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

Jeremy Sylvestre

Chapter 16 Composite and inverse functions

Section 16.1 Composite functions

The examples in Section 14.2 all had a similar pattern to the initial approach: we made a substitution for \(u\) based on recognizing an inner component and an outer component to the function. For example, for
\begin{equation*} q(t) = \sin^3(t) = {\bbrac{\sin(t)}}^3 \end{equation*}
from Example 14.2.4, the inner component is the sine function and the outer component is a power function. For
\begin{equation*} q(t) = \sqrt{t^2 + 1} \end{equation*}
from Example 14.2.5, the inner component is a polynomial function and the outer component is the square root component. It is just a matter of changing perspective on the function from a single input-output process to a chain of two (or more) input-output processes.
Schematic of a function as a chained multi-step input-output process.
Figure 16.1.1. Schematic of a function as a chained multi-step input-output process.

Definition 16.1.2. Composite of two functions.

Given functions \(f\) and \(g\text{,}\) the composite function \(\comp{g}{f}\) is the function defined by
\begin{equation*} (\comp{g}{f})(t) = g\bbrac{f(t)} \text{.} \end{equation*}
If \(f(t)\) and \(g(t)\) are described in terms of formulas, a formula for \((\comp{g}{f})(t)\) can be obtained by replacing every occurrence of \(t\) in the formula for \(g(t)\) with the entire formula for \(f(t)\text{.}\)

Example 16.1.3. Composing a polynomial with a root.

Consider functions
\begin{align*} f(t) \amp = t^3 + t - 1 \amp g(t) \amp = \sqrt{t} \text{.} \end{align*}
Then
\begin{align*} (\comp{g}{f})(t) \amp = g\bbrac{f(t)} \\ \amp = \sqrt{f(t)} \\ \amp = \sqrt{t^3 + t - 1} \text{.} \end{align*}
Notice that if we form the composition in the other order, we don’t get the same result:
\begin{align*} (\comp{f}{g})(t) \amp = f\bbrac{g(t)} \\ \amp = {\bbrac{g(t)}}^3 + g(t) - 1 \\ \amp = (\sqrt{t})^3 + \sqrt{t} - 1 \\ \amp = t^{3/2} + t^{1/2} - 1 \text{.} \end{align*}

Example 16.1.4. Transformations are compositions with a linear function.

Consider functions
\begin{align*} f(t) \amp = 4 t - 5 \amp g(t) \amp = \sqrt{t} \text{.} \end{align*}
Then composing in the order \(\comp{g}{f}\) has the effect of horizontally transforming \(g(t)\text{:}\)
\begin{align*} (\comp{g}{f})(t) \amp = g\bbrac{f(t)} \\ \amp = \sqrt{f(t)} \\ \amp = \sqrt{4 t - 5} \text{.} \end{align*}
On the other hand, composing in the order \(\comp{f}{g}\) has the effect of vertically transforming \(g(t)\text{:}\)
\begin{align*} (\comp{f}{g})(t) \amp = f\bbrac{g(t)} \\ \amp = 4 g(t) - 5 \\ \amp = 4 \sqrt{t} - 5 \text{.} \end{align*}
More important than computing composite functions is recognizing an existing function as a composite.

Example 16.1.5. Recognizing a composite function.

The function \(h(t) = e^{t^2 + 1}\) is a composition of the natural exponential function with a polynomial. Writing \(\exp(t) = e^t\) as usual, and also \(f(t) = t^2 + 1\text{,}\) then
\begin{align*} (\comp{\exp}{f})(t) \amp = \exp\bbrac{f(t)} \\ \amp = e^{f(t)} \\ \amp = e^{t^2 + 1} \text{,} \end{align*}
so we can say
\begin{equation*} h = \comp{\exp}{f} \text{.} \end{equation*}
We can also have compositions of more than two functions.

Example 16.1.6. A composition of four functions.

Consider the function
\begin{equation*} h(t) = e^{\sqrt{\abs{\cos(t)}}} \text{.} \end{equation*}
Working from the outside inward, we can write
\begin{equation*} h = \comp{\exp}{\comp{\sqrtop}{\comp{\absop}{\cos}}} \text{.} \end{equation*}
To verify that this is correct, we can compute:
\begin{align*} (\comp{\exp}{\comp{\sqrtop}{\comp{\absop}{\cos}}})(t) \amp = \exp\bbrac{ (\comp{\sqrtop}{\comp{\absop}{\cos}})(t) }\\ \amp = e^{ \sqrtop\bbrac{(\comp{\absop}{\cos})(t)} } \\ \amp = e^{ \sqrt{ \absop\bbrac{\cos(t)} } } \\ \amp = e^{ \sqrt{ \abs{\cos(t)} } } \text{,} \end{align*}
which is exactly the formula for \(h(t)\text{.}\)
Finally, here is a version of the Chain Rule in terms of composition of functions.

Justification.

If we set \(u = g(t)\text{,}\) then we may regard the composite
\begin{equation*} (\comp{f}{g})(t) = f\bbrac{g(t)} = f(u) \end{equation*}
as a function of \(u\text{.}\) Using our inital version of the Chain Rule, we have
\begin{align*} \opddt \, (\comp{f}{g})(t) \amp = \opddt \, f(u) \\ \amp = \opddu \, f(u) \cdot \dudt \\ \amp = f'(u) \cdot g'(t) \\ \amp = f'\bbrac{g(t)} \cdot g'(t) \text{,} \end{align*}
as required.

Section 16.2 Inverse functions

Example 16.2.1. Outputs as inputs.

Suppose a reservoir is filled with water to a depth of 3 m. A period of steady, heavy rain begins, and it is observed that the depth of water is increasing by 50 cm⁄h. Knowing that a constant variation must mean a linear quantity, we can create quantity model
\begin{equation*} D(t) = 0.5 t + 3 \end{equation*}
for the depth of water in metres at time \(t\) hours after it began raining.
With our model expressed as a function of time, we can directly answer questions like “What will the depth be 1 h after the rain began?” But often we want to answer questions like “If it keeps raining at this rate, how long until the reservoir overflows?” In other words, “How long must it rain at this rate for the water to reach a specific depth \(D_0\text{?}\)” To answer this type of question directly, we need a formula where \(D\) is the input and \(t\) is the output:
\begin{gather*} D = 0.5 t + 3 \\ D - 3 = 0.5 t \\ t = 2 (D - 3) \text{.} \end{gather*}
Not every function can be “inverted” as in Example 16.2.1. To isolate the old independent variable, we need to be able to “undo” whatever operations have been applied. The key is to recognize that in special situations, computing a composition has a simplifying effect.

Example 16.2.2. Undoing cube and cube root.

Consider functions
\begin{align*} f(t) \amp = t^3 \amp g(t) \amp = \sqrt[3]{t} \text{.} \end{align*}
Let’s compute both compositions \(\comp{f}{g}\) and \(\comp{g}{f}\text{:}\)
\begin{align*} (\comp{f}{g})(t) \amp = f\bbrac{g(t)} \amp (\comp{g}{f})(t) \amp = g\bbrac{f(t)}\\ \amp = {\bbrac{g(t)}}^3 \amp \amp = \sqrt[3]{f(t)}\\ \amp = {(\sqrt[3]{t})}^3 \amp \amp = \sqrt[3]{t^3}\\ \amp = t \amp \amp = t\text{.} \end{align*}

Definition 16.2.3. Inverse functions.

If \(f(t)\) and \(g(t)\) are functions so that
\begin{equation*} f\bbrac{g(t)} = t \end{equation*}
for all \(t\) in the domain of \(g\) and
\begin{equation*} g\bbrac{f(t)} = t \end{equation*}
for all \(t\) in the domain of \(f\text{,}\) then \(f\) and \(g\) are a pair of inverse functions.
A function for which a corresponding inverse exists is called intertible.

Remark 16.2.4. Inverses are unique.

Not every function has an corresponding inverse function, but when a function \(f\) does have an inverse, that inverse function is unique in the sense that no other function can be an inverse for \(f\text{.}\)
Because an invertible function \(f\) has only one unique inverse function, we can unambiguously write \(\inv{f}\) to mean the inverse function for \(f\), and the requirements of Definition 16.2.3 become
\begin{align*} \inv{f}\bbrac{f(t)} \amp = t \amp f\bbrac{\inv{f}(t)} \amp = t \text{.} \end{align*}

Warning 16.2.5.

In this case, the negative exponent in \(\inv{f}\) does not mean reciprocal. A reciprocal is a type of inverse, since division is the reverse process to multiplication. But function inverses are more general, as not every function involves a multiplication process. So, in general, \(\inv{f}(t)\) does not mean \(1/f(t)\text{.}\)

Example 16.2.6. Cube and cube root are inverse functions.

We can now say that Example 16.2.2 demonstrates that the cube and cube-root functions are inverses of each other. That is, if \(f(t) = t^3\text{,}\) then \(\inv{f}(t) = \sqrt[3]{t}\text{.}\) Or if we write \(g(t) = \sqrt[3]{t}\text{,}\) then \(\inv{g}(t) = t^3\text{.}\)
Some functions don’t have an inverse function, even when is seems like they should.

Example 16.2.8. Square and square root.

Consider functions
\begin{align*} f(t) \amp = t^2 \amp g(t) \amp = \sqrt{t} \text{.} \end{align*}
Let’s compute both compositions \(\comp{f}{g}\) and \(\comp{g}{f}\text{:}\)
\begin{align*} (\comp{f}{g})(t) \amp = f\bbrac{g(t)} \amp (\comp{g}{f})(t) \amp = g\bbrac{f(t)}\\ \amp = {\bbrac{g(t)}}^2 \amp \amp = \sqrt{f(t)}\\ \amp = {(\sqrt{t})}^2 \amp \amp = \sqrt{t^2}\\ \amp = t \amp \amp = \abs{t}\text{.} \end{align*}
So the square and square root functions are not inverses of each other.
The problem in Example 16.2.8 is that in the first calculation, we must use the domain of the square root function, since \(t \lt 0\) makes the formula \({(\sqrt{t})}^2\) undefined. But in the second calculation, we assume that we are using the domain of the square function, and in that case negative \(t\) are permitted but do not result back at \(t\) when composed. For example,
\begin{equation*} \sqrt{{(-1)}^2} = \sqrt{1} = 1 \neq -1 \text{.} \end{equation*}
We will fix this problem later in this chapter, so that the square root function is the inverse of a version of the square function.
If we suspect a function is invertible but don’t know an inverse function for it, is there a way we can attempt to compute an inverse? Consider again the example of cube and cube root. The two equalities
\begin{align} 2^3 \amp = 8 \amp \sqrt[3]{8} \amp = 2 \tag{✶} \end{align}
both state the same relationship between the numbers \(2\) and \(8\text{.}\) It’s only the point of view that has changed: in the first equality we think of \(2\) as the input into the cube process with \(8\) as the corresponding output, whereas in the second equality we think of \(8\) as the input into the cube-root process with \(2\) as the corresponding output. This reveals why composing the two processes together returns us back to where we started:
\begin{align*} \sqrt[3]{2^3} \amp = 2 \amp {(\sqrt[3]{8})}^3 \amp = 8 \text{.} \end{align*}
Diagram illustrating the input-output and corresponding output-input processes on the cube function’s graph.
(a) Forward and and reverse on the cube graph.
Diagram illustrating the input-output and corresponding output-input processes on the cube-root-graph.
(b) Forward and and reverse on the cube-root graph.
Figure 16.2.9. An input of \(2\) for the cube function produces an output of \(8\text{.}\) Given that output value of \(8\text{,}\) we can also trace that output to the input \(2\) that produced it, which corresponds to an input of \(8\) for the cube-root function.
So in an inverse function, input-to-output relationships are reversed, This tells us how to attempt to compute an inverse function formula: interchange the roles of input and output, just as in Example 16.2.1.

Example 16.2.10. Computing an inverse function formula.

Consider the function
\begin{equation*} f(t) = {(2 t + 3)}^3 \text{.} \end{equation*}
Let’s write \(y\) as the output variable, so that \((t,y)\) is an input-output pair whenever
\begin{equation*} y = (2 t + 3)^3 \text{.} \end{equation*}
If \(f(t)\) is invertible, then, exactly as in (✶), the inverse function should maintain the same relationship between input \(t\) and output \(y\text{,}\) just with their roles reversed:
\begin{gather*} y = {(2 t + 3)}^3 \\ y^{1/3} = {\bbrac{(2 t + 3)^3}}^{1/3} \\ \sqrt[3]{y} = 2 t + 3 \\ \sqrt[3]{y} - 3 = 2 y \\ t = \frac{\sqrt[3]{y} - 3}{2} \text{.} \end{gather*}
In this form, the relationship between \(t\) and \(y\) is preserved, but with \(t\) expressed as a formula in \(y\) we can view \(y\) as the input and \(t\) as the output.
The reversed formula above tells us the formula for the inverse function:
\begin{equation*} \inv{f}(t) = \frac{\sqrt[3]{t} - 3}{2} \text{.} \end{equation*}
If we consider an input-output formula as a sequence of calculation instructions (paying proper mind to BEDMAS), it makes sense that these two formulas are inverse: to compute an output for \(f(t)\text{,}\) one must
  1. multiply the input by \(2\)
  2. then add \(3\) to that result
  3. then compute the cube of that result,
whereas to compute an output for \(\inv{f}(t)\text{,}\) one must
  1. compute the cube-root of the input,
  2. then subtract \(3\) from that result
  3. then divide that result by \(2\text{.}\)
Notice that the two sequences of operations involve the opposite operations in the reverse order, which confirms that if we bothered to compute both compositions, we should find
\begin{align*} \inv{f}\bbrac{f(t)} \amp = t \amp f\bbrac{\inv{f}(t)} \amp = t \text{,} \end{align*}
as required by Definition 16.2.3.

Section 16.3 One-to-one functions

How can we tell whether a function is invertible in the first place?

Example 16.3.1. A function without an inverse.

If we try to apply the same procedure as in Example 16.2.10 to
\begin{equation*} f(t) = {(2 t + 3)}^2 \text{,} \end{equation*}
we run into a problem: there are two different ways we could isolate \(t\text{.}\) If we set
\begin{equation*} y = {(2 t + 3)}^2 \text{,} \end{equation*}
then certainly \(y\) can’t be negative and we can take its square root. But there are two different numbers whose square is the right-hand side above:
\begin{align*} \sqrt{y} \amp = 2 t + 3 \amp \sqrt{y} \amp = -(2 t + 3) \\ \sqrt{y} - 3 \amp = 2 t \amp \sqrt{y} + 3 \amp = -2 t \\ t \amp = \frac{\sqrt{y} - 3}{2} \amp t \amp = \frac{\sqrt{y} + 3}{-2} \text{.} \end{align*}
Neither function
\begin{align*} g_1(t) \amp = \frac{\sqrt{t} - 3}{2} \amp g_2(t) \amp = - \frac{\sqrt{t} + 3}{2} \end{align*}
will universally satisfy
\begin{equation*} g\bbrac{f(t)} = t \end{equation*}
for all \(t\) in the domain of \(f\text{,}\) so we cannot say that either one is the inverse function of \(f\text{.}\)

Example 16.3.2. A function that really, really doesn’t have an inverse.

Consider function \(f(t) = t^4 + 2 t^3 - 7 t^2 + t - 5\text{.}\) The Fundamental Theorem of Algebra says that the equation
\begin{gather*} y = t^4 + 2 t^3 - 7 t^2 + t - 5 \\ \implies \qquad t^4 + 2 t^3 - 7 t^2 + t - (5 + y) = 0 \end{gather*}
could have up to \(4\) solutions for \(t\text{,}\) depending on the value of \(y\text{.}\)
The equality
\begin{equation*} \inv{q}\bbrac{q(t)} = t \end{equation*}
is actually stating the purpose of an inverse function: to answer the question “Given an output \(q(t)\text{,}\) what input \(t\) produced it?” The reason that Example 16.2.10 works but Example 16.3.1 and Example 16.3.2 don’t is that in that first example, every output value \(y\) can be traced back to a single corresponding input value \(t\text{,}\) but in the other two examples some output values could correspond to two or more input values.

Definition 16.3.3. One-to-one function.

A function \(f\) is called one-to-one if for every output value \(y\) in the range of \(f\) there is one and only one corresponding input value \(t\) so that \(f(t) = y\text{.}\)

Warning 16.3.4. Don’t get mixed up!

A common mistake is to mix up the above definition with the statement every input produces exactly one output, but every function does that! (That is the idea behind the Vertical Line Test.) The correct (shortened) version of Definition 16.3.3 is to say that a function is one-to-one if every output is produced by exactly one input.

Example 16.3.5. A shifted cube function is one-to-one.

The range of the function \(f(t) = t^3 + 4\) is all of \(\R\text{,}\) and for each output value \(y = f(t)\) in that range, there is exactly one input \(t = \sqrt[3]{y - 4}\) that produces that output.
Diagram illustrating the result of tracing an output back to the input for a vertically shifted cube function.
Figure 16.3.6. Tracing an output back to the input for a vertically shifted cube function.

Example 16.3.7. The sine function is not one-to-one.

The range of the function \(\sin(t)\) is \(-1 \le y \le 1\text{,}\) but there are output values in that range which have many corresponding input values. For example, \(\sin(t) = 1\) occurs at positive inputs
\begin{equation*} t = \frac{\pi}{2}, \frac{5 \pi}{2}, \frac{9 \pi}{2}, \dotsc \end{equation*}
and also at negative inputs
\begin{equation*} t = - \frac{3 \pi}{2}, - \frac{7 \pi}{2}, - \frac{11 \pi}{2}, \dotsc \text{.} \end{equation*}
Diagram illustrating that many different input values produce the same output value 1 for the sine function the input-output and corresponding output-input processes.
Figure 16.3.8. Many different input values produce the same output value \(1\) for the sine function.
The diagrams in Figure 16.3.6 and Figure 16.3.8 suggest a graphical test for the property of one-to-one.
The graph of a function that fails the Horizontal Line Test.
Figure 16.3.10. The graph of a function that fails the Horizontal Line Test.

Remark 16.3.11.

In the Horizontal Line Test, it’s fine if there are horizontal lines that do not intersect the graph at all — this just indicates that \(y\)-value at the height of that line is not in the range of the function. But if some horizontal line intersects the graph more than once, then those different intersection points represent input-output pairs with the same output value but different input values, as in Figure 16.3.10.

Example 16.3.12. The square function fails the Horizontal Line Test.

The function \(f(t) = t^2\) fails the Horizontal Line Test for many horizontal lines. In fact, every horizontal line at height \(y \gt 0\) will intersect the graph twice.
A graph demonstrating that the square function fails the Horizontal Line Test.
Figure 16.3.13. For all \(y \gt 0\text{,}\) there are two inputs that produce that output level.
The one-to-one property is exactly what lets us form an inverse function.
And because creating an inverse function is essentially interchanging the roles of input and output variables, we have the following further properties.

Section 16.4 Graph of an inverse function

If creating an inverse function interchanges the roles input and output variables, then creating the graph of an inverse function interchanges the horizontal and vertical axes.

Example 16.4.2. Graphs of the natural logarithm and exponential functions.

The graph of the natural logarithm clearly passes the Horizontal Line Test. We have previously explored how the graph of the natural exponential function lines up with the reflection of the graph of the natural logarithm in the line \(y = t\text{.}\) (See Pattern 9.3.7.)
A diagram illustrating that the natural exponential graph is a reflection of the natural logarithm graph.
Figure 16.4.3. The natural exponential graph is a reflection of the natural logarithm graph.
This tells us that
\begin{align*} \inv{\ln}(t) \amp = \exp(t) \amp \inv{\exp}(t) \amp = \ln(t) \text{.} \end{align*}
We already knew that
\begin{equation*} \ln(e^t) = t \ln e = t \text{,} \end{equation*}
and this now also confirms that
\begin{equation*} e^{\ln(t)} = t \text{.} \end{equation*}
When we reflect in the line \(y = t\text{,}\) a horizontal line becomes a vertical line. So when we verify the Horizontal Line Test for an invertible function \(f\text{,}\) it is equivalent to verifying the Vertical Line Test for its inverse function.

Example 16.4.4. Restricting the domain to create an invertible function.

We know that the square function \(f(t) = t^2\) fails the Horizontal Line Test, but we also know that the square root function is sort of like an inverse function for the square function.
Graph of the square function with a typical horizontal line.
(a) The square function.
Diagram illustrating that the graph of the square root function is a reflection of only "half" of the graph of the square function.
(b) The graph of the square root function is a reflection of only “half” of the graph of the square function.
Figure 16.4.5. The square function fails the Horizontal Line Test but the square root function passes the Vertical Line test.
As we see in Figure 16.4.5.(a), the graph of \(g(t) = \sqrt{t}\) is in fact the reflection of part of the graph of \(f(t) = t^2\text{,}\) and the graph of \(g(t)\) passes the Vertical Line Test because the “bottom half” of this sideways parabola is not actually part of the graph. This indicates how we should “fix” the square function so that the square root function is actually its inverse: just remove the “left” half of the upright parabola. In other words, suppose we create a new square function with a restricted domain that cuts off the portion of the graph that is causing the Horizontal Line Test to fail:
\begin{align*} h(t) \amp = t^2 \amp t \amp \ge 0 \text{.} \end{align*}
Then the new graph of this new function becomes one-to-one and hence the function is invertible. And, in fact, \(\inv{h}(t) = \sqrt{t}\text{.}\)
Diagram illustrating that the graph of the square root function is a reflection of one-half of the graph of the square function.
Figure 16.4.6. Restricting the domain of the square function creates a one-to-one graph, and if we reflect this half-a-parabola in the line \(y = t\) the result is the graph of the square root function.
We have fixed also fixed the problem where
\begin{equation*} g\bbrac{f(t)} = \sqrt{t^2} = \abs{t} \neq t \end{equation*}
(as in Example 16.2.8). When we now compute \(g\bbrac{h(t)}\text{,}\) we assume that \(t\) is in the domain of \(h\text{,}\) which has been restricted to \(t \ge 0\text{.}\) In that case,
\begin{equation*} g\bbrac{h(t)} = \sqrt{t^2} = \abs{t} = t \text{.} \end{equation*}

Section 16.5 Derivative of an inverse function

If the graphs of a function and its inverse are related as in the previous section, then slopes on the graphs should also be related.

Example 16.5.1. Derivatives of the cube and cube root functions.

Even though Pattern 12.4.3 allows us to directly compute formulas for the derivatives of both \(q(t) = t^3\) and \(\inv{q}(t) = t^{1/3}\text{,}\) let’s explore the relationship between their derivatives.
To compute a Derivative value like \(q'(2)\) from the definition, we should first approximate that instantaneous rate with an average rate calculation over a short time interval containing \(2\text{.}\) So suppose we choose \(t_1 \le 2\) and \(t_2 \ge 2\) so that \(\Delta t = t_2 - t_1\) is small but not zero. The corresponding average rate value
\begin{equation*} \frac{\text{rise}}{\text{run}} = \slope{q}{t} = \frac{q_2 - q_1}{t_2 - t_1} \end{equation*}
is the slope of the secant line between the points \((t_1,q_1)\) and \((t_2,q_2)\text{.}\)
Diagram illustrating a secant line on the cube graph.
(a) A secant line around \(t = 2\) on the cube graph.
Diagram illustrating a secant line on the cube-root graph.
(b) A secant line around \(q = 8\) on the cube-root graph.
Figure 16.5.2. Interchanging horizontal and vertical axes causes slopes to become reciprocals.
Notice that in Figure 16.5.2.(b), which is the graph of the cube root function, we have labelled the horizontal axis as \(q\) and the vertical axis as \(t\text{,}\) to match our idea that the inverse function interchanges the roles of input and output. So while \(q_1 \le 8 \le q_2\) creates a small range around \(q = 8\) on the vertical axis for the cube graph, it also creates a small domain around \(q = 8\) on the horizontal axis for the cube root graph. This allows us to relate the derivative value \(q'(2)\) not to \((\inv{q})'(2)\) but to \((\inv{q})'(8)\text{.}\) On the second graph, the slope of the reflected secant line must be calculated as
\begin{equation*} \frac{\text{rise}}{\text{run}} = \slope{t}{q} = \frac{t_2 - t_1}{q_2 - q_1} \text{.} \end{equation*}
The result is that an average rate for the inverse function is the reciprocal of a corresponding average rate for the original function. Since this is true about every average rate function calculation, we conclude that there should also be a reciprocal relationship for the derivatives of the two functions.
Let’s test out this conclusion. On the one hand, we know \(q'(t) = 3 t^2\text{,}\) so \(q'(2) = 12\text{.}\) On the other hand, we can also use Pattern 12.4.3 to directly compute that
\begin{equation*} \inv{q}(t) = t^{1/3} \quad\implies\quad (\inv{q})'(t) = \frac{1}{3} \cdot t^{-2/3} = \frac{1}{3 t^{2/3}} \text{,} \end{equation*}
so
\begin{align*} (\inv{q})'(8) \amp = \frac{1}{3 \cdot 8^{2/3}} \\ \amp = \frac{1}{3 \cdot {(\sqrt[3]{8})}^2 } \\ \amp = \frac{1}{3 \cdot 2^2 } \\ \amp = \frac{1}{12} \text{,} \end{align*}
which is indeed the reciprocal of our computed value for \(q'(2)\text{.}\)

Remark 16.5.4.

If we like, we can turn the formula in Pattern 16.5.3 into one involving \(t\) by just replacing \(y\) by \(t\text{:}\)
\begin{equation*} (\inv{f})'(t) = \frac{1}{f'\bbrac{\inv{f}(t)}} \text{.} \end{equation*}
Kep in mind that the reason there is an \(\inv{f}(t)\) inside the \(f'\) in the fraction on the right is because an input for \(f\) is the same as an output for \(\inv{f}\text{,}\) and the derivative function \(f'\) should have the same inputs as \(f\) (except for any points of non-differentiability).

Example 16.5.5. Derivative of the natural exponential function.

We’ve already used logarithmic differentiation to confirm \(d\bbrac{\exp(t)}/dt = \exp(t)\) (see Procedure 14.5.4), so let’s use that knowledge to test the formula of Pattern 16.5.3. (In fact, this calculation is mathematically equivalent to our previous verification via logarithmic differentiation.)
With the knowledge of the Fundamental Theorem of Calculus, we might interpret our definition of the natural logarithm (Definition 8.1.1) as specifically designed so that
\begin{equation*} \opddt \bbrac{\ln(t)} = \frac{1}{t} \text{.} \end{equation*}
Viewing the natural exponential function as the inverse of the natural logarithm, we take \(f = \ln\) in Pattern 16.5.3, so that \(\inv{f} = \exp\text{.}\) Using \(\ln'(t) = 1/t\) in the formula from that pattern, we have
\begin{align*} \exp'(t) \amp = (\inv{\ln})'(t) \\ \amp = \frac{1}{\ln'\bbrac{\inv{\ln}(t)}} \\ \amp = \frac{1}{1 / \inv{\ln}(t)} \\ \amp = \frac{1}{1 / \exp(t)} \\ \amp = \exp(t) \text{,} \end{align*}
as expected.
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