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Derivatives of Inverse Functions

Calculate the derivative of an inverse function.

Introduction

Much of mathematics relies 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 \(f(x) = e^x\), along with its inverse, the natural logarithm \(g(x) = \ln(x)\). 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.

Exploration

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 \(B\). We call \(A\) the domain of \(f\) and \(B\) the codomain of \(f\). 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 \(B\), then we say that \(g\) is the inverse of \(f\).

We often use the notation \(f^{-1}\) (read \(f\)-inverse) to denote the inverse of \(f\). The inverse function undoes the work of \(f\). Indeed, if \(y = f(x)\), then \[\begin{aligned}\end{aligned}\]. 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 \(x\), while the other is solved for \(y\).

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

Note

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

\(f\) has an inverse if and only if \(f\) is one-to-one A function \(f\) is one-to-one provided that no two distinct inputs lead to the same output. A one-to-one function is sometimes called an injection. and onto 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. An onto function is sometimes called a surjection. ;
provided \(f^{-1}\) exists, the domain of \(f^{-1}\) is the codomain of \(f\), and the codomain of \(f^{-1}\) is the domain of \(f\);
\(f^{-1}(f(x)) = x\) for every \(x\) in the domain of \(f\) and \(f(f^{-1}(y)) = y\) for every \(y\) in the codomain of \(f\);
\(y = f(x)\) if and only if \(x = f^{-1}(y)\).

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, \(x = f^{-1}(y) = \ln(y)\). Thus, writing \(y = e^x\) is interchangeable with \(x = \ln(y)\), plus \(\ln(e^x) = x\) for every real number \(x\) and \(e^{\ln(y)} = y\) for every positive real number \(y\).

Condensed — the full section is in Boelkins, Active Calculus.

The derivative of the natural logarithm function

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

The righthand side is simply \(1\); by applying the chain rule to the left side, we find that \[\begin{aligned}\end{aligned}\].

Next we solve for \(g'(x)\), to get \[\begin{aligned}\end{aligned}\].

Finally, we recall that \(g(x) = \ln(x)\), so \(e^{g(x)} = e^{\ln(x)} = x\), and thus \[\begin{aligned}\end{aligned}\].

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

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 \(x\), 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 \(f(t) = \ln(t^2 + 1)\), then by the chain rule, \(f'(t) = \frac{1}{t^2 + 1} \cdot 2t\).

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

In Figure, 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 \(A = (\ln(0.5), 0.5)\), the slope of the tangent line is \(m_A = 0.5\), and at \(B = (\ln(5), 5)\), the tangent line's slope is \(m_B = 5\). 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 \(A' = (0.5, \ln(0.5))\), we have \(m_{A'} = \frac{1}{0.5} = 2\), and at \(B' = (5, \ln(5))\), \(m_{B'} = \frac{1}{5}\).

Condensed — the full section is in Boelkins, Active Calculus.

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 \([-\frac{\pi}{2}, \frac{\pi}{2}]\). 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 \[\begin{aligned}\end{aligned}\], as we see in Figure. We call \(f^{-1}\) the arcsine (or inverse sine) function and write \(f^{-1}(y) = \arcsin(y)\). The arcsine of \(y\) means the angle whose sine is \(y\). It is especially important to remember that \[\begin{aligned}\end{aligned}\] say the same thing.

For example, \(\arcsin(\frac{1}{2}) = \frac{\pi}{6}\) means that \(\frac{\pi}{6}\) is the angle whose sine is \(\frac{1}{2}\), which is equivalent to writing \(\sin(\frac{\pi}{6}) = \frac{1}{2}\).

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

Differentiating both sides of the previous equation, we have \[\begin{aligned}\end{aligned}\]. The righthand side is simply \(1\), and by applying the chain rule applied to the left side, \[\begin{aligned}\end{aligned}\].

Solving for \(h'(x)\), it follows that \[\begin{aligned}\end{aligned}\].

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

Substituting this most recent expression for \(\cos(\arcsin(x))\) into Equation, we have now shown that \[\begin{aligned}\end{aligned}\].

For all real numbers \(x\) such that \(-1 \lt x \lt 1\), \[\begin{aligned}\end{aligned}\].

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

Condensed — the full section is in Boelkins, Active Calculus.

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

In Figure, 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 \(y = x\). In particular, we observed that at the point \((\ln(2), 2)\) on the graph of \(f(x) = e^x\), the slope of the tangent line is \(f'(\ln(2)) = 2\), while at the corresponding point \((2, \ln(2))\) on the graph of \(f^{-1}(x) = \ln(x)\), the slope of the tangent line is \((f^{-1})'(2) = \frac{1}{2}\), which is the reciprocal of \(f'(\ln(2))\).

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 \(x = g(y)\), so \(f(g(x)) = x\) for every \(x\) in the domain of \(f^{-1}\). Differentiating both sides of this equation, we have \[\begin{aligned}\end{aligned}\], and by the chain rule, \[\begin{aligned}\end{aligned}\].

Solving for \(g'(x)\), we have \(g'(x) = \frac{1}{f'(g(x))}\). 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 \((g(x),f(g(x))) = (g(x),x)\).

To see this more clearly, consider the graph of the function \(y = f(x)\) shown in Figure, along with its inverse \(y = g(x)\). Given a point \((a,b)\) that lies on the graph of \(f\), we know that \((b,a)\) lies on the graph of \(g\); because \(f(a) = b\) and \(g(b) = a\).

Now, applying the rule that \(g'(x) = 1/f'(g(x))\) to the value \(x = b\), we have \[\begin{aligned}\end{aligned}\], 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 \((a,b)\), since these two lines are reflections of one another across the line \(y = x\).

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 \(f'(a) \ne 0\). Then \[\begin{aligned}\end{aligned}\].

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

The rules we derived for \(\ln(x)\), \(\arcsin(x)\), and \(\arctan(x)\) are all examples of this general property of the derivative of an inverse function. For instance, with \(g(x) = \ln(x)\) and \(f(x) = e^x\), it follows that \[\begin{aligned}\end{aligned}\].

Summary

For all positive real numbers \(x\), \(\frac{d}{dx}[\ln(x)] = \frac{1}{x}\).
For all real numbers \(x\) such that \(-1 \lt x \lt 1\), \(\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}\). In addition, for all real numbers \(x\), \(\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}\).
If \(g\) is the inverse of a differentiable function \(f\), then for any point \(x\) in the domain of \(g'\), \(g'(x) = \frac{1}{f'(g(x))}\).

The Derivative of an Inverse Function

We begin by considering a function and its inverse. If \(f(x)\) is both invertible and differentiable, it seems reasonable that the inverse of \(f(x)\) is also differentiable. shows the relationship between a function \(f(x)\) and its inverse \({f}^{-1}(x).\) Look at the point \((a,{f}^{-1}(a))\) on the graph of \({f}^{-1}(x)\) having a tangent line with a slope of \({({f}^{-1})}^{'}(a)=\frac{p}{q}.\) This point corresponds to a point \(({f}^{-1}(a),a)\) on the graph of \(f(x)\) having a tangent line with a slope of \({f}^{'}({f}^{-1}(a))=\frac{q}{p}.\) Thus, if \({f}^{-1}(x)\) is differentiable at \(a,\) then it must be the case that

\[{({f}^{-1})}^{'}(a)=\frac{1}{{f}^{'}({f}^{-1}(a))}.\]

We may also derive the formula for the derivative of the inverse by first recalling that \(x=f({f}^{-1}(x)).\) Then by differentiating both sides of this equation (using the chain rule on the right), we obtain

\[1={f}^{'}({f}^{-1}(x))({f}^{-1}{)}^{'}(x).\]

Solving for \(({f}^{-1}{)}^{'}(x),\) we obtain

\[{({f}^{-1})}^{'}(x)=\frac{1}{{f}^{'}({f}^{-1}(x))}.\]

We summarize this result in the following theorem.

Example

Try it.

Use the inverse function theorem to find the derivative of \(g(x)=\frac{x+2}{x}.\) Compare the resulting derivative to that obtained by differentiating the function directly.

Solution

The inverse of \(g(x)=\frac{x+2}{x}\) is \(f(x)=\frac{2}{x-1}.\) Since \({g}^{'}(x)=\frac{1}{{f}^{'}(g(x))},\) begin by finding \({f}^{'}(x).\) Thus,

\[{f}^{'}(x)=\frac{-2}{{(x-1)}^{2}}\ \text{and}\ {f}^{'}(g(x))=\frac{-2}{{(g(x)-1)}^{2}}=\frac{-2}{{(\frac{x+2}{x}-1)}^{2}}=-\frac{{x}^{2}}{2}.\]

Finally,

\[{g}^{'}(x)=\frac{1}{{f}^{'}(g(x))}=-\frac{2}{{x}^{2}}.\]

We can verify that this is the correct derivative by applying the quotient rule to \(g(x)\) to obtain

\[{g}^{'}(x)=-\frac{2}{{x}^{2}}.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Derivatives of Inverse Trigonometric Functions

We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time, we see that the derivative of a function need not be of the same type as the original function.

Example

Try it.

Use the inverse function theorem to find the derivative of \(g(x)={\text{sin}}^{-1}x.\)

Solution

Since for \(x\) in the interval \([-\frac{\pi }{2},\frac{\pi }{2}],f(x)=\text{sin}\ x\) is the inverse of \(g(x)={\text{sin}}^{-1}x,\) begin by finding \({f}^{'}(x).\) Since

\[{f}^{'}(x)=\text{cos}\ x\ \text{and}\ {f}^{'}(g(x))=\text{cos}\ ({\text{sin}}^{-1}x)=\sqrt{1-{x}^{2}},\]

we see that

\[{g}^{'}(x)=\frac{d}{dx}({\text{sin}}^{-1}x)=\frac{1}{{f}^{'}(g(x))}=\frac{1}{\sqrt{1-{x}^{2}}}.\]

Example

Try it.

Apply the chain rule to the formula derived in to find the derivative of \(h(x)={\text{sin}}^{-1}(g(x))\) and use this result to find the derivative of \(h(x)={\text{sin}}^{-1}(2{x}^{3}).\)

Solution

Applying the chain rule to \(h(x)={\text{sin}}^{-1}(g(x)),\) we have

\[{h}^{'}(x)=\frac{1}{\sqrt{1-{(g(x))}^{2}}}{g}^{'}(x).\]

Now let \(g(x)=2{x}^{3},\) so \({g}^{'}(x)=6{x}^{2}.\) Substituting into the previous result, we obtain

\[\begin{array}{ll}{h}^{'}(x) & =\frac{1}{\sqrt{1-4{x}^{6}}}\cdot 6{x}^{2} \\ & =\frac{6{x}^{2}}{\sqrt{1-4{x}^{6}}}.\end{array}\]

The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.

Example

Try it.

Find the derivative of \(f(x)={\text{tan}}^{-1}({x}^{2}).\)

Solution

Let \(g(x)={x}^{2},\) so \({g}^{'}(x)=2x.\) Substituting into , we obtain

\[{f}^{'}(x)=\frac{1}{1+{({x}^{2})}^{2}}\cdot (2x).\]

Simplifying, we have

\[{f}^{'}(x)=\frac{2x}{1+{x}^{4}}.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

The inverse function theorem allows us to compute derivatives of inverse functions without using the limit definition of the derivative.
We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions.

Key Equations

Inverse function theorem	\({({f}^{-1})}^{'}(x)=\frac{1}{{f}^{'}({f}^{-1}(x))}\) whenever \({f}^{'}({f}^{-1}(x))\ne 0\) and \(f(x)\) is differentiable.
Power rule with rational exponents	\(\frac{d}{dx}({x}^{m\text{/}n})=\frac{m}{n}{x}^{(m\text{/}n)-1}.\)
Derivative of inverse sine function	\(\frac{d}{dx}\ {\text{sin}}^{-1}x=\frac{1}{\sqrt{1-{(x)}^{2}}}\)
Derivative of inverse cosine function	\(\frac{d}{dx}\ {\text{cos}}^{-1}x=\frac{-1}{\sqrt{1-{(x)}^{2}}}\)
Derivative of inverse tangent function	\(\frac{d}{dx}\ {\text{tan}}^{-1}x=\frac{1}{1+{(x)}^{2}}\)
Derivative of inverse cotangent function	\(\frac{d}{dx}\ {\text{cot}}^{-1}x=\frac{-1}{1+{(x)}^{2}}\)
Derivative of inverse secant function	\(\frac{d}{dx}\ {\text{sec}}^{-1}x=\frac{1}{\|x\|\sqrt{{(x)}^{2}-1}}\)
Derivative of inverse cosecant function	\(\frac{d}{dx}\ {\text{csc}}^{-1}x=\frac{-1}{\|x\|\sqrt{{(x)}^{2}-1}}\)

Derivatives of Inverse Functions

For the following exercises, use the graph of \(y=f(x)\) to

sketch the graph of \(y={f}^{-1}(x),\) and
use part a. to estimate \({({f}^{-1})}^{'}(1).\)

For the following exercises, use the functions \(y=f(x)\) to find

\(\frac{df}{dx}\) at \(x=a\) and
\(x={f}^{-1}(y).\)
Then use part b. to find \(\frac{d{f}^{-1}}{dy}\) at \(y=f(a).\)

For each of the following functions, find \({({f}^{-1})}^{'}(a).\)

For each of the given functions \(y=f(x),\)

find the slope of the tangent line to its inverse function \({f}^{-1}\) at the indicated point \(P,\) and
find an equation of the tangent line to the graph of \({f}^{-1}\) at the indicated point.

For the following exercises, find \(\frac{dy}{dx}\) for the given function.

For the following exercises, use the given values to find \({({f}^{-1})}^{'}(a).\)

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Recall that when we write \(\ln(5)\), this means the power to which we raise \(e\) in order to get 5. That is, writing \(z = \ln(5)\) is saying the exact same thing as \(e^z = 5\).
Let \(f(x) = e^x\) and \(g(y)=\ln(y)\). For each of the expressions on the left below, drag the expression onto the equivalent simpler expression on the right.
Consider the graph of \(y = f(x)\) provided in Figure and use it to answer the following questions.
1. Use the provided graph to estimate the value of \(f'(1)\).
2. Sketch an approximate graph of \(y = f^{-1}(x)\). Label at least three distinct points on the graph that correspond to three points on the graph of \(f\).
3. Based on your work in (a), what is the value of \((f^{-1})'(-1)\)? Why?
Révèle la réponse
1. By plotting the tangent line as shown in the figure below, it appears that \(f'(1) \approx 2\).
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 \(y = x\), we see the following figure.
3. Because the original function passes through the point \((1,-1)\), the inverse function passes through \((-1,1)\). Moreover, since the tangent line to the original function at \((1,-1)\) has slope \(f'(1) \approx 2\), and the tangent line to the inverse function at \((-1,1)\) corresponds to the reflection of the original tangent line across the line \(y=x\), it follows that \((f^{-1})'(-1) \approx 1/2\). This can also be deduced from the derivative of an inverse function rule, whereby \((f^{-1})'(-1) = \frac{1}{f'(1)}\).
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. \(g(v) = \ln\left( \frac{\arctan(v)}{\arcsin(v) + v^2} \right)\)
Révèle la réponse
1. By the chain rule, \[\begin{aligned}\end{aligned}\], and by the sum and constant multiple rules, it follows that \[\begin{aligned}\end{aligned}\].
2. By the chain rule applied to the outermost function, \[\begin{aligned}\end{aligned}\]. Applying the chain rule again to evaluate \(\frac{d}{dz} \left( \ln(\arcsin(z)) \right)\), \[\begin{aligned}\end{aligned}\]. Finally, using the basic rule for \(\frac{d}{dz}\left(\arcsin(z)\right)\), \[\begin{aligned}\end{aligned}\].
3. First, by the product rule, \[\begin{aligned}\end{aligned}\]. Next, we recall that \(\arcsin^4(7t) = \left(\arcsin(7t)\right)^4\) and \(\arctan^2(3t) = \left( \arctan(3t) \right)^2\), and apply the chain rule to find the derivative of each of these functions to get \[\begin{aligned}\end{aligned}\]. Finally, applying the chain rule one more time to evaluate the remaining derivatives, \[\begin{aligned}\end{aligned}\].
4. By the chain rule, \[\begin{aligned}\end{aligned}\] Thus, by the quotient rule, \[\begin{aligned}\end{aligned}\]
Let \(f(x) = \frac{1}{4}x^3 + 4\).
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 \(g\).
3. Compute \(f'(x)\), \(g'(x)\), \(f'(2)\), and \(g'(6)\). What is the special relationship between \(f'(2)\) and \(g'(6)\)? Why?
Révèle la réponse
1. \(f\) is an invertible function because its graph passes the horizontal line test.
2. To find a formula for \(g = f^{-1}\), we take the equation \(y = \frac{1}{4}x^3 + 4\) and solve for \(x\) in terms of \(y\). Doing so, we find that \(y - 4 = \frac{1}{4}x^3\), so \(x^3 = 4(y-4)\), and thus \[\begin{aligned}\end{aligned}\]. Writing the inverse as a function of \(x\), we have found that \(f^{-1}(x) = g(x) = \sqrt[3]{4x-16}\).
3. First, \(f'(x) = \frac{3}{4}x^2\), and thus \(f'(2) = 3\). Next, \(g'(x) = \frac{1}{3}(4x-16)^{-2/3} \cdot 4\), so that \(g'(6) = \frac{1}{3} \cdot 8^{-2/3} \cdot 4 = \frac{1}{3} \cdot \frac{1}{4} \cdot 4 = \frac{1}{3}\). What is special about these two derivative values is that they happen to be reciprocals. Noting that \(f(2) = 3\) and \(g(3) = 2\), 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 \(g'(3) = \frac{1}{3}\).
Let \(h(x) = x + \sin(x)\).
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 \(h^{-1}\).
3. Observe that the point \((\frac{\pi}{2}, \frac{\pi}{2} + 1)\) lies on the graph of \(y = h(x)\). Determine the value of \((h^{-1})'(\frac{\pi}{2} + 1)\).
Révèle la réponse
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 \(y\).
3. By the rule for the derivative of an invertible function, we have \((h^{-1})'(\frac{\pi}{2} + 1) = \frac{1}{h'\left(\frac{\pi}{2} \right)}\). Thus, we first compute \(h'(x)\) and find that \(h'(x) = 1 + \cos(x)\). Since \(h'\left(\frac{\pi}{2} \right) = 1 + \cos\left(\frac{\pi}{2} \right) = 1\), it follows that \((h^{-1})'(\frac{\pi}{2} + 1) = \frac{1}{1} = 1\).
Consider the graph of \(y = f(x)\) provided in Figure and use it to answer the following questions.
1. Use the provided graph to estimate the value of \(f'(1)\).
2. Sketch an approximate graph of \(y = f^{-1}(x)\). Label at least three distinct points on the graph that correspond to three points on the graph of \(f\).
3. Based on your work in (a), what is the value of \((f^{-1})'(-1)\)? Why?
Révèle la réponse
1. By plotting the tangent line as shown in the figure below, it appears that \(f'(1) \approx 2\).
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 \(y = x\), we see the following figure.
3. Because the original function passes through the point \((1,-1)\), the inverse function passes through \((-1,1)\). Moreover, since the tangent line to the original function at \((1,-1)\) has slope \(f'(1) \approx 2\), and the tangent line to the inverse function at \((-1,1)\) corresponds to the reflection of the original tangent line across the line \(y=x\), it follows that \((f^{-1})'(-1) \approx 1/2\). This can also be deduced from the derivative of an inverse function rule, whereby \((f^{-1})'(-1) = \frac{1}{f'(1)}\).
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. \(g(v) = \ln\left( \frac{\arctan(v)}{\arcsin(v) + v^2} \right)\)
Révèle la réponse
1. By the chain rule, \[\begin{aligned}\end{aligned}\], and by the sum and constant multiple rules, it follows that \[\begin{aligned}\end{aligned}\].
2. By the chain rule applied to the outermost function, \[\begin{aligned}\end{aligned}\]. Applying the chain rule again to evaluate \(\frac{d}{dz} \left( \ln(\arcsin(z)) \right)\), \[\begin{aligned}\end{aligned}\]. Finally, using the basic rule for \(\frac{d}{dz}\left(\arcsin(z)\right)\), \[\begin{aligned}\end{aligned}\].
3. First, by the product rule, \[\begin{aligned}\end{aligned}\]. Next, we recall that \(\arcsin^4(7t) = \left(\arcsin(7t)\right)^4\) and \(\arctan^2(3t) = \left( \arctan(3t) \right)^2\), and apply the chain rule to find the derivative of each of these functions to get \[\begin{aligned}\end{aligned}\]. Finally, applying the chain rule one more time to evaluate the remaining derivatives, \[\begin{aligned}\end{aligned}\].
4. By the chain rule, \[\begin{aligned}\end{aligned}\] Thus, by the quotient rule, \[\begin{aligned}\end{aligned}\]
Let \(f(x) = \frac{1}{4}x^3 + 4\).
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 \(g\).
3. Compute \(f'(x)\), \(g'(x)\), \(f'(2)\), and \(g'(6)\). What is the special relationship between \(f'(2)\) and \(g'(6)\)? Why?
Révèle la réponse
1. \(f\) is an invertible function because its graph passes the horizontal line test.
2. To find a formula for \(g = f^{-1}\), we take the equation \(y = \frac{1}{4}x^3 + 4\) and solve for \(x\) in terms of \(y\). Doing so, we find that \(y - 4 = \frac{1}{4}x^3\), so \(x^3 = 4(y-4)\), and thus \[\begin{aligned}\end{aligned}\]. Writing the inverse as a function of \(x\), we have found that \(f^{-1}(x) = g(x) = \sqrt[3]{4x-16}\).
3. First, \(f'(x) = \frac{3}{4}x^2\), and thus \(f'(2) = 3\). Next, \(g'(x) = \frac{1}{3}(4x-16)^{-2/3} \cdot 4\), so that \(g'(6) = \frac{1}{3} \cdot 8^{-2/3} \cdot 4 = \frac{1}{3} \cdot \frac{1}{4} \cdot 4 = \frac{1}{3}\). What is special about these two derivative values is that they happen to be reciprocals. Noting that \(f(2) = 3\) and \(g(3) = 2\), 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 \(g'(3) = \frac{1}{3}\).
Let \(h(x) = x + \sin(x)\).
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 \(h^{-1}\).
3. Observe that the point \((\frac{\pi}{2}, \frac{\pi}{2} + 1)\) lies on the graph of \(y = h(x)\). Determine the value of \((h^{-1})'(\frac{\pi}{2} + 1)\).
Révèle la réponse
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 \(y\).
3. By the rule for the derivative of an invertible function, we have \((h^{-1})'(\frac{\pi}{2} + 1) = \frac{1}{h'\left(\frac{\pi}{2} \right)}\). Thus, we first compute \(h'(x)\) and find that \(h'(x) = 1 + \cos(x)\). Since \(h'\left(\frac{\pi}{2} \right) = 1 + \cos\left(\frac{\pi}{2} \right) = 1\), it follows that \((h^{-1})'(\frac{\pi}{2} + 1) = \frac{1}{1} = 1\).
Use the inverse function theorem to find the derivative of \(g(x)=\frac{x+2}{x}.\) Compare the resulting derivative to that obtained by differentiating the function directly.

Révèle la réponse

The inverse of \(g(x)=\frac{x+2}{x}\) is \(f(x)=\frac{2}{x-1}.\) Since \({g}^{'}(x)=\frac{1}{{f}^{'}(g(x))},\) begin by finding \({f}^{'}(x).\) Thus,
\[{f}^{'}(x)=\frac{-2}{{(x-1)}^{2}}\ \text{and}\ {f}^{'}(g(x))=\frac{-2}{{(g(x)-1)}^{2}}=\frac{-2}{{(\frac{x+2}{x}-1)}^{2}}=-\frac{{x}^{2}}{2}.\]
Finally,
\[{g}^{'}(x)=\frac{1}{{f}^{'}(g(x))}=-\frac{2}{{x}^{2}}.\]
We can verify that this is the correct derivative by applying the quotient rule to \(g(x)\) to obtain
\[{g}^{'}(x)=-\frac{2}{{x}^{2}}.\]
Use the inverse function theorem to find the derivative of \(g(x)=\frac{1}{x+2}.\) Compare the result obtained by differentiating \(g(x)\) directly.

Révèle la réponse

\({g}^{'}(x)=-\frac{1}{{(x+2)}^{2}}\)
Use the inverse function theorem to find the derivative of \(g(x)=\sqrt[3]{x}.\)

Révèle la réponse

The function \(g(x)=\sqrt[3]{x}\) is the inverse of the function \(f(x)={x}^{3}.\) Since \({g}^{'}(x)=\frac{1}{{f}^{'}(g(x))},\) begin by finding \({f}^{'}(x).\) Thus,
\[{f}^{'}(x)=3{x}^{2}\ \text{and}\ {f}^{'}(g(x))=3{(\sqrt[3]{x})}^{2}=3{x}^{2\text{/}3}.\]
Finally,
\[{g}^{'}(x)=\frac{1}{3{x}^{2\text{/}3}}=\frac{1}{3}{x}^{-2\text{/}3}.\]
Find the derivative of \(g(x)=\sqrt[5]{x}\) by applying the inverse function theorem.

Révèle la réponse

\(g(x)=\frac{1}{5}{x}^{\text{-}4\text{/}5}\)
Find an equation of the line tangent to the graph of \(y={x}^{2\text{/}3}\) at \(x=8.\)

Révèle la réponse

First find \(\frac{dy}{dx}\) and evaluate it at \(x=8.\) Since
\[\frac{dy}{dx}=\frac{2}{3}{x}^{-1\text{/}3}\ \text{and}\ \frac{dy}{dx}|\begin{array}{l} \\ {}_{x=8}\end{array}=\frac{1}{3}\]
the slope of the tangent line to the graph at \(x=8\) is \(\frac{1}{3}.\)
Substituting \(x=8\) into the original function, we obtain \(y=4.\) Thus, the tangent line passes through the point \((8,4).\) Substituting into the point-slope formula for a line, we obtain the tangent line
\[y=\frac{1}{3}x+\frac{4}{3}.\]
Find the derivative of \(s(t)=\sqrt{2t+1}.\)

Révèle la réponse

\({s}^{'}(t)={(2t+1)}^{\text{-}1\text{/}2}\)
Use the inverse function theorem to find the derivative of \(g(x)={\text{sin}}^{-1}x.\)

Révèle la réponse

Since for \(x\) in the interval \([-\frac{\pi }{2},\frac{\pi }{2}],f(x)=\text{sin}\ x\) is the inverse of \(g(x)={\text{sin}}^{-1}x,\) begin by finding \({f}^{'}(x).\) Since
\[{f}^{'}(x)=\text{cos}\ x\ \text{and}\ {f}^{'}(g(x))=\text{cos}\ ({\text{sin}}^{-1}x)=\sqrt{1-{x}^{2}},\]
we see that
\[{g}^{'}(x)=\frac{d}{dx}({\text{sin}}^{-1}x)=\frac{1}{{f}^{'}(g(x))}=\frac{1}{\sqrt{1-{x}^{2}}}.\]
Apply the chain rule to the formula derived in to find the derivative of \(h(x)={\text{sin}}^{-1}(g(x))\) and use this result to find the derivative of \(h(x)={\text{sin}}^{-1}(2{x}^{3}).\)

Révèle la réponse

Applying the chain rule to \(h(x)={\text{sin}}^{-1}(g(x)),\) we have
\[{h}^{'}(x)=\frac{1}{\sqrt{1-{(g(x))}^{2}}}{g}^{'}(x).\]
Now let \(g(x)=2{x}^{3},\) so \({g}^{'}(x)=6{x}^{2}.\) Substituting into the previous result, we obtain
\[\begin{array}{ll}{h}^{'}(x) & =\frac{1}{\sqrt{1-4{x}^{6}}}\cdot 6{x}^{2} \\ & =\frac{6{x}^{2}}{\sqrt{1-4{x}^{6}}}.\end{array}\]
Use the inverse function theorem to find the derivative of \(g(x)={\text{tan}}^{-1}x.\)

Révèle la réponse

\({g}^{'}(x)=\frac{1}{1+{x}^{2}}\)
Find the derivative of \(f(x)={\text{tan}}^{-1}({x}^{2}).\)

Révèle la réponse

Let \(g(x)={x}^{2},\) so \({g}^{'}(x)=2x.\) Substituting into , we obtain
\[{f}^{'}(x)=\frac{1}{1+{({x}^{2})}^{2}}\cdot (2x).\]
Simplifying, we have
\[{f}^{'}(x)=\frac{2x}{1+{x}^{4}}.\]
Find the derivative of \(h(x)={x}^{2}{\text{sin}}^{-1}x.\)

Révèle la réponse

By applying the product rule, we have
\[{h}^{'}(x)=2x\ {\text{sin}}^{-1}x+\frac{1}{\sqrt{1-{x}^{2}}}\cdot {x}^{2}.\]
Find the derivative of \(h(x)={\text{cos}}^{-1}(3x-1).\)

Révèle la réponse

\({h}^{'}(x)=\frac{-3}{\sqrt{6x-9{x}^{2}}}\)
The position of a particle at time \(t\) is given by \(s(t)={\text{tan}}^{-1}(\frac{1}{t})\) for \(t\ge \frac{1}{2}.\) Find the velocity of the particle at time \(t=1.\)

Révèle la réponse

Begin by differentiating \(s(t)\) in order to find \(v(t).\) Thus,
\[v(t)={s}^{'}(t)=\frac{1}{1+{(\frac{1}{t})}^{2}}\cdot \frac{-1}{{t}^{2}}.\]
Simplifying, we have
\[v(t)=-\frac{1}{{t}^{2}+1}.\]
Thus, \(v(1)=-\frac{1}{2}.\)
Find an equation of the line tangent to the graph of \(f(x)={\text{sin}}^{-1}x\) at \(x=0.\)

Révèle la réponse

\(y=x\)
\(f(x)=6x-1,x=-2\)
\(f(x)=2{x}^{3}-3,x=1\)

Révèle la réponse

a. 6, b. \(x={f}^{-1}(y)={(\frac{y+3}{2})}^{1\text{/}3},\) c. \(\frac{1}{6}\)
\(f(x)=9-{x}^{2},0\le x\le 3,x=2\)
\(f(x)=\text{sin}\ x,x=0\)

Révèle la réponse

a. \(1,\) b. \(x={f}^{-1}(y)={\text{sin}}^{-1}y,\) c. \(1\)
\(f(x)={x}^{2}+3x+2,x\ge -\frac{3}{2},a=2\)
\(f(x)={x}^{3}+2x+3,a=0\)

Révèle la réponse

\(\frac{1}{5}\)
\(f(x)=x+\sqrt{x},a=2\)
\(f(x)=x-\frac{2}{x},x<0,a=1\)

Révèle la réponse

\(\frac{1}{3}\)
\(f(x)=x+\text{sin}\ x,a=0\)
\(f(x)=\text{tan}\ x+3{x}^{2},a=0;0

Révèle la réponse

\(1\)
\(f(x)=\frac{4}{1+{x}^{2}},P(2,1)\)
\(f(x)=\sqrt{x-4},P(2,8)\)

Révèle la réponse

a. \(4,\) b. \(y=4x\)
\(f(x)={({x}^{3}+1)}^{4},P(16,1)\)
\(f(x)=\text{-}{x}^{3}-x+2,P(-8,2)\)

Révèle la réponse

a. \(-\frac{1}{13},\) b. \(y=-\frac{1}{13}x+\frac{18}{13}\)
\(f(x)={x}^{5}+3{x}^{3}-4x-8,P(-8,1)\)
\(y={\text{sin}}^{-1}({x}^{2})\)

Révèle la réponse

\(\frac{2x}{\sqrt{1-{x}^{4}}}\)
\(y={\text{cos}}^{-1}(\sqrt{x})\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\arcsin,\ \sin^{-1}

inverse sine
The angle whose sine is the given value (and likewise arccos, arctan).

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Derivatives of Inverse Functions

Calculate the derivative of an inverse function.
Recognize the derivatives of the standard inverse trigonometric functions.
The inverse function theorem allows us to compute derivatives of inverse functions without using the limit definition of the derivative.
We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions.
sketch the graph of
use part a. to estimate
Then use part b. to find
find the slope of the tangent line to its inverse function

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0), OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Plus en Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests