Derivative of an Inverse Function

Why the Derivative of an Inverse Is a Reciprocal

The derivative measures how a function changes its output when its input changes. An inverse function reverses that relationship: it takes a change in output and recovers the corresponding change in input. This suggests that, when the rates of change are nonzero, the inverse rate should be the reciprocal of the original one.

There is a point to settle before taking a derivative of the inverse. The value $f(a)$ must be an interior point of the image, and inverse inputs near $f(a)$ must correspond to original inputs near $a$. Neither fact follows just from writing down a reciprocal difference quotient. We will establish both facts from continuity, injectivity, and a nonzero derivative.

Definition (Inverse Function): If $f:I\to\mathbb{R}$ is injective, its inverse $f^{-1}:f(I)\to I$ is defined by the condition $f^{-1}(y)=x$ exactly when $f(x)=y$. In particular, $f^{-1}(f(x))=x$ for every $x\in I$.

The formula we will prove is local: it concerns the inverse at $b=f(a)$, and its hypotheses are imposed on $f$ near $a$. The assumption that $f$ is injective on its interval ensures that this local inverse agrees with the inverse $f^{-1}$ on the full image $f(I)$.

Opposite-Side Values and a Local Inverse

Let $L=f'(a)\ne0$. By the definition of the derivative, for all sufficiently small nonzero $h$, the difference quotient is close to $L$. If $L>0$, that quotient is positive, so $f(a+h)-f(a)$ has the same sign as $h$. If $L<0$, the signs are opposite. Thus the values immediately to the left and right of $a$ lie on opposite sides of $b=f(a)$.

This observation, together with the Intermediate Value Theorem, provides an interval around $b$ contained in $f(I)$. It also gives the control needed to show that inverse inputs approach $a$ as their outputs approach $b$. We record both facts in one lemma.

Lemma (Local Image and Continuity of the Inverse): Let $I$ be an interval, let $a$ be an interior point of $I$, and suppose $f:I\to\mathbb{R}$ is continuous and injective. If $f$ is differentiable at $a$ and $f'(a)\ne0$, then $b=f(a)$ is an interior point of $f(I)$, and $f^{-1}$ is continuous at $b$.

Proof. Write $L=f'(a)$, so $L\ne0$. Since $a$ is interior to $I$, there is a positive radius on which $a+h\in I$. Differentiability at $a$ gives a possibly smaller radius $\rho>0$ such that, whenever $0<|h|<\rho$,

\left|\frac{f(a+h)-f(a)}{h}-L\right|<\frac{|L|}{2}.

The quotient in this inequality has the same sign as $L$: if $L>0$, it is greater than $L/2>0$; if $L<0$, it is less than $L/2<0$. Choose $r$ with $0<r<\rho$. Applying this sign conclusion at $h=-r$ and $h=r$ shows that $f(a-r)-b$ and $f(a+r)-b$ have opposite signs. In particular, neither endpoint value equals $b$.

Set

\alpha=\min\bigl(|f(a-r)-b|,\ |f(a+r)-b|\bigr)>0.

If $|y-b|<\alpha$, then $y$ lies strictly between $f(a-r)$ and $f(a+r)$, because those endpoint values are on opposite sides of $b$ and $y$ is closer to $b$ than either endpoint. The Intermediate Value Theorem applied to $f$ on $[a-r,a+r]$ gives a point $x\in(a-r,a+r)$ with $f(x)=y$. Therefore $(b-\alpha,b+\alpha)\subseteq f(I)$, so $b$ is interior to $f(I)$.

It remains to prove continuity of the inverse at $b$. Let $\varepsilon>0$. Choose $r$ again, small enough that $0<r<\varepsilon$, both $a-r$ and $a+r$ lie in $I$, and the derivative sign argument above applies. Define $\alpha$ from these endpoints as before. For any $y\in f(I)$ with $|y-b|<\alpha$, let $x=f^{-1}(y)$. If $y=b$, injectivity gives $x=a$. We show that $x$ cannot lie outside $(a-r,a+r)$.

Suppose first that $x\geq a+r$. Put $c=f(a+r)$, so $c\ne b$. If $y-b$ and $c-b$ have the same sign, then $y$ lies strictly between $b=f(a)$ and $c=f(a+r)$. The Intermediate Value Theorem on $[a,a+r]$ gives $z\in(a,a+r)$ with $f(z)=y=f(x)$, contradicting injectivity. If $y-b$ and $c-b$ have opposite signs, then $b$ lies strictly between $c=f(a+r)$ and $y=f(x)$. Here $x>a+r$, since $x=a+r$ would imply $y=c$, contrary to $|y-b|<|c-b|$. The Intermediate Value Theorem on $[a+r,x]$ then gives $z\in(a+r,x)$ with $f(z)=b=f(a)$, again contradicting injectivity.

If $x\leq a-r$, apply the same reasoning at the left endpoint. Put $c=f(a-r)$. When $y-b$ and $c-b$ have the same sign, $y$ lies strictly between $f(a-r)$ and $f(a)$, giving a second point with value $y$ by the Intermediate Value Theorem on $[a-r,a]$. When their signs are opposite, $b$ lies strictly between $f(x)=y$ and $f(a-r)$; the Intermediate Value Theorem on $[x,a-r]$ gives a point distinct from $a$ with value $b$. Both alternatives contradict injectivity. Thus $a-r<x<a+r$, and consequently

|f^{-1}(y)-a|=|x-a|<r<\varepsilon.

This proves that $f^{-1}(y)\to a$ as $y\to b$ within $f(I)$, which is continuity of $f^{-1}$ at $b$. $\square$

The Derivative Formula

The inverse derivative is obtained by comparing two secant slopes. For $x\ne a$, the slope of $f$ between $a$ and $x$ is $(f(x)-f(a))/(x-a)$. The corresponding inverse slope between $b=f(a)$ and $f(x)$ is its reciprocal. The lemma ensures that as the inverse input approaches $b$, the corresponding $x$ approaches $a$.

Theorem (Derivative of an Inverse Function): Let $I$ be an interval, let $a$ be an interior point of $I$, and suppose $f:I\to\mathbb{R}$ is continuous and injective. If $f$ is differentiable at $a$ and $f'(a)\ne0$, then $b=f(a)$ is interior to $f(I)$, $f^{-1}$ is differentiable at $b$, and $$ (f^{-1})'(b)=\frac{1}{f'(a)}=\frac{1}{f'(f^{-1}(b))}. $$

Proof. The lemma shows that $b$ is interior to $f(I)$ and that $f^{-1}$ is continuous at $b$. For $y\in f(I)$ with $y\ne b$, set $x=f^{-1}(y)$. Injectivity gives $x\ne a$, and $f(x)=y$. Hence

\frac{f^{-1}(y)-f^{-1}(b)}{y-b} = \frac{x-a}{f(x)-f(a)} = \frac{1}{\dfrac{f(x)-f(a)}{x-a}}.

As $y\to b$, the lemma gives $x=f^{-1}(y)\to a$. The definition of $f'(a)$ therefore gives

\frac{f(x)-f(a)}{x-a}\longrightarrow f'(a)=L.

Since $L\ne0$, these quotients are bounded away from zero when $x$ is sufficiently close to $a$. More explicitly, sufficiently near $a$ their distance from $L$ is less than $|L|/2$, so their absolute values are greater than $|L|/2$. Their reciprocals consequently converge to $1/L$; for a quotient $Q$ in this neighborhood,

\left|\frac{1}{Q}-\frac{1}{L}\right| = \frac{|Q-L|}{|Q||L|} \leq \frac{2|Q-L|}{|L|^2} \longrightarrow 0.

The inverse difference quotient thus tends to $1/L$. This is precisely differentiability of $f^{-1}$ at $b$, with the stated derivative. Finally, $f^{-1}(b)=a$, giving the second form of the formula. $\square$

Worked Examples

Worked Example: A Cubic with a Globally Defined Inverse

Let $f(x)=x^3+x$ on $\mathbb{R}$. For $u\ne v$,

$$ f(v)-f(u) =(v-u)(v^2+vu+u^2+1). $$

The second factor is positive: $v^2+vu+u^2\geq0$, so adding $1$ gives a strictly positive number. Thus $f(v)\ne f(u)$ whenever $v\ne u$, and $f$ is injective. It is continuous and differentiable, with $f'(x)=3x^2+1$. At $a=1$, $b=f(1)=2$ and $f'(1)=4\ne0$. The theorem gives

(f^{-1})'(2)=\frac{1}{4}.

No explicit formula for $f^{-1}$ is needed. The value of its derivative follows from the local change in $f$ at the input $1$.

Worked Example: The Square Function on the Positive Half-Line

Let $f(x)=x^2$ on $I=(0,\infty)$. If $u,v>0$ and $u\ne v$, then

v^2-u^2=(v-u)(v+u)\ne0,

so $f$ is injective; it is also continuous. Its image is $(0,\infty)$, and its inverse is $f^{-1}(y)=\sqrt{y}$. At $a=2$, $b=f(2)=4$, and the power rule gives $f'(2)=2(2)=4\ne0$. Therefore

(f^{-1})'(4)=\frac{1}{f'(2)}=\frac14.

This agrees with differentiating the explicit inverse: the derivative of $\sqrt{y}$ at $y=4$ is $1/(2\sqrt{4})=1/4$. The inverse theorem gives the same result without requiring an explicit inverse formula.

Worked Example: A Zero Derivative Can Produce a Nondifferentiable Inverse

Let $f(x)=x^3$ on $\mathbb{R}$. For $u\ne v$,

$$ v^3-u^3=(v-u)(v^2+vu+u^2). $$

The second factor is positive whenever $u\ne v$: it equals $(v+u/2)^2+3u^2/4$, which can be zero only if $u=v=0$, contrary to $u\ne v$. Thus $f$ is injective. Its inverse is $f^{-1}(y)=\sqrt[3]{y}$. At $a=0$, however, $f'(0)=0$, so the inverse derivative theorem does not apply.

In fact, the inverse is not differentiable at $b=0$. For $y\ne0$, its difference quotient there is

\frac{f^{-1}(y)-f^{-1}(0)}{y-0} = \frac{\sqrt[3]{y}}{y} = \frac{1}{(\sqrt[3]{y})^2} = \frac{1}{|y|^{2/3}}.

This quotient tends to $+\infty$ as $y\to0$, so it has no finite limit. The example shows why the nonzero-derivative hypothesis matters; it does not assert that every inverse associated with a zero derivative must fail to be differentiable.

How to Use the Formula—and Its Limits

The formula can be written in either of two equivalent ways:

(f^{-1})'(b)=\frac{1}{f'(a)} \qquad\text{when }b=f(a),

(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}

at points $y$ where the theorem's hypotheses hold. The first form is often convenient when a particular input $a$ is known. The second makes the dependence on the inverse input explicit.

A common error is to apply the reciprocal rule without checking that the inverse exists and is differentiable at the value in question. Injectivity supplies a well-defined inverse, continuity and the Intermediate Value Theorem place an open interval of output values in the image, and the nonzero derivative ensures the reciprocal is finite. The lemma also prevents another hidden gap: it proves that inverse inputs near $b$ really do approach $a$.

Geometrically, when $f'(a)\ne0$, the tangent slope of the original graph is nonzero. Reversing the roles of input and output reciprocates that slope. If the original slope is zero, the reciprocal expression is undefined, and the inverse may have no finite derivative—as the cube-root example demonstrates.

Check Your Understanding

Use the inverse derivative theorem and its proof to answer the following.

Why must $f$ be injective for the notation $f^{-1}$ to define a function on $f(I)$?
How does $f'(a)\ne0$ show that $f(a-r)$ and $f(a+r)$ lie on opposite sides of $f(a)$ for sufficiently small $r>0$?
In the theorem's proof, why does $x=f^{-1}(y)$ approach $a$ when $y$ approaches $b=f(a)$?
If $f(3)=7$ and $f'(3)=-5$, what does the theorem give for $(f^{-1})'(7)$, assuming its other hypotheses hold?
For $f(x)=x^3$ at $a=0$, what feature of the inverse difference quotient shows that the inverse has no finite derivative there?