2.6: Derivatives of Inverse Functions

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

$$f^{-1}(y) = f^{-1}(f(x)) = x\text{.}$$

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 $y = x\text{.}$

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 $y\text{.}$

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

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

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

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

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 $f^{-1}(x) = \ln(x)\text{.}$

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. $\displaystyle h(x) = x^2\ln(x)$
2. $\displaystyle p(t) = \frac{\ln(t)}{e^t + 1}$
3. $\displaystyle s(y) = \ln(\cos(y) + 2)$
4. $\displaystyle z(x) = \tan(\ln(x))$
5. $\displaystyle m(z) = \ln(\ln(z))$

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

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

Differentiating both sides of the previous equation, we have

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

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

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.

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

Inverse sine

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

$$\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}\text{.}$$

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 $r'(x)\text{.}$

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. $\displaystyle \displaystyle f(x) = x^3 \arctan(x) + e^x \ln(x)$
2. $\displaystyle \displaystyle p(t) = 2^{t\arcsin(t)}$
3. $\displaystyle \displaystyle h(z) = (\arcsin(5z) + \arctan(4-z))^{27}$
4. $\displaystyle \displaystyle s(y) = \cot(\arctan(y))$
5. $\displaystyle \displaystyle m(v) = \ln(\sin^2(v)+1)$
6. $\displaystyle \displaystyle g(w) = \arctan\left( \frac{\ln(w)}{1+w^2} \right)$

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

and by the chain rule,

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 $(g(x),f(g(x))) = (g(x),x)\text{.}$

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

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

$$g'(b) = \frac{1}{f'(a)}\text{.}$$

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

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 $g'\text{,}$ $g'(x) = \frac{1}{f'(g(x))}\text{.}$