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Section 2.6 Derivatives of Inverse Functions

Much of mathematics centers on the notion of function. Indeed, throughout our study of calculus, we are investigating the behavior of functions, with particular emphasis on how fast the output of the function changes in response to changes in the input. Because each function represents a process, a natural question to ask is whether or not the particular process can be reversed. That is, if we know the output that results from the function, can we determine the input that led to it? And if we know how fast a particular process is changing, can we determine how fast the inverse process is changing?

One of the most important functions in all of mathematics is the natural exponential function \(f(x) = e^x\text{.}\) Its inverse, the natural logarithm, \(g(x) = \ln(x)\text{,}\) is similarly important. One of our goals in this section is to learn how to differentiate the logarithm function. First, we review some of the basic concepts surrounding functions and their inverses.

Preview Activity 2.6.1.

The equation \(y = \frac{5}{9}(x-32)\) relates a temperature given in \(x\) degrees Fahrenheit to the corresponding temperature \(y\) measured in degrees Celcius.

  1. Solve the equation \(y = \frac{5}{9}(x-32)\) for \(x\) to write \(x\) (Fahrenheit temperature) in terms of \(y\) (Celcius temperature).

  2. Let \(C(x) = \frac{5}{9}(x-32)\) be the function that takes a Fahrenheit temperature as input and produces the Celcius temperature as output. In addition, let \(F(y)\) be the function that converts a temperature given in \(y\) degrees Celcius to the temperature \(F(y)\) measured in degrees Fahrenheit. Use your work in (a) to write a formula for \(F(y)\text{.}\)

  3. Next consider the new function defined by \(p(x) = F(C(x))\text{.}\) Use the formulas for \(F\) and \(C\) to determine an expression for \(p(x)\) and simplify this expression as much as possible. What do you observe?

  4. Now, let \(r(y) = C(F(y))\text{.}\) Use the formulas for \(F\) and \(C\) to determine an expression for \(r(y)\) and simplify this expression as much as possible. What do you observe?

  5. What is the value of \(C'(x)\text{?}\) of \(F'(y)\text{?}\) How do these values appear to be related?

Subsection 2.6.1 Basic facts about inverse functions

A function \(f : A \to B\) is a rule that associates each element in the set \(A\) to one and only one element in the set \(B\text{.}\) We call \(A\) the domain of \(f\) and \(B\) the codomain of \(f\text{.}\) If there exists a function \(g : B \to A\) such that \(g(f(a)) = a\) for every possible choice of \(a\) in the set \(A\) and \(f(g(b)) = b\) for every \(b\) in the set \(B\text{,}\) then we say that \(g\) is the inverse of \(f\text{.}\)

We often use the notation \(f^{-1}\) (read “\(f\)-inverse”) to denote the inverse of \(f\text{.}\) The inverse function undoes the work of \(f\text{.}\) Indeed, if \(y = f(x)\text{,}\) then

\begin{equation*} f^{-1}(y) = f^{-1}(f(x)) = x\text{.} \end{equation*}

Thus, the equations \(y = f(x)\) and \(x = f^{-1}(y)\) say the same thing. The only difference between the two equations is one of perspective — one is solved for \(x\text{,}\) while the other is solved for \(y\text{.}\)

Here we briefly remind ourselves of some key facts about inverse functions.

Note 2.6.1.

For a function \(f : A \to B\text{,}\)

  • \(f\) has an inverse if and only if \(f\) is one-to-one  1  and onto  2 ;

  • provided \(f^{-1}\) exists, the domain of \(f^{-1}\) is the codomain of \(f\text{,}\) and the codomain of \(f^{-1}\) is the domain of \(f\text{;}\)

  • \(f^{-1}(f(x)) = x\) for every \(x\) in the domain of \(f\) and \(f(f^{-1}(y)) = y\) for every \(y\) in the codomain of \(f\text{;}\)

  • \(y = f(x)\) if and only if \(x = f^{-1}(y)\text{.}\)

A function \(f\) is one-to-one provided that no two distinct inputs lead to the same output.
A function \(f\) is onto provided that every possible element of the codomain can be realized as an output of the function for some choice of input from the domain.

The last fact reveals a special relationship between the graphs of \(f\) and \(f^{-1}\text{.}\) If a point \((x,y)\) that lies on the graph of \(y = f(x)\text{,}\) then it is also true that \(x = f^{-1}(y)\text{,}\) which means that the point \((y,x)\) lies on the graph of \(f^{-1}\text{.}\) This shows us that the graphs of \(f\) and \(f^{-1}\) are the reflections of each other across the line \(y = x\text{,}\) because this reflection is precisely the geometric action that swaps the coordinates in an ordered pair. In Figure 2.6.2, we see this illustrated by the function \(y = f(x) = 2^x\) and its inverse, with the points \((-1,\frac{1}{2})\) and \((\frac{1}{2},-1)\) highlighting the reflection of the curves across \(y = x\text{.}\)

Figure 2.6.2. A graph of a function \(y = f(x)\) along with its inverse, \(y = f^{-1}(x)\text{.}\)

To close our review of important facts about inverses, we recall that the natural exponential function \(y = f(x) = e^x\) has an inverse function, namely the natural logarithm, \(x = f^{-1}(y) = \ln(y)\text{.}\) Thus, writing \(y = e^x\) is interchangeable with \(x = \ln(y)\text{,}\) plus \(\ln(e^x) = x\) for every real number \(x\) and \(e^{\ln(y)} = y\) for every positive real number \(y\text{.}\)

Subsection 2.6.2 The derivative of the natural logarithm function

In what follows, we find a formula for the derivative of \(g(x) = \ln(x)\text{.}\) To do so, we take advantage of the fact that we know the derivative of the natural exponential function, the inverse of \(g\text{.}\) In particular, we know that writing \(g(x) = \ln(x)\) is equivalent to writing \(e^{g(x)} = x\text{.}\) Now we differentiate both sides of this equation and observe that

\begin{equation*} \frac{d}{dx}\left[e^{g(x)}\right] = \frac{d}{dx}[x]\text{.} \end{equation*}

The righthand side is simply \(1\text{;}\) by applying the chain rule to the left side, we find that

\begin{equation*} e^{g(x)} g'(x) = 1\text{.} \end{equation*}

Next we solve for \(g'(x)\text{,}\) to get

\begin{equation*} g'(x) = \frac{1}{e^{g(x)}}\text{.} \end{equation*}

Finally, we recall that \(g(x) = \ln(x)\text{,}\) so \(e^{g(x)} = e^{\ln(x)} = x\text{,}\) and thus

\begin{equation*} g'(x) = \frac{1}{x}\text{.} \end{equation*}
Natural Logarithm.

For all positive real numbers \(x\text{,}\) \(\frac{d}{dx}[\ln(x)] = \frac{1}{x}\text{.}\)

This rule for the natural logarithm function now joins our list of basic derivative rules. Note that this rule applies only to positive values of \(x\text{,}\) as these are the only values for which \(\ln(x)\) is defined.

Also notice that for the first time in our work, differentiating a basic function of a particular type has led to a function of a very different nature: the derivative of the natural logarithm is not another logarithm, nor even an exponential function, but rather a rational one.

Derivatives of logarithms may now be computed in concert with all of the rules known to date. For instance, if \(f(t) = \ln(t^2 + 1)\text{,}\) then by the chain rule, \(f'(t) = \frac{1}{t^2 + 1} \cdot 2t\text{.}\)

There are interesting connections between the graphs of \(f(x) = e^x\) and \(f^{-1}(x) = \ln(x)\text{.}\)

In Figure 2.6.3, we are reminded that since the natural exponential function has the property that its derivative is itself, the slope of the tangent to \(y = e^x\) is equal to the height of the curve at that point. For instance, at the point \(A = (\ln(0.5), 0.5)\text{,}\) the slope of the tangent line is \(m_A = 0.5\text{,}\) and at \(B = (\ln(5), 5)\text{,}\) the tangent line's slope is \(m_B = 5\text{.}\)

Figure 2.6.3. A graph of the function \(y = e^x\) along with its inverse, \(y = \ln(x)\text{,}\) where both functions are viewed using the input variable \(x\text{.}\)

At the corresponding points \(A'\) and \(B'\) on the graph of the natural logarithm function (which come from reflecting \(A\) and \(B\) across the line \(y = x\)), we know that the slope of the tangent line is the reciprocal of the \(x\)-coordinate of the point (since \(\frac{d}{dx}[\ln(x)] = \frac{1}{x}\)). Thus, at \(A' = (0.5, \ln(0.5))\text{,}\) we have \(m_{A'} = \frac{1}{0.5} = 2\text{,}\) and at \(B' = (5, \ln(5))\text{,}\) \(m_{B'} = \frac{1}{5}\text{.}\)

In particular, we observe that \(m_{A'} = \frac{1}{m_A}\) and \(m_{B'} = \frac{1}{m_B}\text{.}\) This is not a coincidence, but in fact holds for any curve \(y = f(x)\) and its inverse, provided the inverse exists. This is due to the reflection across \(y = x\text{.}\) It changes the roles of \(x\) and \(y\text{,}\) thus reversing the rise and run, so the slope of the inverse function at the reflected point is the reciprocal of the slope of the original function.

Activity 2.6.2.

For each function given below, find its derivative.

  1. \(h(x) = x^2\ln(x)\)

  2. \(p(t) = \frac{\ln(t)}{e^t + 1}\)

  3. \(s(y) = \ln(\cos(y) + 2)\)

  4. \(z(x) = \tan(\ln(x))\)

  5. \(m(z) = \ln(\ln(z))\)

Hint

  1. Is \(h\) a product, quotient, or composition of basic functions?

  2. Is \(p\) a product, quotient, or composition of basic functions?

  3. Is \(s\) a product, quotient, or composition of basic functions?

  4. Is \(z\) a product, quotient, or composition of basic functions?

  5. Is \(m\) a product, quotient, or composition of basic functions?

Answer

  1. \(h'(x) = x + 2x\ln(x)\text{.}\)

  2. \(p'(t) = \frac{(e^t + 1) \frac{1}{t} - \ln(t) \cdot e^t}{(e^t + 1)^2}\text{.}\)

  3. \(s'(y) = \frac{1}{\cos(y) + 2)} \cdot (-\sin(y))\text{.}\)

  4. \(z'(x) = \sec^2(\ln(x)) \cdot \frac{1}{x}\text{.}\)

  5. \(m'(z) = \frac{1}{\ln(z)} \cdot \frac{1}{z}\text{.}\)

Solution

  1. By the product rule,

    \begin{equation*} h'(x) = x^2\cdot \frac{1}{x} + \ln(x) \cdot 2x = x + 2x\ln(x)\text{.} \end{equation*}

  2. By the quotient rule,

    \begin{equation*} p'(t) = \frac{(e^t + 1) \frac{1}{t} - \ln(t) \cdot e^t}{(e^t + 1)^2}\text{.} \end{equation*}

  3. The chain rule tells us that

    \begin{equation*} s'(y) = \frac{1}{\cos(y) + 2)} \cdot (-\sin(y))\text{.} \end{equation*}

  4. Again using the chain rule,

    \begin{equation*} z'(x) = \sec^2(\ln(x)) \cdot \frac{1}{x}\text{.} \end{equation*}

  5. Noting that \(m\) is composite with the natural logarithm function serving as both the inner and outer function, we find that

    \begin{equation*} m'(z) = \frac{1}{\ln(z)} \cdot \frac{1}{z}\text{.} \end{equation*}

Subsection 2.6.3 Inverse trigonometric functions and their derivatives

Trigonometric functions are periodic, so they fail to be one-to-one, and thus do not have inverse functions. However, we can restrict the domain of each trigonometric function so that it is one-to-one on that domain.

For instance, consider the sine function on the domain \([-\frac{\pi}{2}, \frac{\pi}{2}]\text{.}\) Because no output of the sine function is repeated on this interval, the function is one-to-one and thus has an inverse. Thus, the function \(f(x) = \sin(x)\) with \([-\frac{\pi}{2}, \frac{\pi}{2}]\) and codomain \([-1,1]\) has an inverse function \(f^{-1}\) such that

\begin{equation*} f^{-1} : [-1,1] \to [-\frac{\pi}{2}, \frac{\pi}{2}]\text{.} \end{equation*}

We call \(f^{-1}\) the arcsine (or inverse sine) function and write \(f^{-1}(y) = \arcsin(y)\text{.}\) It is especially important to remember that

\begin{equation*} y = \sin(x) \ \ \text{and} \ \ x = \arcsin(y) \end{equation*}

say the same thing. “The arcsine of \(y\)” means “the angle whose sine is \(y\text{.}\)” For example, \(\arcsin(\frac{1}{2}) = \frac{\pi}{6}\) means that \(\frac{\pi}{6}\) is the angle whose sine is \(\frac{1}{2}\text{,}\) which is equivalent to writing \(\sin(\frac{\pi}{6}) = \frac{1}{2}\text{.}\)

Figure 2.6.4. A graph of \(f(x) = \sin(x)\) (in blue), restricted to the domain \([-\frac{\pi}{2}, \frac{\pi}{2}]\text{,}\) along with its inverse, \(f^{-1}(x) = \arcsin(x)\) (in magenta).

Next, we determine the derivative of the arcsine function. Letting \(h(x) = \arcsin(x)\text{,}\) our goal is to find \(h'(x)\text{.}\) Since \(h(x)\) is the angle whose sine is \(x\text{,}\) it is equivalent to write

\begin{equation*} \sin(h(x)) = x\text{.} \end{equation*}

Differentiating both sides of the previous equation, we have

\begin{equation*} \frac{d}{dx}[\sin(h(x))] = \frac{d}{dx}[x]\text{.} \end{equation*}

The righthand side is simply \(1\text{,}\) and by applying the chain rule applied to the left side,

\begin{equation*} \cos(h(x)) h'(x) = 1\text{.} \end{equation*}

Solving for \(h'(x)\text{,}\) it follows that

\begin{equation*} h'(x) = \frac{1}{\cos(h(x))}\text{.} \end{equation*}

Finally, we recall that \(h(x) = \arcsin(x)\text{,}\) so the denominator of \(h'(x)\) is the function \(\cos(\arcsin(x))\text{,}\) or in other words, “the cosine of the angle whose sine is \(x\text{.}\)” A bit of right triangle trigonometry allows us to simplify this expression considerably.

Let's say that \(\theta = \arcsin(x)\text{,}\) so that \(\theta\) is the angle whose sine is \(x\text{.}\) We can picture \(\theta\) as an angle in a right triangle with hypotenuse \(1\) and a vertical leg of length \(x\text{,}\) as shown in Figure 2.6.5. The horizontal leg must be \(\sqrt{1-x^2}\text{,}\) by the Pythagorean Theorem.

Figure 2.6.5. The right triangle that corresponds to the angle \(\theta = \arcsin(x)\text{.}\)

Now, because \(\theta = \arcsin(x)\text{,}\) the expression for \(\cos(\arcsin(x))\) is equivalent to \(\cos(\theta)\text{.}\) From the figure, \(\cos(\arcsin(x)) = \cos(\theta) = \sqrt{1-x^2}\text{.}\)

Substituting this expression into our formula, \(h'(x) = \frac{1}{\cos(\arcsin(x))}\text{,}\) we have now shown that

\begin{equation*} h'(x) = \frac{1}{\sqrt{1-x^2}}\text{.} \end{equation*}
Inverse sine.

For all real numbers \(x\) such that \(-1 \lt x \lt 1\text{,}\)

\begin{equation*} \frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}\text{.} \end{equation*}
Activity 2.6.3.

The following prompts in this activity will lead you to develop the derivative of the inverse tangent function.

  1. Let \(r(x) = \arctan(x)\text{.}\) Use the relationship between the arctangent and tangent functions to rewrite this equation using only the tangent function.

  2. Differentiate both sides of the equation you found in (a). Solve the resulting equation for \(r'(x)\text{,}\) writing \(r'(x)\) as simply as possible in terms of a trigonometric function evaluated at \(r(x)\text{.}\)

  3. Recall that \(r(x) = \arctan(x)\text{.}\) Update your expression for \(r'(x)\) so that it only involves trigonometric functions and the independent variable \(x\text{.}\)

  4. Introduce a right triangle with angle \(\theta\) so that \(\theta = \arctan(x)\text{.}\) What are the three sides of the triangle?

  5. In terms of only \(x\) and \(1\text{,}\) what is the value of \(\cos(\arctan(x))\text{?}\)

  6. Use the results of your work above to find an expression involving only \(1\) and \(x\) for \(r'(x)\text{.}\)

Hint

  1. Recall that for any function and its inverse, writing \(y = f^{-1}(x)\) is equivalent to writing \(f(y) = x\text{.}\)

  2. Apply the chain rule to differentiate \(\tan(r(x))\text{.}\)

  3. This question is only asking you to substitute the expression for \(r(x)\) into what you found in (b).

  4. Let the vertical leg of the triangle be \(x\text{.}\) What must the horizontal leg be?

  5. Recall that the cosine of an angle is the length of the adjacent leg over the length of they hypotenuse.

  6. Note that \(\cos^2(\alpha) = (\cos(\alpha))^2\text{.}\)

Answer

  1. \(\tan(r(x)) = x\text{.}\)

  2. \(r'(x) = \cos^2(r(x))\text{.}\)

  3. \(r'(x) = \cos^2(\arctan(x))\text{.}\)

  4. With \(\theta = \arctan(x)\text{,}\)

  5. \(\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}\text{.}\)

  6. \(r'(x) = \frac{1}{1+x^2}\text{.}\)

Solution

  1. Since \(r(x) = \arctan(x)\text{,}\) it is equivalent to write \(\tan(r(x)) = x\text{.}\)

  2. Differentiating, we have \(\frac{d}{dx}[\tan(r(x))] = \frac{d}{dx}[x]\text{,}\) so

    \begin{equation*} \sec^2(r(x)) r'(x) = 1\text{,} \end{equation*}

    and thus

    \begin{equation*} r'(x) = \frac{1}{\sec^2(r(x))} = \cos^2(r(x))\text{,} \end{equation*}

    since \(\frac{1}{\sec(\alpha)} = \cos(\alpha)\text{.}\)

  3. Since \(r(x) = \arctan(x)\text{,}\) we now have that \(r'(x) = \cos^2(\arctan(x))\text{.}\)

  4. Letting \(\theta = \arctan(x)\text{,}\) it follows that we can view \(\theta\) as an angle in a right triangle with legs \(1\) and \(x\) (so that \(\tan(\theta) = \frac{x}{1}\text{,}\) and thus by the Pythagorean Theorem, the triangle's hypotenuse is \(\sqrt{1+x^2}\text{,}\) as shown below.

  5. To evaluate \(\cos(\arctan(x))\text{,}\) we use the right triangle developed in (d) and observe that

    \begin{equation*} \cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}\text{.} \end{equation*}

  6. Finally, we recall that we know \(r'(x) = \cos^2(\arctan(x)) = \left( \cos(\arctan(x)) \right)^2\text{.}\) Having established that \(\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}\text{,}\) we now have that

    \begin{equation*} r'(x) = \frac{1}{1+x^2}\text{.} \end{equation*}

While derivatives for other inverse trigonometric functions can be established similarly, for now we limit ourselves to the arcsine and arctangent functions.

Activity 2.6.4.

Determine the derivative of each of the following functions.

  1. \(\displaystyle f(x) = x^3 \arctan(x) + e^x \ln(x)\)

  2. \(\displaystyle p(t) = 2^{t\arcsin(t)}\)

  3. \(\displaystyle h(z) = (\arcsin(5z) + \arctan(4-z))^{27}\)

  4. \(\displaystyle s(y) = \cot(\arctan(y))\)

  5. \(\displaystyle m(v) = \ln(\sin^2(v)+1)\)

  6. \(\displaystyle g(w) = \arctan\left( \frac{\ln(w)}{1+w^2} \right)\)

Hint

  1. Use the sum rule followed by the product rule on each term in the sum.

  2. Use the chain rule first. What rule is needed to differentiate \(t\arcsin(t)\text{?}\)

  3. Note that the chain rule is needed to differentiate \(\arcsin(5z)\text{.}\)

  4. You can use right triangle trigonometry to simplify the function first.

  5. Use the chain rule twice.

  6. Think about the overall structure of this function.

Answer

  1. \(f'(x) = \left[x^3 \cdot \frac{1}{1+x^2} + \arctan(x) \cdot 3x^2 \right] + \left[e^x \cdot \frac{1}{x} + \ln(x) \cdot e^x\right]\text{.}\)

  2. \(p'(t) = 2^{t\arcsin(t)} \ln(2) [t \cdot \frac{1}{\sqrt{1-t^2}} + \arcsin(t) \cdot 1]\text{.}\)

  3. \(h'(z) = 27(\arcsin(5z) + \arctan(4-z))^{26} \left[\frac{1}{1-(5z)^2} \cdot 5 + \frac{1}{1+(4-z)^2} \cdot (-1) \right]\text{.}\)

  4. \(s'(y) = -\frac{1}{y^2}\text{.}\)

  5. \(m'(v) = \frac{1}{\sin^2(v)+1} \cdot \left[ 2\sin(v)\cos(v) \right]\text{.}\)

  6. \(g'(w) = \frac{1}{1+ \left( \frac{\ln(w)}{1+w^2} \right)^2} \cdot \left[ \frac{(1+w^2) \frac{1}{w} - \ln(w) \cdot 2w}{(1+w^2)^2} \right] \)

Solution

  1. By the sum rule followed by two applications of the product rule,

    \begin{align*} f'(x) =\mathstrut \amp \frac{d}{dx}[x^3 \arctan(x)] + \frac{d}{dx}[e^x \ln(x)]\\ =\mathstrut \amp \left[x^3 \cdot \frac{1}{1+x^2} + \arctan(x) \cdot 3x^2 \right] + \left[e^x \cdot \frac{1}{x} + \ln(x) \cdot e^x\right] \end{align*}

  2. Using the chain rule followed by the product rule,

    \begin{align*} p'(t) =\mathstrut \amp 2^{t\arcsin(t)} \ln(2) \frac{d}{dt}[t\arcsin(t)]\\ =\mathstrut \amp 2^{t\arcsin(t)} \ln(2) [t \cdot \frac{1}{\sqrt{1-t^2}} + \arcsin(t) \cdot 1] \end{align*}

  3. Applying the chain rule twice,

    \begin{align*} h'(z) =\mathstrut \amp 27(\arcsin(5z) + \arctan(4-z))^{26} \frac{d}{dz}[\arcsin(5z) + \arctan(4-z)]\\ =\mathstrut \amp 27(\arcsin(5z) + \arctan(4-z))^{26} \left[\frac{1}{1-(5z)^2} \cdot \frac{d}{dz}[5z] + \frac{1}{1+(4-z)^2} \cdot \frac{d}{dz}[4-z] \right]\\ =\mathstrut \amp 27(\arcsin(5z) + \arctan(4-z))^{26} \left[\frac{1}{1-(5z)^2} \cdot 5 + \frac{1}{1+(4-z)^2} \cdot (-1) \right] \end{align*}

  4. Using right triangle trigonometry, it is straightforward to show that \(\cot(\arctan(y)) = \frac{1}{y}\text{.}\) As such, \(s'(y) = -\frac{1}{y^2}\text{.}\)

  5. By two applications of the chain rule (noting particularly that \(\sin^2(v) = (\sin(v))^2\text{,}\) we have

    \begin{align*} m'(v) =\mathstrut \amp \frac{1}{\sin^2(v)+1} \cdot \frac{d}{dv} \left[ \sin^2(v) + 1 \right]\\ =\mathstrut \amp \frac{1}{\sin^2(v)+1} \cdot \left[ 2\sin(v)\cos(v) \right] \end{align*}

  6. By the chain rule followed by the quotient rule,

    \begin{align*} g'(w) =\mathstrut \amp \frac{1}{1+ \left( \frac{\ln(w)}{1+w^2} \right)^2} \cdot \frac{d}{dw} \left[ \frac{\ln(w)}{1+w^2} \right]\\ =\mathstrut \amp \frac{1}{1+ \left( \frac{\ln(w)}{1+w^2} \right)^2} \cdot \left[ \frac{(1+w^2) \frac{1}{w} - \ln(w) \cdot 2w}{(1+w^2)^2} \right] \end{align*}

Subsection 2.6.4 The link between the derivative of a function and the derivative of its inverse

In Figure 2.6.3, we saw an interesting relationship between the slopes of tangent lines to the natural exponential and natural logarithm functions at points reflected across the line \(y = x\text{.}\) In particular, we observed that at the point \((\ln(2), 2)\) on the graph of \(f(x) = e^x\text{,}\) the slope of the tangent line is \(f'(\ln(2)) = 2\text{,}\) while at the corresponding point \((2, \ln(2))\) on the graph of \(f^{-1}(x) = \ln(x)\text{,}\) the slope of the tangent line is \((f^{-1})'(2) = \frac{1}{2}\text{,}\) which is the reciprocal of \(f'(\ln(2))\text{.}\)

That the two corresponding tangent lines have reciprocal slopes is not a coincidence. If \(f\) and \(g\) are differentiable inverse functions, then \(y = f(x)\) if and only if \(x = g(y)\text{,}\) then\(f(g(x)) = x\) for every \(x\) in the domain of \(f^{-1}\text{.}\) Differentiating both sides of this equation, we have

\begin{equation*} \frac{d}{dx} [f(g(x))] = \frac{d}{dx} [x]\text{,} \end{equation*}

and by the chain rule,

\begin{equation*} f'(g(x)) g'(x) = 1\text{.} \end{equation*}

Solving for \(g'(x)\text{,}\) we have \(g'(x) = \frac{1}{f'(g(x))}\text{.}\) Here we see that the slope of the tangent line to the inverse function \(g\) at the point \((x,g(x))\) is precisely the reciprocal of the slope of the tangent line to the original function \(f\) at the point \((g(x),f(g(x))) = (g(x),x)\text{.}\)

Figure 2.6.6. A graph of function \(y = f(x)\) along with its inverse, \(y = g(x) = f^{-1}(x)\text{.}\) Observe that the slopes of the two tangent lines are reciprocals of one another.

To see this more clearly, consider the graph of the function \(y = f(x)\) shown in Figure 2.6.6, along with its inverse \(y = g(x)\text{.}\) Given a point \((a,b)\) that lies on the graph of \(f\text{,}\) we know that \((b,a)\) lies on the graph of \(g\text{;}\) because \(f(a) = b\) and \(g(b) = a\text{.}\) Now, applying the rule that \(g'(x) = 1/f'(g(x))\) to the value \(x = b\text{,}\) we have

\begin{equation*} g'(b) = \frac{1}{f'(g(b))} = \frac{1}{f'(a)}\text{,} \end{equation*}

which is precisely what we see in the figure: the slope of the tangent line to \(g\) at \((b,a)\) is the reciprocal of the slope of the tangent line to \(f\) at \((a,b)\text{,}\) since these two lines are reflections of one another across the line \(y = x\text{.}\)

Derivative of an inverse function.

Suppose that \(f\) is a differentiable function with inverse \(g\) and that \((a,b)\) is a point that lies on the graph of \(f\) at which \(f'(a) \ne 0\text{.}\) Then

\begin{equation*} g'(b) = \frac{1}{f'(a)}\text{.} \end{equation*}

More generally, for any \(x\) in the domain of \(g'\text{,}\) we have \(g'(x) = 1/f'(g(x))\text{.}\)

The rules we derived for \(\ln(x)\text{,}\) \(\arcsin(x)\text{,}\) and \(\arctan(x)\) are all just specific examples of this general property of the derivative of an inverse function. For example, with \(g(x) = \ln(x)\) and \(f(x) = e^x\text{,}\) it follows that

\begin{equation*} g'(x) = \frac{1}{f'(g(x))} = \frac{1}{e^{\ln(x)}} = \frac{1}{x}\text{.} \end{equation*}

Subsection 2.6.5 Summary

  • For all positive real numbers \(x\text{,}\) \(\frac{d}{dx}[\ln(x)] = \frac{1}{x}\text{.}\)

  • For all real numbers \(x\) such that \(-1 \lt x \lt 1\text{,}\) \(\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}\text{.}\) In addition, for all real numbers \(x\text{,}\) \(\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}\text{.}\)

  • If \(g\) is the inverse of a differentiable function \(f\text{,}\) then for any point \(x\) in the domain of \(g'\text{,}\) \(g'(x) = \frac{1}{f'(g(x))}\text{.}\)

Exercises 2.6.6 Exercises

1. Composite function involving logarithms and polynomials.
2. Composite function involving trigonometric functions and logarithms.
3. Product involving \(\arcsin(w)\).
4. Derivative involving \(\arctan(x)\).
5. Composite function from a graph.
6. Composite function involving an inverse trigonometric function.
7. Mixing rules: product, chain, and inverse trig.
8. Mixing rules: product and inverse trig.
9.

Determine the derivative of each of the following functions. Use proper notation and clearly identify the derivative rules you use.

  1. \(f(x) = \ln(2\arctan(x) + 3\arcsin(x) + 5)\)

  2. \(r(z) = \arctan(\ln(\arcsin(z)))\)

  3. \(q(t) = \arctan^2(3t) \arcsin^4(7t)\)

  4. \(g(v) = \ln\left( \frac{\arctan(v)}{\arcsin(v) + v^2} \right)\)

Answer

  1. \(f'(x) = \frac{1}{2\arctan(x) + 3\arcsin(x) + 5} \cdot \left(\frac{2}{1+x^2} + \frac{3}{\sqrt{1-x^2}}\right)\text{.}\)

  2. \(r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \left( \frac{1}{\arcsin(z)} \right) \cdot \frac{1}{\sqrt{1-z^2}}\text{.}\)

  3. \(q'(t) = \arctan^2(3t) \cdot \left[4\arcsin^3(7t) \left( \frac{7}{\sqrt{1+(7t)^2}} \right)\right] + \arcsin^4(7t) \cdot \left[2\arctan(3t) \left(\frac{3}{1+(3t)^2}\right) \right]\text{.}\)

  4. \(g'(v) = \frac{1}{\frac{\arctan(v)}{\arcsin(v) + v^2}} \cdot \frac{(\arcsin(v) + v^2) \cdot \frac{1}{1+v^2} - \arctan(v) \cdot \left(\frac{1}{\sqrt{1-v^2}} + 2v \right)}{(\arcsin(v) + v^2)^2} \)

Solution

  1. By the chain rule,

    \begin{equation*} f'(x) = \frac{1}{2\arctan(x) + 3\arcsin(x) + 5} \cdot \frac{d}{dx}\left( 2\arctan(x) + 3\arcsin(x) + 5\right)\text{,} \end{equation*}

    and by the sum and constant multiple rules, it follows that

    \begin{equation*} f'(x) = \frac{1}{2\arctan(x) + 3\arcsin(x) + 5} \cdot \left(\frac{2}{1+x^2} + \frac{3}{\sqrt{1-x^2}}\right)\text{.} \end{equation*}

  2. By the chain rule applied to the outermost function,

    \begin{equation*} r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \frac{d}{dz} \left( \ln(\arcsin(z)) \right)\text{.} \end{equation*}

    Applying the chain rule again to evaluate \(\frac{d}{dz} \left( \ln(\arcsin(z)) \right)\text{,}\)

    \begin{equation*} r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \left( \frac{1}{\arcsin(z)} \right) \cdot \frac{d}{dz}\left(\arcsin(z)\right)\text{.} \end{equation*}

    Finally, using the basic rule for \(\frac{d}{dz}\left(\arcsin(z)\right)\text{,}\)

    \begin{equation*} r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \left( \frac{1}{\arcsin(z)} \right) \cdot \frac{1}{\sqrt{1-z^2}}\text{.} \end{equation*}

  3. First, by the product rule,

    \begin{equation*} q'(t) = \arctan^2(3t) \cdot \frac{d}{dt}\left(\arcsin^4(7t)\right) + \arcsin^4(7t) \cdot \frac{d}{dt}\left(\arctan^2(3t)\right)\text{.} \end{equation*}

    Next, we recall that \(\arcsin^4(7t) = \left(\arcsin(7t)\right)^4\) and \(\arctan^2(3t) = \left( \arctan(3t) \right)^2\text{,}\) and apply the chain rule to find the derivative of each of these functions to get

    \begin{equation*} q'(t) = \arctan^2(3t) \cdot \left[4\arcsin^3(7t) \frac{d}{dt}\left( \arcsin(7t) \right)\right] + \arcsin^4(7t) \cdot \left[2\arctan(3t) \frac{d}{dt}\left(\arctan(3t)\right) \right] \text{.} \end{equation*}

    Finally, applying the chain rule one more time to evaluate the remaining derivatives,

    \begin{equation*} q'(t) = \arctan^2(3t) \cdot \left[4\arcsin^3(7t) \left( \frac{7}{\sqrt{1+(7t)^2}} \right)\right] + \arcsin^4(7t) \cdot \left[2\arctan(3t) \left(\frac{3}{1+(3t)^2}\right) \right]\text{.} \end{equation*}

  4. By the chain rule,

    \begin{equation*} g'(v) = \frac{1}{ \frac{\arctan(v)}{\arcsin(v) + v^2} } \cdot \frac{d}{dv}\left[ \frac{\arctan(v)}{\arcsin(v) + v^2} \right] \end{equation*}

    Thus, by the quotient rule,

    \begin{equation*} g'(v) = \frac{1}{\frac{\arctan(v)}{\arcsin(v) + v^2}} \cdot \frac{(\arcsin(v) + v^2) \cdot \frac{1}{1+v^2} - \arctan(v) \cdot \left(\frac{1}{\sqrt{1-v^2}} + 2v \right)}{(\arcsin(v) + v^2)^2} \end{equation*}
10.

Consider the graph of \(y = f(x)\) provided in Figure 2.6.7 and use it to answer the following questions.

  1. Use the provided graph to estimate the value of \(f'(1)\text{.}\)

  2. Sketch an approximate graph of \(y = f^{-1}(x)\text{.}\) Label at least three distinct points on the graph that correspond to three points on the graph of \(f\text{.}\) Recall that if \(y=f(x)\) then \(x=f^{-1}(y)\text{.}\)

  3. Based on your work in (a), what is the value of \((f^{-1})'(-1)\text{?}\) Why?

Figure 2.6.7. A function \(y = f(x)\)
Answer

  1. \(f'(1) \approx 2\text{.}\)

  3. \(f'(-1) \approx 1/2\text{.}\)

Solution

  1. By plotting the tangent line as shown in the figure below, it appears that \(f'(1) \approx 2\text{.}\)

  2. One way to estimate the graph of \(y = f^{-1}(x)\) is to plot several \((x,y)\) points on the graph of \(y = f(x)\) and then plot the corresponding points \((y,x)\) with the coordinates reversed. Doing so and recalling that the graph of \(y = f^{-1}(x)\) is the reflection of the graph of \(y = f(x)\) across the line \(y = x\text{,}\) we see the following figure.

  3. Because the original function passes through the point \((1,-1)\text{,}\) the inverse function passes through \((-1,1)\text{.}\) Moreover, since the tangent line to the original function at \((1,-1)\) has slope \(f'(1) \approx 2\text{,}\) and the tangent line to the inverse function at \((-1,1)\) corresponds to the reflection of the original tangent line across the line \(y=x\text{,}\) it follows that \(f'(-1) \approx 1/2\text{.}\) This can also be deduced from the ``derivative of an inverse function'' rule, whereby \(f'(-1) = \frac{1}{f'(1)}\text{.}\)

11.

Let \(f(x) = \frac{1}{4}x^3 + 4\text{.}\)

  1. Sketch a graph of \(y = f(x)\) and explain why \(f\) is an invertible function.

  2. Let \(g\) be the inverse of \(f\) and determine a formula for \(g\text{.}\)

  3. Compute \(f'(x)\text{,}\) \(g'(x)\text{,}\) \(f'(2)\text{,}\) and \(g'(6)\text{.}\) What is the special relationship between \(f'(2)\) and \(g'(6)\text{?}\) Why?

Answer

  1. \(f\) passes the horizontal line test.

  2. \(f^{-1}(x) = g(x) = \sqrt[3]{4x-16}\text{.}\)

  3. \(f'(x) = \frac{3}{4}x^2\text{;}\) \(f'(2) = 3\text{.}\) \(g'(x) = \frac{1}{3}(4x-16)^{-2/3} \cdot 4\text{;}\) \(g'(6) = \frac{1}{3}\text{.}\) These two derivative values are reciprocals.

Solution

  1. \(f\) is an invertible function because its graph passes the horizontal line test.

  2. To find a formula for \(g = f^{-1}\text{,}\) we take the equation \(y = \frac{1}{4}x^3 + 4\) and solve for \(x\) in terms of \(y\text{.}\) Doing so, we find that \(y - 4 = \frac{1}{4}x^3\text{,}\) so \(x^3 = 4(y-4)\text{,}\) and thus

    \begin{equation*} x = \sqrt[3]{4(y-4)} = f^{-1}(y)\text{.} \end{equation*}

    Writing the inverse as a function of \(x\text{,}\) we have found that \(f^{-1}(x) = g(x) = \sqrt[3]{4x-16}\text{.}\)

  3. First, \(f'(x) = \frac{3}{4}x^2\text{,}\) and thus \(f'(2) = 3\text{.}\) Next, \(g'(x) = \frac{1}{3}(4x-16)^{-2/3} \cdot 4\text{,}\) so that \(g'(6) = \frac{1}{3} \cdot 8^{-2/3} \cdot 4 = \frac{1}{3} \cdot \frac{1}{4} \cdot 4 = \frac{1}{3}\text{.}\) What is special about these two derivative values is that they happen to be reciprocals. Noting that \(f(2) = 3\) and \(g(3) = 2\text{,}\) we know that \(g'(3) = \frac{1}{f'(2)}\) by the rule for the derivative of an inverse function, and this matches our finding above that \(g'(3) = \frac{1}{3}\text{.}\)

12.

Let \(h(x) = x + \sin(x)\text{.}\)

  1. Sketch a graph of \(y = h(x)\) and explain why \(h\) must be invertible.

  2. Explain why it does not appear to be algebraically possible to determine a formula for \(h^{-1}\text{.}\)

  3. Observe that the point \((\frac{\pi}{2}, \frac{\pi}{2} + 1)\) lies on the graph of \(y = h(x)\text{.}\) Determine the value of \((h^{-1})'(\frac{\pi}{2} + 1)\text{.}\)

Answer

  1. \(h\) passes the horizontal line test.

  2. The equation \(y = x + \sin(x)\) can't be solved for \(x\) in terms of \(y\text{.}\)

  3. \((h^{-1})'(\frac{\pi}{2} + 1) = 1\text{.}\)

Solution

  1. The function \(h\) is invertible because its graph passes the horizontal line test.

  2. It does not seem possible to solve the equation \(y = x + \sin(x)\) for \(x\) in terms of \(y\text{.}\)

  3. By the rule for the derivative of an invertible function, we have \((h^{-1})'(\frac{\pi}{2} + 1) = \frac{1}{h'\left(\frac{\pi}{2} \right)}\text{.}\) Thus, we first compute \(h'(x)\) and find that \(h'(x) = 1 + \cos(x)\text{.}\) Since \(h'\left(\frac{\pi}{2} \right) = 1 + \cos\left(\frac{\pi}{2} \right) = 1\text{,}\) it follows that \((h^{-1})'(\frac{\pi}{2} + 1) = \frac{1}{1} = 1\text{.}\)

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