Why the Derivative of an Inverse Is a Reciprocal
If a function \(f\) sends an input \(b\) to an output \(a\), its inverse sends \(a\) back to \(b\). The derivative of the inverse describes how the returning input changes as the output changes. The key idea is that the inverse reverses the input-output relationship, so its local rate of change is the reciprocal of the original rate—provided that this reciprocal is defined and the inverse is being taken on a suitable branch.
The formula is easy to misapply if the two inputs are confused. We will derive it carefully, distinguish the point \(a\) from its preimage \(b\), and explain why both invertibility and a nonzero derivative matter.
Inverse Functions and Local Branches
A function \(f:I\to J\) is one-to-one if distinct inputs in \(I\) always have distinct outputs. If \(f\) is one-to-one and maps \(I\) onto \(J\), then it has an inverse function \(f^{-1}:J\to I\), characterized by
The inverse reverses the roles of input and output. In this tutorial, let \(b\in I\) and set \(a=f(b)\). Thus \(f^{-1}(a)=b\). The inverse derivative formula, when its conditions hold, is
A function need not be one-to-one on its entire domain to have an inverse on a smaller interval. Such an inverse on a restricted interval is called a local inverse branch. The choice of interval matters: different branches can give different inverse values and derivatives.
The Inverse Derivative Rule
Here is a precise version that separates the derivative calculation from the existence of an inverse. Suppose \(f\) is one-to-one on an interval \(I\), \(b\) is an interior point of \(I\), \(a=f(b)\), and the inverse \(g=f^{-1}\) is continuous at \(a\). If \(f\) is differentiable at \(b\) and \(f'(b)\ne 0\), then \(g\) is differentiable at \(a\), and
A standard sufficient condition for an appropriate local inverse branch is that \(f'\) is continuous near \(b\) and \(f'(b)\ne 0\). Then \(f'\) keeps the same nonzero sign on a sufficiently small interval around \(b\). The Mean Value Theorem implies that \(f\) is strictly monotone on that interval, so it is one-to-one there. Its inverse on that interval is continuous. In applications, a strictly monotone branch can also be specified directly.
Deriving the Formula from the Inverse Identity
Let \(g=f^{-1}\), so that \(f(g(x))=x\) for \(x\) near \(a\). If we already know that \(g\) is differentiable, the chain rule gives \(f'(g(a))g'(a)=1\). Since \(g(a)=b\), this suggests the reciprocal formula. But this chain-rule observation alone does not establish that \(g\) is differentiable. The difference quotient supplies that missing justification.
For \(x\) in the inverse’s domain, \(f(g(x))=x\), and \(g(a)=b\).
For \(x\ne a\), injectivity gives \(g(x)\ne b\), so we may divide by \(g(x)-b\).
As \(x\to a\), continuity gives \(g(x)\to b\), and the resulting quotient tends to \(1/f'(b)\).
In detail, because \(f(g(x))=x\) and \(f(g(a))=a\), we have
As \(x\to a\), continuity of \(g\) at \(a\) implies \(g(x)\to b\). Also \(g(x)\ne b\) when \(x\ne a\), since \(f\) is one-to-one. Therefore the denominator inside the final reciprocal tends, by differentiability of \(f\) at \(b\), to \(f'(b)\). Since \(f'(b)\ne0\), taking the reciprocal is valid and gives
This limit is exactly the definition of \(g'(a)\), proving the rule. The proof explains both the reciprocal and the conditions: an inverse branch lets us use \(g\), continuity ensures its inputs approach \(b\), and a nonzero \(f'(b)\) permits division by the limiting difference quotient.
Worked Examples
Worked Example: The Positive Branch of a Square Function
Let \(f(x)=x^2\) on \(I=(0,\infty)\). Find the derivative of its inverse at \(a=9\). On this interval \(f\) is strictly increasing, so the inverse branch is \(g(x)=\sqrt{x}\) for \(x>0\).
First find the point \(b\) such that \(f(b)=a\):
where \(b=3\) because the chosen branch has positive inputs. Next, differentiate the original function and evaluate at \(b\):
The inverse derivative rule gives
Check using the explicit inverse: \(g(x)=\sqrt{x}\), so \(g'(x)=1/(2\sqrt{x})\), and \(g'(9)=1/(2\cdot3)=1/6\). This agrees with the rule.
Worked Example: A Polynomial Inverse Without an Explicit Formula
Let \(f(x)=x^3+x\), and find the derivative of its inverse at \(a=2\). Since \(f'(x)=3x^2+1>0\) for every real \(x\), the function is strictly increasing and has an inverse on its image.
Find the preimage of \(a\):
Differentiate and evaluate at that preimage:
Therefore,
As a check, write \(g=f^{-1}\). The identity \(f(g(x))=x\) gives \(g(x)^3+g(x)=x\). Differentiating this identity at \(x=2\), where \(g(2)=1\), yields \((3g(2)^2+1)g'(2)=1\), or \(4g'(2)=1\). Thus \(g'(2)=1/4\), as obtained above. This check uses the chain rule after the inverse branch and its differentiability have been established.
Worked Example: A Zero Derivative and a Nondifferentiable Inverse
Consider \(f(x)=x^3\) at \(b=0\). This function is strictly increasing, so it has an inverse \(g(x)=\sqrt[3]{x}\), and \(a=f(0)=0\). However,
The reciprocal formula cannot be used because it would require division by zero. Check directly whether the inverse has a finite derivative at \(0\):
As \(h\to0\), this quotient grows without bound, so it has no finite limit. Thus \(g\) is not differentiable at \(0\), even though the inverse branch exists. The example shows that invertibility alone is not enough for the derivative rule.
Why the Nonzero Derivative Is Necessary
The nonzero condition is not merely a technical convenience. If \(f'(b)=0\) and an inverse \(g\) were differentiable at \(a=f(b)\), then differentiating \(f(g(x))=x\) at \(a\) by the chain rule would give
Since \(g(a)=b\), the left side would be \(f'(b)g'(a)=0\), which cannot equal \(1\) for a finite value of \(g'(a)\). Hence, when \(f'(b)=0\), the inverse cannot have a finite derivative at \(a\) under these differentiability assumptions. It may still have an inverse, as the cube example demonstrates.
Common Pitfalls
The first common error is evaluating the original derivative at \(a\) rather than at \(b=f^{-1}(a)\). For the positive square branch at \(a=9\), the incorrect calculation would be
But \(9\) is the input to the inverse, while the corresponding input to \(f\) is \(b=f^{-1}(9)=3\). The correct calculation is \(1/f'(3)=1/6\), as verified by differentiating \(\sqrt{x}\).
A second error is assuming that an inverse exists on the whole original domain. The function \(f(x)=x^2\) on \(\mathbb{R}\) is not one-to-one: \(f(2)=f(-2)=4\). It therefore has no inverse function on all of \(\mathbb{R}\). Restricting to \(x>0\) gives the positive branch \(f^{-1}(x)=\sqrt{x}\); restricting to \(x<0\) gives the negative branch \(f^{-1}(x)=-\sqrt{x}\). The branch must be specified before calculating an inverse derivative.
When the Rule Is Useful
The formula is especially useful when an inverse exists but cannot be written conveniently in an explicit form. One can still find its derivative by locating the corresponding point on the original function and evaluating \(f'\) there. This method also clarifies the local meaning of reciprocal slopes: near a point where \(f\) changes rapidly, its inverse changes more slowly, while a small nonzero slope of \(f\) corresponds to a large inverse slope.
The rule does not remove the need to check the branch or the derivative condition. It is a conclusion about a differentiable inverse near a specified point, not a formula that automatically applies to every function called an inverse.
Check Your Understanding
For each question, identify the original-function input and check the relevant conditions before using the formula.
- If \(f(b)=a\), at which point must \(f'\) be evaluated to compute \((f^{-1})'(a)\)?
- For \(f(x)=x^2\) restricted to \(x>0\), find \((f^{-1})'(16)\).
- Why does the identity \(f(f^{-1}(x))=x\) suggest a reciprocal derivative?
- Give a sufficient local condition that guarantees \(f\) has an inverse branch near \(b\).
- What can fail for the derivative of the inverse when \(f'(b)=0\), even if \(f\) is one-to-one?
- Why does \(x^2\) have no inverse function on all of \(\mathbb{R}\)?
