Recall that a function \(y=f(x)\) is said to be one-to-one if it passes the horizontal line test; that is, for two different \(x\)-values \(x_1\) and \(x_2\text{,}\) we do not have \(f\mathopen{}\left(x_1\right)\mathclose{}=f\mathopen{}\left(x_2\right)\mathclose{}\text{.}\) In some cases the domain of \(f\) must be restricted so that it is one-to-one. For instance, consider \(f(x)=x^2\text{.}\) Clearly, \(f(-1)= f(1)\text{,}\) so \(f\) is not one-to-one on its regular domain, but by restricting \(f\) to \((0,\infty)\text{,}\) \(f\) is one-to-one.

Now recall that one-to-one functions have inverses. That is, if \(f\) is one-to-one, it has an inverse function, denoted by \(f^{-1}\text{,}\) such that if \(f(a)=b\text{,}\) then \(f^{-1}(b) = a\text{.}\) The domain of \(f^{-1}\) is the range of \(f\text{,}\) and vice-versa. For ease of notation, we set \(g=f^{-1}\) and treat \(g\) as a function of \(x\text{.}\)

Since \(f(a)=b\) implies \(g(b)=a\text{,}\) when we compose \(f\) and \(g\) we get a nice result: \begin{equation*} f\big(g(b)\big) = f(a) = b. \end{equation*}

In general, \(f\big(g(x)\big) = x\) and \(g\big(f(x)\big) = x\text{.}\) This gives us a convenient way to check if two functions are inverses of each other: compose them and if the result is \(x\) (on the appropriate domains), then they are inverses.

When the point \((a,b)\) lies on the graph of \(f\text{,}\) the point \((b,a)\) lies on the graph of \(g\text{.}\) This leads us to discover that the graph of \(g\) is the reflection of \(f\) across the line \(y=x\text{.}\) In Figure 2.7.1 we see a function graphed along with its inverse. See how the point \((1,1.5)\) lies on one graph, whereas \((1.5,1)\) lies on the other. Because of this relationship, whatever we know about \(f\) can quickly be transferred into knowledge about \(g\text{.}\)

For example, consider Figure 2.7.2 where the tangent line to \(f\) at the point \((1,1.5)\) is drawn. That line has slope \(3\text{.}\) Through reflection across \(y=x\text{,}\) we can see that the tangent line to \(g\) at the point \((1.5,1)\) has slope \(1/3\text{.}\) Their slopes are reciprocals. This should make sense since reflecting a line (such as a tangent line) across the line \(y=x\) switches the \(x\) and \(y\) values. Also consider the point \((0,0.5)\) on the graph of \(f\text{,}\) where the tangent line is horizontal. At the point \((0.5,0)\) on \(g\text{,}\) the tangent line is vertical.

More generally, consider the tangent line to \(f\) at the point \((a,b)\text{.}\) That line has slope \(\fp(a)\text{.}\) Through reflection across \(y=x\text{,}\) we can extend our above observation to say that the tangent line to \(g\) at the point \((b,a)\) should have slope \(1/\fp(a)\text{.}\) This then tells us that \(g'(b)=1/\fp(a)\text{.}\)

The information from these two graphs is summarized in Table 2.7.3 below:

We have discovered a relationship between \(\fp\) and \(g'\) in a mostly graphical way. We can realize this relationship analytically as well. Let \(y = g(x)\text{,}\) where again \(g = f^{-1}\text{.}\) We want to find \(y'\text{.}\) Since \(y = g(x)\text{,}\) we know that \(f(y) = x\text{.}\) Using the <<Unresolved xref, reference "thm_Chain_Rule"; check spelling or use "provisional" attribute>> and Implicit Differentiation, take the derivative of both sides of this last equality. \begin{align*} \lzoo{x}{f(y)} \amp = \lzoo{x}{x}\\ \fp(y)\cdot y' \amp = 1\\ y' \amp = \frac{1}{\fp(y)}\\ y' \amp = \frac{1}{\fp(g(x))} \end{align*}

This leads us to the following theorem.

The results of Theorem 2.7.4 are not trivial; the notation may seem confusing at first. Careful consideration, along with examples, should earn understanding.

In the next example we apply Theorem 2.7.4 to the arcsine function.

Remember that the input \(x\) of the arcsine function is a ratio of a side of a right triangle to its hypotenuse; the absolute value of this ratio will never be greater than \(1\text{.}\) Therefore the inside of the square root will never be negative.

In order to make \(y=\sin(x)\) one-to-one, we restrict its domain to \([-\pi/2,\pi/2]\text{;}\) on this domain, the range is \([-1,1]\text{.}\) Therefore the domain of \(y=\arcsin(x)\) is \([-1,1]\) and the range is \([-\pi/2,\pi/2]\text{.}\) When \(x=\pm 1\text{,}\) note how the derivative of the arcsine function is undefined; this corresponds to the fact that as \(x\to \pm1\text{,}\) the tangent lines to arcsine approach vertical lines with undefined slopes.

In Figure 2.7.7 we see \(f(x) = \sin(x)\) and \(f^{-1} = \sin^{-1}(x)\) graphed on their respective domains. The line tangent to \(\sin(x)\) at the point \(\left(\pi/3, \sqrt{3}/2\right)\) has slope \(\cos(\pi)/3 = 1/2\text{.}\) The slope of the corresponding point on \(\sin^{-1}(x)\text{,}\) the point \(\left(\sqrt{3}/2,\pi/3\right)\text{,}\) is \begin{align*} \frac{1}{\sqrt{1-\left(\sqrt{3}/2\right)^2}} \amp = \frac{1}{\sqrt{1-3/4}}\\ \amp = \frac{1}{\sqrt{1/4}}\\ \amp = \frac{1}{1/2}=2\text{,} \end{align*} verifying yet again that at corresponding points, a function and its inverse have reciprocal slopes.

Using similar techniques, we can find the derivatives of all the inverse trigonometric functions. In Table 2.7.8 we show the restrictions of the domains of the standard trigonometric functions that allow them to be invertible.

Note how each derivative is the negative of the derivative of its “co” function. Because of this, derivatives of \(\sin^{-1}(x)\text{,}\) \(\tan^{-1}(x)\text{,}\) and \(\sec^{-1}(x)\) are used almost exclusively throughout this text.

In Section 2.3, we stated without proof or explanation that \(\lzoo{\ln(x)}=\frac{1}{x}\text{.}\) We can justify that now using Theorem 2.7.4, as shown in the example.

In this chapter we have defined the derivative, given rules to facilitate its computation, and given the derivatives of a number of standard functions. We restate the most important of these in the following theorem, intended to be a reference for further work.