examining the graphs of these trigonometric functions, we clearly see that for \(t \approx 0\) we have \(\cos(t) \approx 0\text{,}\) and it also appears that \(\sin(t)\) because very close to the line \(y = t\text{.}\) But in each of these cases we can say something much stronger about how close these approximations are relative to how small \(t\) is.

Justification.

We will verify the statement for positive values of \(t \approx 0\text{;}\) the justification for negative values of \(t\) is similar.
Recall that values of \(\sin(t)\) represent the \(y\)-coordinate of a position on the unit circle. That is, in Figure 12.6.3, the value of \(\sin(t)\) is the equal to the triangle side length marked \(y\text{.}\)
Diagram comparing the values of the sine and cosine functions with the angle measure.
Figure 12.6.3. Diagram comparing the values of \(\sin(t)\text{,}\) \(\cos(t)\text{,}\) and \(t\text{.}\)
Now, we have \(y \lt a\) because \(a\) is the length of the hypotenuse in a right triangle and \(y\) is the length of one of the legs of that triangle. We also have \(a \lt t\) because the arc that \(t\) measures has the same endpoints as the segment measured by \(a\text{,}\) but takes the segment is a straight path between those points. Going out further, for a similar reason we can say \(t \lt b + d\text{.}\) But also \(d \lt c\) because \(c\) is a hypotenuse and \(d\) is a leg. And we have \(b + c = \tan(t)\) by using opposite-over-adjacent in the large right triangle. So we finally have
\begin{equation*} \sin(t) = y \lt a \lt t \lt b + d \lt b + c = \tan(t) \text{.} \end{equation*}
Splitting in two at \(t\) and doing a little re-arranging, we have
\begin{align*} \sin(t) \amp\lt t \amp t \amp \lt \tan(t) \\ \frac{\sin(t)}{t} \amp\lt 1 \amp t \amp \lt \frac{\sin(t)}{\cos(t)} \\ \amp\amp \cos(t) \amp \lt \frac{\sin(t)}{t} \text{.} \end{align*}
(Note that for \(t \approx 0\) but positive, the value of \(\cos(t)\) is also positive, so multiplying or dividing both sides of an inequality by either \(t\) or \(\cos(t)\) will not reverse the inequality.)
Combining these inequalities, we have
\begin{equation*} \cos(t) \lt \frac{\sin(t)}{t} \lt 1 \qquad \text{for} \qquad t \approx 0^+\text{,} \end{equation*}
where the superscript \(+\) indicates that \(t\) is assumed to be positive. This inequality says that the graph of \(\sin(t)/t\text{,}\) whatever it might look like, must be confined to the shaded region in Figure 12.6.4 below. For \(t \approx 0^+\text{,}\) this graph must be squeezed into the little wedge in the upper-left corner of the shaded region, very close to a height of \(1\text{,}\) and so we conclude
\begin{equation*} \frac{\sin(t)}{t} \approx 1 \qquad \text{for} \qquad t \approx 0^+\text{.} \end{equation*}
Diagram illustrating a region to which the graph of sine divided by independent variable must be confined.
Figure 12.6.4. For \(0 \lt t \lt \pi/2\text{,}\) the graph of \(\sin(t)/t\) must be confined to the shaded region.
The following corresponding property of the cosine function can be justified in a similar manner.

Remark 12.6.6.

The statement that
\begin{equation*} \frac{\cos(t) - 1}{t} \approx 0 \end{equation*}
for \(t \approx 0\) says something stronger than just \(\cos(t) \approx 1\text{.}\) In the ratio
\begin{equation*} \frac{\cos(t) - 1}{t} \text{,} \end{equation*}
both the numerator and denominator are approximately \(0\) for small values of \(t\text{.}\) But similarly to Section 3.2, it becomes a “race” to \(0\text{.}\) The fact that
\begin{equation*} \frac{\cos(t) - 1}{t} \approx 0 \end{equation*}
says that the numerator “wins” the race, so that the approximation \(\cos(t) \approx 1\) must be very tight compared to \(t \approx 0\) since if it were the other way around then we would expect the expression
\begin{equation*} \frac{\cos(t) - 1}{t} \end{equation*}
to have a singularity at \(t = 0\text{.}\)

Justification.

As usual, we start with
\begin{equation*} \opddt \cos(t) = \frac{ \cos(t + dt) - \cos(t) }{ dt } \text{.} \end{equation*}
Using the addition identity for cosine, we have
\begin{equation*} \cos(t + dt) = \cos(t) \cos(dt) - \sin(t) \sin(dt) \text{.} \end{equation*}
Therefore,
\begin{align*} \opddt \cos(t) \amp = \frac{ \cos(t + dt) - \cos(t) }{ dt } \\ \amp = \frac{ \cos(t) \cos(dt) - \sin(t) \sin(dt) - \cos(t) }{ dt } \\ \amp = \cos(t) \cdot \frac{ \cos(dt) - 1 }{dt} - \sin(t) \cdot \frac{ \sin(dt) }{ dt } \text{.} \end{align*}
For \(dt \approx 0\text{,}\) using Pattern 12.6.2 and Pattern 12.6.5 we effectively have
\begin{align*} \frac{ \cos(dt) - 1 }{dt} \amp = 0 \amp \frac{ \sin(dt) }{dt} \amp = 1\text{,} \end{align*}
so
\begin{align*} \opddt \cos(t) \amp = \cos(t) \cdot 0 - \sin(t) \cdot 1 \\ \amp = - \sin(t) \text{,} \end{align*}
as desired.
Similarly, combining Pattern 10.5.2 with Principle 12.1.2 tells us to expect that
\begin{equation*} \opddt \sin(t) = \cos(t) \text{,} \end{equation*}
and this is in fact the case.

Justification idea.

Apply the same sort of reasoning as in the justification for Pattern 12.6.7, beginning by applying an addition identity to \(\sin(t + dt)\) in
\begin{equation*} \opddt \sin(t) = \frac{ \sin(t + dt) - \sin(t) }{ dt } \text{,} \end{equation*}
and then applying Pattern 12.6.2 and Pattern 12.6.5 as appropriate.

Section 12.7 Calculating derivatives with Sage

Of course, Sage can symbolically calculate derivatives. Here is the syntax for computing a derivative formula in Sage.
	derivative(formula, variable)
Alternatively, you can define a function and then apply the derivative method to it.
 q(t) = some_formula 
 q.derivative()
If you need a specific derivative value, you can use the substitute method with the first syntax option above. In the second syntax option, a function is returned to that can be evaluated with function notation.
 q(t) = some_formula 
 r = q.derivative() 
 r(some_number)
Here is an example using both syntax variations.

Example 12.7.1. Using Sage to compute the equation of a tangent line.

Consider function
\begin{equation*} q(t) = t^3 - 2 t + 5 \text{.} \end{equation*}
What is the equation of the tangent line to the graph of this function at the point
\begin{equation*} q(1) = 4 \text{?} \end{equation*}
It is not too difficult to use Pattern 12.4.8 to carry this out with pencil-and-paper, but let’s use Sage.
Now we apply Pattern 11.6.4 to get tangent line equation
\begin{align*} y \amp = 1 t + (4 - 1 \cdot 1) \\ \amp = t + 3 \text{.} \end{align*}
Here is the same calculation, but using function notation.
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