2.1: Inverse Functions

The derivatives calculated in the previous chapter were mostly for polynomials and a few trigonometric functions. This chapter will show how to find the derivatives of other types of functions, beginning in this section with inverse functions. The idea here is that if a function is differentiable and has an inverse then that inverse function is also differentiable.

Recall that a function is a rule that assigns a single object $y$ from one set (the range) to each object $x$ from another set (the domain). That rule can be written as $y = f(x)$, where $f$ is the function (see Figure [fig:function]). There is a simple vertical rule for determining whether a rule $y=f(x)$ is a function: $f$ is a function if and only if every vertical line intersects the graph of $y=f(x)$ in the $xy$-coordinate plane at most once (see Figure [fig:verticalrule]).

Recall that a function $f$ is one-to-one (often written as $1-1$) if it assigns distinct values of $y$ to distinct values of $x$. In other words, if $x_1 \ne x_2$ then $f(x_1 ) \ne f(x_2 )$. Equivalently, $f$ is one-to-one if $f(x_1 ) = f(x_2 )$ implies $x_1 = x_2$. There is a simple horizontal rule for determining whether a function $y=f(x)$ is one-to-one: $f$ is one-to-one if and only if every horizontal line intersects the graph of $y=f(x)$ in the $xy$-coordinate plane at most once (see Figure [fig:horizontalrule]).

If a function $f$ is one-to-one on its domain, then $f$ has an inverse function, denoted by $f^{-1}$, such that $y=f(x)$ if and only if $f^{-1}(y) = x$. The domain of $f^{-1}$ is the range of $f$.

The basic idea is that $f^{-1}$ “undoes” what $f$ does, and vice versa. In other words,

$$\begin{aligned} {3} f^{-1}(f(x)) ~&=~ x \quad&&\text{for all $x$ in the domain of $f$, and}\\ f(f^{-1}(y)) ~&=~ y \quad&&\text{for all $y$ in the range of $f$.}\end{aligned} \nonumber$$

Intuitively it is clear that a function is one-to-one (and hence invertible) when it is either strictly increasing or strictly decreasing; if the function, say, increases and then decreases (as in Figure [fig:horizontalrule](b)) then the horizontal rule would be violated around that “turning point.” For differentiable functions a positive derivative means the function is increasing, while a negative derivative means the function is decreasing (this will be proved in Chapter 3).

However, a function can still be one-to-one even if its derivative is zero only at isolated points (i.e. not identically zero over an entire interval of points) and either positive everywhere else or negative everywhere else. For example, the function $f(x) = x^3$ has derivative $f'(x) = 3x^2$, which is zero only at the isolated point $x = 0$ and positive for all other values of $x$. Clearly $f$ is one-to-one over the set of all real numbers (why?) and hence it has an inverse function $x = f^{-1}(y) = \sqrt[3]{y}$ defined for all real numbers $y$ (i.e. the range of $f$).

Thus, having a derivative that is either always positive or always negative is sufficient for a function to be one-to-one but not necessary. Having a nonzero derivative is necessary, though, for the inverse function to be differentiable.

In algebra you learned that $\frac{a}{b} = \frac{1}{\frac{b}{a}}$ for all real numbers $a \ne 0$ and $b \ne 0$ (and hence $\frac{b}{a} \ne 0$). The same holds true for the infinitesimals $\dy$ and $\dx$ (nonzero by definition) since they can be treated like numbers, which immediately yields a formula for the derivative of an inverse function:

The inverse of a function would still exist at a point where $\dydx = 0$ but it would not be differentiable there, since its derivative would be the undefined quantity $\frac{1}{0}$.

Since $y$ is a function of $x$, $\dydx$ will be in terms of $x$ and hence $\frac{1}{\dydx}$ will be in terms of $x$. However, since (by invertibility) $x$ is a function of $y$, $\dxdy$ would normally be in terms of $y$, not $x$, so that the two sides of the equation $\dxdy = \frac{1}{\dydx}$ are not in the same terms! One way to handle this discrepancy is to use the formula $y = f(x)$ to solve for $x$ in terms of $y$ then substitute that expression into $\dydx$, so that $\dxdy = \frac{1}{\dydx}$ is now in terms of $y$. That might not always be possible, however (e.g. try solving for $x$ in the formula $y = x \sin x$).

Example $\PageIndex{1}$: invderiv

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Solution

Find the inverse $f^{-1}$ of the function $f(x) = x^3$ then find the derivative of $f^{-1}$.

Solution: The function $y = f(x) = x^3$ is one-to-one over the set of all real numbers (why?) so it has an inverse function $x = f^{-1}(y)$ defined for all real numbers, namely $x = f^{-1}(y) = \sqrt[3]{y}$.

The derivative of $f^{-1}$ is

\[\begin{aligned} \dxdy ~&=~ \frac{1}{\dydx} ~=~ \frac{1}{3x^2} \quad\text{, which is in terms of $x$, so putting it in terms of $y$ yields}\

$$6pt] &=~ \frac{1}{3 \left(\sqrt[3]{y}\right)^2} ~=~ \frac{1}{3 y^{2/3}} \end{aligned} \nonumber$$

which agrees with the derivative obtained by differentiating $x = \sqrt[3]{y}$ directly. Note that this derivative is defined for all $y$ except $y = 0$, which occurs when $x = \sqrt[3]{0} = 0$, i.e. at the point $(x,y) = (0,0)$.

Functions are often expressed in terms of $x$, so it is common to see an inverse function also expressed in terms of $x$: writing the inverse of $f(x) = x^3$ as $f^{-1}(x) = \sqrt[3]{x}$ (not as $f^{-1}(y) = \sqrt[3]{y}$)), as confusing as that might be. In that case, the idea is to switch the roles of $x$ and $y$ in the original function $y = f(x)$, making it $x = f(y)$, and then write $y = f^{-1}(x)$ and use

$$\dydx ~=~ \frac{1}{\dxdy} \nonumber$$

to put the derivative of $f^{-1}$ in terms of $x$, following the same procedure mentioned earlier.

Example $\PageIndex{1}$: invderivalt

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Solution

Find the inverse $f^{-1}$ of the function $f(x) = x^3$ then find the derivative of $f^{-1}$.

Solution: Rewrite $y = f(x) = x^3$ as $x = f(y) = y^3$, so that its inverse function $y = f^{-1}(x)= \sqrt[3]{x}$ has derivative

\[\begin{aligned} \dydx ~&=~ \frac{1}{\dxdy} ~=~ \frac{1}{3y^2} \quad\text{, which is in terms of $y$, so putting it in terms of $x$ yields}\

$$6pt] &=~ \frac{1}{3 \left(\sqrt[3]{x}\right)^2} ~=~ \frac{1}{3 x^{2/3}} \end{aligned} \nonumber$$

which agrees with the derivative obtained by differentiating $y = \sqrt[3]{x}$ directly.

To obtain a formula in prime notation for the derivative of an inverse function, notice that for all $x$ in the domain of an invertible differentiable function $f$,

$$f^{-1}(f(x)) ~=~ x \quad\Rightarrow\quad \ddx\,\left(f^{-1}(f(x))\right) ~=~ \ddx\,(x) \quad\Rightarrow\quad \left(f^{-1}\right)'(f(x)) \;\cdot\; f'(x) ~=~ 1 \nonumber$$

by the Chain Rule, and hence:

[sec2dot1]

For Exercises 1-8, show that the given function $y = f(x)$ is one-to-one over the given interval, then find the formulas for the inverse function $f^{-1}$ and its derivative. Use Example

Example $\PageIndex{1}$: invderivalt

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Solution

as a guide, including putting $f^{-1}$ and its derivative in terms of $x$.

2

$f(x) = x$, for all $x$

$f(x) = 3x$, for all $x$

2

$f(x) = x^2$, for all $x \ge 0$

$f(x) = \sqrt{x}$, for all $x \ge 0$

2

$f(x) = \frac{1}{x}$, for all $x > 0$

$f(x) = \frac{1}{x}$, for all $x < 0$

2

$f(x) = \frac{1}{x^2}$, for all $x > 0$

$f(x) = x^5$, for all $x\vphantom{\frac{1}{x^2}}$

The unit circle $x^2 + y^2 = 1$ does not define $y$ as a single function of $x$, since $y = \pm \sqrt{1 - x^2}$ defines two separate functions. But the part of the unit circle in the first quadrant, i.e. for $0 \le x \le 1$ and $0 \le y \le 1$, does define $y = f(x) = \sqrt{1 - x^2}$ as a single function of $x$ that is one-to-one on the interval $\ival{0}{1}$. Find the formulas for its inverse function $f^{-1}$ and its derivative. [[1.]]

Show that if $f$ is differentiable and invertible, and if $f^{-1}$ is twice-differentiable, then

$$\left(f^{-1}\right)''(x) ~=~ -\frac{f''(f^{-1}(x))}{\left(f'(f^{-1}(x))\right)^3} ~. \nonumber$$

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1. The arc notation $\arcsin\,x$, $\arccos\,x$, $\arctan\,x$, $\arccsc\,x$, $\arcsec\,x$, $\arccot\,x$ is often used in place of $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$, $\csc^{-1}x$, $\sec^{-1}x$, $\cot^{-1}x$, respectively.↩
2. For the general case see pp.61-63 in Franklin, P., A Treatise on Advanced Calculus, New York: Dover Publications, Inc., 1964.↩
3. For example, to see how square roots and cube roots are calculated see Chapter 2 in Fike, C.T., Computer Evaluation of Mathematical Functions, Englewood Cliffs, New Jersey: Prentice-Hall, Inc., 1968.↩
4. It will be shown in Chapter 9 that this limit does in fact exist.↩
5. This is admittedly a “hand waving” argument. In Chapter 3 a more exact method will be discussed for proving limits like this one.↩
6. This text almost used $\log\,x$ as well, prevented only by the desire for compatibility with other mathematics texts.↩
7. Note that when using the formula $\ln\,\left(\frac{a}{b}\right) = \ln\,a - \ln\,b$ in numerical computations—especially on hand-held calculators—it is preferable to use the left side of the equation, i.e. $\ln\,\left(\frac{a}{b}\right)$, since the right side $\ln\,a - \ln\,b$ is vulnerable to the problem of subtractive cancellation, which can give an incorrect answer of 0 if $a$ and $b$ are nearly equal. For a discussion of subtractive cancellation see § 1.3 in Henrici, P., Essentials of Numerical Analysis, with Pocket Calculator Demonstrations, New York: John Wiley & Sons, Inc., 1982.↩
8. Binary notation leads to the joke “There are 10 kinds of people in the world: those who understand binary and those who do not.”↩