2.4: Inverse Functions

Two real functions $f$ and $g$ are called inverse functions if the two equations $$y = f(x), \quad x = g(y) \nonumber$$

have the same graphs in the $(x, y)$ plane. That is, a point $(x, y)$ is on the curve $y = f(x)$ if, and only if, it is on the curve $x = g(y)$. (In general, the graph of the equation $x = g(y)$ is different from the graph of $y = g(x)$, but is the same as the graph of $y =f(x)$; see Figure $\PageIndex{1}$.)

Figure $\PageIndex{1}$: Inverse functions.

For example, the function $y = x^{2}$, $x \geq 0$ has the inverse function $x = \sqrt{y}$; the function $y = x^{3}$ has the inverse function $x = \sqrt[3]{y}$.

If we think off as a black box operating on an input $x$ to produce an output $f(x)$, the inverse function $g$ is a black box operating on the output $f(x)$ to undo the
work of $f$ and produce the original input $x$ (see Figure $\PageIndex{2}$).

Figure $\PageIndex{2}$: The inverse function applied to a function produces the original function input.

Many functions, such as $y = x^{2}$, do not have inverse functions: In Figure $\PageIndex{3}$, we see that $x$ is not a function of $y$ because at $y = 1$, $x$ has the two values $x = 1$ and $x = -1$.

Often one can tell whether a function $f$ has an inverse by looking at its graph. If there is a horizontal line $y = c$ which cuts the graph at more than one point, the function $f$ has no inverse. (See Figure $\PageIndex{3}$.) If no horizontal line cuts the graph at more than one point, then $f$ has an inverse function $g$. Using this rule, we can see in Figure $\PageIndex{4}$ that the functions $y = |x|$ and $y = \sqrt{1 - x^{2}}$ do not have inverses.

Figure $\PageIndex{3}$: Horizontal line test for inverse function.

Figure $\PageIndex{4}$: No inverse functions.

Table $\PageIndex{1}$shows some familiar functions which do have inverses. Note that in each case, $\dfrac{dx}{dy} = \dfrac{1}{dy/dx}$.

Table $\PageIndex{1}$. Inverses of some common functions.

function $y = f(x)$	$\dfrac{dy}{dx}$	inverse function $x = g(y)$	$\dfrac{dx}{dy} = \dfrac{1}{dy/dx}$
$\quad y = x+c$	$1$	$\quad x = y-c$	$1$
$\quad y = kx$	$k$	$\quad x = y/k$	$1/k$
$\quad y = x^{2}, \quad x \geq 0$	$2x$	$\quad x = \sqrt{y}$	$\dfrac{1}{2 \sqrt{y}} = \dfrac{1}{2x}$
$\quad y = x^{2}, \quad x \leq 0$	$2x$	$\quad x = -\sqrt{y}$	$-\dfrac{1}{2 \sqrt{y}} = \dfrac{1}{2x}$
$\quad y = 1/x$	$- \dfrac{1}{x^{2}}$	$\quad x = 1/y$	$-\dfrac{1}{y^{2}} = -x^{2}$

Suppose the $(x, y)$ plane is flipped over about the diagonal line $y = x$. This will make the $x$- and $y$-axes change places, forming the $(y, x)$ plane. If $f$ has an inverse function $g$, the graph of the function $y = f(x)$ will become the graph of the inverse function $x = g(y)$ in the $(y, x)$ plane, as shown in Figure $\PageIndex{5}$.

Figure $\PageIndex{5}$: Graphs of some common functions and their inverses.

The following rule shows that the derivatives of inverse functions are always reciprocals of each other.

Inverse Function Rule

Suppose $f$ and $g$ are inverse functions, so that the two equations $$y = f(x) \text{ and } x= g(y) \nonumber$$ have the same graphs. If both derivatives $f'(x)$ and $g'(y)$ exist and are nonzero, then $$f'(x) = \frac{1}{g'(y)}. \nonumber$$

That is, $$\frac{dy}{dx} = \frac{1}{dx/dy}. \nonumber$$

Proof

Let $\Delta x$ be a nonzero infinitesimal and let $\Delta y$ be the corresponding change in $y$. Then $\Delta y$ is also infinitesimal because $f'(x)$ exists and is nonzero because $f(x)$ has an inverse function. By the rules for standard parts, $$\begin{align*} f'(x) \cdot g'(y) &= st \left(\frac{\Delta y}{\Delta x}\right) \cdot st \left(\frac{\Delta x}{\Delta y}\right) \\ &= st \left(\frac{\Delta y}{\Delta x} \cdot \frac{\Delta x}{\Delta y}\right) = st(1) = 1. \end{align*}$$

Therefore, $$f'(x) = \frac{1}{g'(y)}. \nonumber$$

The formula $$\frac{dy}{dx} = \frac{1}{dx/dy}. \nonumber$$ in the Inverse Function Rule is not as trivial as it looks. A more complete statement is $$\frac{dy}{dx} \text{ computed with } x \text{ the independent variable} = \frac{1}{dx/dy} \text{ computed with } y \text{ the independent variable.} \nonumber$$

Sometimes it is easier to compute $dx/dy$ than $dy/dx$, and in such cases the Inverse Function Rule is a useful method.

Example $\PageIndex{1}$

Find $dy/dx$ where $x = 1 + y^{-3}$.

Before solving the problem we note that \[y = \frac{1}{\sqrt[3]{x-1}, \nonumber\]

so $x$ and $y$ are inverse functions of each other. We want to find $$\frac{dy}{dx} = \frac{d \left(1/\sqrt[3]{x-1}\right)}{dx} \nonumber$$

with $x$ the independent variable. This looks hard, but it is easy to compute $$\frac{dx}{dy} = \frac{d \left(1 + y^{-3}\right)}{dy} \nonumber$$ with $y$ the independent variable.

Solution

$\dfrac{dx}{dy} = -3y^{-4},$

$\dfrac{dy}{dx} = \dfrac{1}{-3y^{-4}} = \frac{1}{3} y^{4}.$

We can write $dy/dx$ in terms of $x$ by substituting, $$\frac{dy}{dx} = -\frac{1}{3} (x-1)^{-4/3}. \nonumber$$

Example $\PageIndex{2}$

Find $dy/dx$ where $x = y^{5} + y^{3} + y$. Compute $dy/dx$ at the point $(1, 3)$.

Solution

Although we cannot solve the equation explicitly for $y$ as a function of $x$, we can see from the graph in Figure $\PageIndex{6}$ that there is an inverse function $y = f(x)$. 

Figure $\PageIndex{6}$: Copy and Paste Caption here. (Copyright; author via source)

By the Inverse Function Rule, $$\begin{align*} \frac{dx}{dy} &= 5y^{4} + 3y^{2} + 1, \\[4pt] \frac{dy}{dx} &= \frac{1}{5y^{4} + 3y^{2} + 1}. \end{align*}$$

This time we must leave the answer in terms of $y$. At the point $(1, 3)$, we substitute $3$ for $y$ and get $dy/dx = 1/9$.

For $y \geq 0$, the function $x = y^{n}$ has the inverse function $y = x^{1/n}$. In the next theorem, we use the Inverse Function Rule to find a new derivative, that of $y = x^{1/n}$.

Theorem $\PageIndex{1}$

If $n$ is a positive integer and $y = x^{1/n}$, then $$\frac{dy}{dx} = \frac{1}{n} x^{(1/n)-1}. \nonumber$$

Remember that $y = x^{1/n}$ is defined for all $x$ if $n$ is odd and for $x > 0$ if $n$ is even. The derivative $\dfrac{1}{n} x^{(1/n)-1}$ is defined for $x \neq 0$ if $n$ is odd and for $x > 0$ if $n$ is even. 

If we are willing to assume that $dy/dx$ exists, then we can quickly find $dy/dx$ by the Inverse Function Rule. $$\begin{align*} x &= y^{n}, \\ \frac{dx}{dy} &= ny^{n-1}, \\ \frac{dy}{dx} &= \frac{1}{dx/dy} = \frac{1}{ny^{n-1}} = \frac{1}{n} y^{1-n} \\ &= \frac{1}{n} \left(x^{1/n}\right)^{1-n} = \frac{1}{n} x^{(1-n)/n} = \frac{1}{n} x^{(1/n)-1}. \end{align*}$$

Here is a longer but complete proof which shows that $dy/dx$ exists and computes its value.

Proof

Let $x \neq 0$ and let $\Delta x$ be nonzero infinitesimal. We first show that $$\Delta y = (x + \Delta x)^{1/n} - x^{1/n} \nonumber$$ is a nonzero infinitesimal. $\Delta y \neq 0$ because $x + \Delta x \neq x$. The standard part of $\Delta y$ is $$\begin{align*} st (\Delta y) &= st \left( (x + \Delta x)^{1/n}\right) - st \left(x^{1/n}\right) \\ &= x^{1/n} - x^{1/n} = 0. \end{align*}$$ Therefore $\Delta y$ is nonzero infinitesimal.

Now $$\begin{align*} x &= y^{n}, \\[4pt] \frac{dx}{dy} &= ny^{n-1}, \\[4pt] \frac{\Delta x}{\Delta y} &\approx ny^{n-1}. \end{align*}$$

Therefore $$\begin{align*} \frac{\Delta y}{\Delta x} &\approx \frac{1}{ny^{n-1}} = \frac{1}{n} x^{(1/n) - 1}, \\[4pt] \frac{dy}{dx} &= \frac{1}{n} x^{(1/n) - 1}. \end{align*}$$  

Figure $\PageIndex{7}$: Graphs of $y = x^{1/3}$ and $y = x^{1/4}$.

Figure $\PageIndex{7}$ shows the graphs of $y = x^{1/3}$ and $y= x^{1/4}$. At $x = 0$, the curves are vertical and have no slope.

Example $\PageIndex{3}$

Find the derivatives of $y = x^{1/n}$ for $n = 2, 3, 4$.

Solution

$$\begin{align*} \frac{d \left(x^{1/2}\right)}{dx} &= \frac{1}{2} x^{-1/2}, \quad x > 0, \\ \frac{d \left(x^{1/3}\right)}{dx} &= \frac{1}{3} x^{-2/3}, \quad x \neq 0, \\ \frac{d \left(x^{1/4}\right)}{dx} &= \frac{1}{4} x^{-3/4}, \quad x > 0. \end{align*}$$

Using Theorem $\PageIndex{1}$ we can show that the Power Rule holds when the exponent is any rational number.

Power Rule for Rational Exponents

Let $y = x^{r}$ where $r$ is a rational number. Then whenever $x > 0$, $$\frac{dy}{dx} = rx^{r - 1}. \nonumber$$

Proof

Let $r = m/n$ where $m$ and $n$ are integers, $n > 0$. Let $$u = x^{1/n}, \quad y = u^{m}. \nonumber$$

Then $$\frac{du}{dx} = \frac{1}{n} x^{(1/n) - 1} \nonumber$$

and $$\begin{align*} \frac{dy}{dx} &= mu^{m-1} \frac{du}{dx} \\[4pt] &= m \left(x^{1/n}\right)^{m-1} \left(\frac{1}{n} x^{(1/n) - 1}\right) \\[4pt] &= \frac{m}{n} x^{(m/n) - 1} = rx^{r-1}. \end{align*}$$

Example $\PageIndex{4}$

Find \(dy/dx) where $$y = x^{-3/7}. \nonumber$$

Solution

$$\frac{dy}{dx} = -\frac{3}{7} x^{(-3/7) - 1} = -\frac{3}{7} x^{-10/7}. \nonumber$$

Example $\PageIndex{5}$

Find \(dy/dx) where $$y = \frac{1}{2 + x^{3/2}}. \nonumber$$

Solution

Let $$u = 2 + x^{3/2}, \quad y = u^{-1}. \nonumber$$

Then $$\begin{align*} \frac{du}{dx} &= \frac{3}{2} x^{1/2}, \\[4pt] \frac{dy}{dx} &= -u^{-2} \frac{du}{dx} \\[4pt] -u^{-2} \left(\frac{3}{2} x^{1/2}\right) = -\frac{3}{2} \frac{x^{1/2}}{\left(2 + x^{3/2}\right)^{2}}. \end{align*}$$

Problems for Section 2.4

In Problems 1-16, find $dy/dx$.

1.	$x= 3y^{3} + 2y$	2.	$x = y^{2} + 1$
3.	$x = 1 - 2y^{2}, \quad y > 0$	4.	$x = 2y^{5} + y^{3} + 4$
5.	$x = \left(y^{2} + 2\right)^{-1}, \quad y > 0$	6.	$y = 1 + \sqrt{x}$
7.	$y = x^{4/3}$	8.	$y = \sqrt{2x}$
9.	$y = \left(\sqrt{x} + 1\right) \left(\sqrt{x} - 1\right)$	10.	$y = \left(2x^{1/3} + 1\right)^{3}$
11.	$y = 1 + 2x^{1/3} + 4x^{2/3} + 6x$	12.	$y = x^{-1/4} + 3x^{-3/4}$
13.	$y = \left(x^{5/3} - x\right)^{-2}$	14.	$x = y + 2 \sqrt{y}$
15.	$x = 3y^{1/3} + 2y, \quad y > 0$	16.	$x = 1/\left(1 + \sqrt{y}\right)$

In Problems 17-25, find the inverse function $y$ and its derivative $dy/dx$ as functions of $x$.

17.	$x = ky + c, \quad k \neq 0$	18.	$x = y^{3} + 1$
19.	$x = 2y^{2} + 1, \quad y \geq 0$	20.	$x = 2y^{2} + 1, \quad y \leq 0$
21.	$x = y^{4} - 3, \quad y \geq 0$	22.	$x = y^{2} + 3y - 1, \quad y \geq -\frac{3}{2}$
23.	$x = y^{4} + y^{2} + 1, \quad y \geq 0$	24.	$x = 1/y^{2} + 1/y - 1, \quad y > 0$
25.	$x = \sqrt{y} + 2y, \quad y > 0$
$\square$ 26.	Show that no second degree polynomial $x = ay^{2} + by + c$ has an inverse function.
$\square$ 27.	Show that $x = ay^{2} + by + c, y \geq -b/2a$, has an inverse function. What does its graph look like?
$\square$ 28.	Prove that a function $y = f(x)$ has an inverse function if and only if whenever $x_{1} \neq x_{2}$, $f \left(x_{1}\right) \neq f \left(x_{2}\right)$.