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APEX Calculus: for University of Lethbridge

Section 2.7 Derivatives of Inverse Functions

Figure 2.7.1. Video introduction to Section 2.7
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\text{.} \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.
Figure 2.7.2. Properties of inverse functions
For a refresher on how to determine the inverse of a given function, watch the video in Figure 2.7.3.
Figure 2.7.3. Finding the inverse of a one-to-one function
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 2.7.4 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 \(g\text{.}\)
A function f along with its inverse. Since they are inverses of each other It doesn’t matter which one we call f.
The graph of \(f\) starts at \((-1,-0.5)\) below the negative \(x\) axis then slopes upwards while moving to the right. A little after passing throught he point \((-0.5,0.375)\) it slopes further and faster to the right and starts moving horizontally parallel to the \(x\) axis. Then it starts to move upagain continuuing through \((1, 1.5)\text{.}\) The graph of \(g\) is a similar curve starting below the \(x\) axis close to the \(y\) axis in the third quadrant it moves to the right while moving up and then after passing through \((0.375,-0.5)\) it moves upwards vertically for a short while and then starts moving to the right while also moving upwards passing through \((1.5,1)\text{.}\)
Figure 2.7.4. A function \(f\) along with its inverse \(f^{-1}\text{.}\) (Note how it does not matter which function we refer to as \(f\text{;}\) the other is \(f^{-1}\text{.}\))
For example, consider Figure 2.7.5 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 \(g'(b)=1/\fp(a)\text{.}\)
Corresponding tangent lines drawn to f and its inverse
The graph of \(f\) starts close to \((-1,-0.5)\) below the negative \(x\) axis and slopes upwards and to the right. After passing through \((-0.5,0.375)\) it starts moving to the right. After intersecting the \(y\) axis at \((0,0.5)\) it continues to move horizontally for a short while then starts moving upwards to the right and continues upwards through the point \((1,1.5)\text{.}\) The tangent line of \(f\) is drawn at \((1,1.5)\text{,}\) it is a straight line that slightly touches the the curve at \((1,1.5)\) and keeps moving straight. \(g\) starts close to the negative \(y\) axis and starts moving to the right while sloping upwards. Once it moves through the point \((0.375,-0.5)\) then starts moving vertically upward and eventually starts moving again upwards and to the right. The tangent line of the curve is drawn at \((1.5,1)\text{.}\)
Figure 2.7.5. Corresponding tangent lines drawn to \(f\) and \(f^{-1}\)
The information from these two graphs is summarized in Table 2.7.6 below:
Table 2.7.6.
Information about \(f\) Information about \(g=f^{-1}\)
\((1,1.5)\) lies on \(f\) \((1.5,1)\) lies on \(g\)
Slope of tangent line to \(f\) at \(x=1\) is \(3\)
Slope of tangent line to \(g\) at \(x=1.5\) is \(1/3\)
\(\fp(1) = 3\) \(g'(1.5) = 1/3\)
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 The Chain Rule 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))}\text{.} \end{align*}
This leads us to the following theorem.
Figure 2.7.8. Video presentation of Theorem 2.7.7
The results of Theorem 2.7.7 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 2.7.7 to the arcsine function.
A word of caution is required here. The function \(\sin(x)\) is clearly not one-to-one. How can we say that \(\arcsin(x)\) is the inverse of \(\sin(x)\text{?}\) To make sense of this, we employ a technique known as restriction of domain: instead of considering the entire domain of the sine function, we consider a portion of it, on which the function is one-to-one, as explained in Figure 2.7.9.
Figure 2.7.9. Restricting the domain of \(\sin(x)\)

Example 2.7.10. Finding the derivative of an inverse trigonometric function.

Let \(y = \arcsin(x) = \sin^{-1}(x)\text{.}\) Find \(y'\) using Theorem 2.7.7.
Solution 1. Video solution
Solution 2.
Adopting our previously defined notation, let \(g(x) = \arcsin(x) \) and \(f(x) = \sin(x)\text{.}\) Thus \(\fp(x) = \cos(x)\text{.}\) Applying the theorem, we have
\begin{align*} g'(x) \amp = \frac{1}{\fp(g(x))}\\ \amp = \frac{1}{\cos(\arcsin(x))}\text{.} \end{align*}
This last expression is not immediately illuminating. Drawing a figure will help, as shown in Figure 2.7.11. Recall that the sine function can be viewed as taking in an angle and returning a ratio of sides of a right triangle, specifically, the ratio “opposite over hypotenuse.” This means that the arcsine function takes as input a ratio of sides and returns an angle. The equation \(y=\arcsin(x)\) can be rewritten as \(y=\arcsin(x/1)\text{;}\) that is, consider a right triangle where the hypotenuse has length \(1\) and the side opposite of the angle with measure \(y\) has length \(x\text{.}\) This means the final side has length \(\sqrt{1-x^2}\text{,}\) using the Pythagorean Theorem.
The right triangle defined by the equation sin(y)=x/1
The right angle triangle is defined by \(y=\sin^{-1}(x/1)\text{.}\) The length of the base is \(\sqrt{1-x^2}\) and the length of the perpendicular is \(x\text{.}\) The length of the hypotenuse is \(1\text{.}\) The angle between the base and the hypotenuse is \(y\text{.}\)
Figure 2.7.11. A right triangle defined by \(y=\sin^{-1}(x/1)\) with the length of the third leg found using the Pythagorean Theorem
\begin{align*} \cos\mathopen{}\left(\sin^{-1}(x)\right)\mathclose{} \amp= \cos(y)\\ \amp = \frac{\sqrt{1-x^2}}{1}\\ \amp = \sqrt{1-x^2}\text{,} \end{align*}
resulting in
\begin{equation*} \lzoo{x}{\arcsin(x)} = \frac{1}{\sqrt{1-x^2}}\text{.} \end{equation*}
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.
The graph of sin(x) and arcsin(x) with their corresponding tangent line
The graph of \(\sin(x)\) starts at \((-\pi/2,-1)\text{.}\) It first curves upwards to the right and then moves towards the origin making a slope. Once the slope passes through the origin it continues upwards to the right in the first quadrant. Once it reaches the point \((\pi/2,1)\text{,}\) the graph starts to move downwards again. The tangent line of the graph is drawn at \((\pi/3,\sqrt{3/2})\text{.}\)
Graphs of sin(x) and arcsin (x) with corresponding tangent line
Starting at the third quadrant at \((-1,-\pi/2)\) the graph of \(\arcsin(x)\) begins with a nearly vertical slope. The slope decreases to \(1\) as the graph passes through the origin. The graph then moves toward its end at \((1, \pi/2)\) with increasing slope. The tangent line is drawn at \((\sqrt{3/2},\pi/3)\text{.}\) The tangent line of \(\sin^{-1}(x)\) is steeper than the tangent line of \(\sin(x)\text{.}\)
Figure 2.7.12. Graphs of \(\sin(x)\) and \(\sin^{-1}(x)\) along with corresponding tangent lines
In Figure 2.7.12 we see \(f(x) = \sin(x)\) and \(f^{-1}(x) = \sin^{-1}(x)\) 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 2.7.13 we show the restrictions of the domains of the standard trigonometric functions that allow them to be invertible.
Table 2.7.13. Domains and ranges of the trigonometric and inverse trigonometric functions
Function Domain Range
\(\sin(x)\) \([-\pi/2, \pi/2]\) \([-1,1]\)
\(\sin^{-1}(x)\) \([-1,1]\) \([-\pi/2, \pi/2]\)
\(\cos(x)\) \([0,\pi]\) \([-1,1]\)
\(\cos^{-1}(x)\) \([-1,1]\) \([0,\pi]\)
\(\tan(x)\) \((-\pi/2,\pi/2)\) \((-\infty,\infty)\)
\(\tan^{-1}(x)\) \((-\infty,\infty)\) \((-\pi/2,\pi/2)\)
\(\csc(x)\) \([-\pi/2,0)\cup (0, \pi/2]\) \((-\infty,-1]\cup [1,\infty)\)
\(\csc^{-1}(x)\) \((-\infty,-1]\cup [1,\infty)\) \([-\pi/2,0)\cup (0, \pi/2]\)
\(\sec(x)\) \([0,\pi/2)\cup (\pi/2,\pi]\) \((-\infty,-1]\cup [1,\infty)\)
\(\sec^{-1}(x)\) \((-\infty,-1]\cup [1,\infty)\) \([0,\pi/2)\cup (\pi/2,\pi]\)
\(\cot(x)\) \((0,\pi)\) \((-\infty,\infty)\)
\(\cot^{-1}(x)\) \((-\infty,\infty)\) \((0,\pi)\)
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.
Figure 2.7.15. Computing the derivative of \(\arctan(x)\)
In Section 2.3, we stated without proof or explanation that \(\lzoo{x}{\ln(x)}=\frac{1}{x}\text{.}\) We can justify that now using Theorem 2.7.7, as shown in the example.

Example 2.7.16. Finding the derivative of \(y=\ln(x)\).

Use Theorem 2.7.7 to compute \(\lzoo{x}{\ln(x)}\text{.}\)
View \(y= \ln(x)\) as the inverse of \(y = e^x\text{.}\) Therefore, using our standard notation, let \(f(x) = e^x\) and \(g(x) = \ln(x)\text{.}\) We wish to find \(g'(x)\text{.}\) Theorem 2.7.7 gives:
\begin{align*} g'(x) \amp = \frac{1}{\fp(g(x))}\\ \amp = \frac{1}{e^{\ln(x) }}\\ \amp = \frac{1}{x}\text{.} \end{align*}
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.

Exercises Exercises

Terms and Concepts

  • True
  • False
Every function has an inverse.
In your own words explain what it means for a function to be “one-to-one.”
If \((1,10)\) lies on the graph of \(y=f(x)\text{,}\) what can be said about the graph of \(y=f^{-1}(x)\text{?}\)
If \((1,10)\) lies on the graph of \(y=f(x)\) and \(\fp(1) = 5\text{,}\) what can be said about \(y=f^{-1}(x)\text{?}\)


Exercise Group.
Verify that the given functions are inverses.
\(f(x) = 2x+6\) and \(g(x) = \frac{1}{2}x-3\)
\(f(x) = x^2+6x+11\text{,}\) \(x\geq 3\) and \(g(x) = \sqrt{x-2}-3\text{,}\) \(x\geq 2\)
\(f(x) = \frac{3}{x-5}\text{,}\) \(x\neq 5\) and \(g(x) = \frac{3+5x}{x}\text{,}\) \(x\neq 0\)
\(f(x) = \frac{x+1}{x-1}\text{,}\) \(x\neq 1\) and \(g(x) = f(x)\)
Exercise Group.
An invertible function \(f(x)\) is given along with a point that lies on its graph. Using Theorem 2.7.7, evaluate \(\left(f^{-1}\right)'(x)\) at the indicated value.
The point \({\left(9,65\right)}\) is on the graph of \(f(x) = {7x+2}\text{.}\) Find \(\left(f^{-1}\right)'({65})\text{.}\)
The point \({\left(-6,51\right)}\) is on the graph of \(f(x) = {x^{2}-2x+3}, x\geq 1\text{.}\) Find \(\left(f^{-1}\right)'({51})\text{.}\)
The point \(\left({\frac{\pi }{24}},{\frac{\sqrt{3}}{2}}\right)\) is on the graph of \(f(x) = {\cos\mathopen{}\left(4x\right)}, {0}\leq x\leq {\frac{\pi }{4}}\text{.}\) Find \(\left(f^{-1}\right)'\left({\frac{\sqrt{3}}{2}}\right)\text{.}\)
The point \({\left(3,576\right)}\) is on the graph of \(f(x) = {x^{3}-27x^{2}+267x-9}\text{.}\) Find \(\left(f^{-1}\right)'({576})\text{.}\)
The point \({\left(2,{\frac{1}{5}}\right)}\) is on the graph of \(f(x) = {\frac{1}{1+x^{2}}}, x\geq 0\text{.}\) Find \(\left(f^{-1}\right)'\left({{\frac{1}{5}}}\right)\text{.}\)
The point \({\left(0,3\right)}\) is on the graph of \(f(x) = {3e^{4x}}\text{.}\) Find \(\left(f^{-1}\right)'({3})\text{.}\)
Exercise Group.
Compute the derivative of the given function.
\(h(w) = {\cos^{-1}\mathopen{}\left(4w\right)}\)
\(h(x) = {\csc^{-1}\mathopen{}\left(7x\right)}\)
\(j(r) = {\tan^{-1}\mathopen{}\left(2r\right)}\)
\(k(w) = {w\cos^{-1}\mathopen{}\left(w\right)}\)
\(p(x) = {\tan\mathopen{}\left(x\right)\cos^{-1}\mathopen{}\left(x\right)}\)
\(f(t) = \ln(t) e^t\)
\(m(z) = {\frac{\tan^{-1}\mathopen{}\left(z\right)}{\sin^{-1}\mathopen{}\left(z\right)}}\)
\(f(x) = {\tan\mathopen{}\left(\sqrt[4]{x}\right)}\)
\(g(q) = {\csc\mathopen{}\left(\frac{1}{q^{3}}\right)}\)
\(g(z) = {\sin\mathopen{}\left(\sin^{-1}\mathopen{}\left(z\right)\right)}\)
Exercise Group.
Compute the derivative of the given function in two ways:
  1. By simplifying first, then taking the derivative, and
  2. by using the Chain Rule first then simplifying.
Verify that the two answers are the same.
Exercise Group.
Find the equation of the line tangent to the graph of \(f\) at the indicated \(x\) value.
\(f(x)=\sin^{-1}(x)\) at \(x={\frac{-\sqrt{3}}{2}}\)
\(f(x)=\cos^{-1}(2x)\) at \(x={\frac{\sqrt{3}}{4}}\)