Why an Inverse Needs the Correct Branch
A function can send different inputs to the same output. When that happens, the function does not have a single inverse on its full domain: the output alone does not tell us which input to recover.
A common example is \(f(x)=x^2\). Both \(3\) and \(-3\) map to \(9\). To define an inverse, we first restrict the domain so that each output comes from exactly one input. The restriction determines which of these two possibilities the inverse returns.
This choice matters for derivatives as well. The inverse-derivative formula uses the derivative of the original function at the input selected by the branch. Different inputs giving the same output can have different derivatives, so choosing the wrong input can give the wrong answer.
Restriction Determines the Inverse’s Domain and Range
Let \(f\) be restricted to a domain \(I\), and suppose the restricted function is one-to-one. Its range is the set of outputs produced by inputs in \(I\). The inverse reverses the input-output pairs, so its domain is that range and its range is \(I\).
Thus, an inverse’s domain is not chosen independently: it is the range of the restricted original function. Its range is the domain chosen for that branch. Keeping both sets visible helps prevent a common source of confusion.
For a continuous function, an interval on which the function is strictly increasing or strictly decreasing is a common choice: either kind of strict monotonicity guarantees that distinct inputs have distinct outputs. The horizontal line test gives a graphical version of the same idea: every horizontal line should meet the restricted graph at most once.
How the Branch Enters the Derivative
Suppose \(g\) is an inverse branch of \(f\), and \(y_0=f(x_0)\), where \(x_0\) belongs to the restricted domain defining that branch. When the inverse is differentiable at \(y_0\), the inverse-derivative formula is
The crucial detail is that \(x_0\) must be the input on the chosen branch for which \(f(x_0)=y_0\). If another branch contains a different input with the same output, that other input gives the derivative of the other inverse, not of \(g\).
Why the Reciprocal Formula Works
The branch choice is part of the reasoning, not just a notation preference. The following difference-quotient argument shows both why the derivative is a reciprocal and why we must use the input \(x_0\) on the selected branch.
Let \(g\) be the chosen inverse branch, let \(y_0=f(x_0)\), and let \(y=f(x)\) for \(x\) on that same branch. Then \(g(y)=x\) and \(g(y_0)=x_0\).
For \(y\ne y_0\), one-to-one behavior ensures \(x\ne x_0\), and the inverse relations give the equality below.
As \(y\) approaches \(y_0\) through the inverse’s domain, \(x=g(y)\) approaches \(x_0\) when the selected inverse is continuous there. The resulting limit is the reciprocal of \(f'(x_0)\), provided that derivative is nonzero.
This derivation explains the branch dependence directly: the limit uses \(x\) approaching the particular \(x_0\) selected by the branch. If a different branch uses a different input for the same output, its difference quotient approaches the derivative at that other input.
Worked Example: Two Inverses of the Square Function
Worked Example: Two Inverses of the Square Function
Restrict \(f(x)=x^2\) first to \(I_+=[0,\infty)\), and then to \(I_-=(-\infty,0]\). Find the derivative of each inverse at \(y=9\).
On either restricted domain the function is one-to-one, and its range is \([0,\infty)\). For the nonnegative branch, \(f(3)=9\), so the inverse value at \(9\) is \(3\). The original derivative is
Therefore the derivative of the inverse branch with range \([0,\infty)\) is
On the nonpositive branch, \(f(-3)=9\), so the inverse value at \(9\) is \(-3\). This time,
Check: the inverse functions are \(g_+(y)=\sqrt{y}\) and \(g_-(y)=-\sqrt{y}\), for \(y\ge 0\). Differentiating at \(y=9\) gives \(g_+'(9)=1/(2\sqrt{9})=1/6\) and \(g_-'(9)=-1/(2\sqrt{9})=-1/6\), matching the inverse-derivative calculations.
Worked Example: Selecting a Cosine Branch
Worked Example: Selecting a Cosine Branch
Let \(f(x)=\cos x\) restricted to \([0,\pi]\). Find the derivative of its inverse at \(y=-\frac12\).
On \([0,\pi]\), cosine is strictly decreasing, so this restriction is one-to-one. Its range is \([-1,1]\); therefore this inverse has domain \([-1,1]\) and range \([0,\pi]\).
We need the input \(x_0\) in the chosen interval satisfying \(\cos x_0=-\frac12\). Within \([0,\pi]\), that input is \(x_0=\frac{2\pi}{3}\). Differentiating the original function gives \(f'(x)=-\sin x\), so
The derivative of the inverse at \(-\frac12\) is consequently
Check: multiplying the original slope by the inverse slope gives \(\left(-\frac{\sqrt{3}}{2}\right)\left(-\frac{2\sqrt{3}}{3}\right)=\frac{6}{6}=1\), as required by the reciprocal relationship.
Worked Example: The Same Output on Two Sine Branches
Worked Example: The Same Output on Two Sine Branches
Consider \(f(x)=\sin x\) on each of the intervals \([0,\frac{\pi}{2}]\) and \([\frac{\pi}{2},\pi]\). Find the inverse derivative at \(y=\frac12\) for each branch.
On the first interval, sine increases from \(0\) to \(1\); on the second, it decreases from \(1\) to \(0\). Each restriction is one-to-one, and each has range \([0,1]\). Thus, each inverse has domain \([0,1]\), but the two inverses have different ranges: one is \([0,\frac{\pi}{2}]\), the other \([\frac{\pi}{2},\pi]\).
For the first branch, \(x_0=\frac{\pi}{6}\), since \(\sin(\frac{\pi}{6})=\frac12\). Because \(f'(x)=\cos x\),
For the second branch, the matching input is \(x_0=\frac{5\pi}{6}\), also satisfying \(\sin(\frac{5\pi}{6})=\frac12\). Thus,
Check: each inverse slope multiplied by its matching original slope equals \(1\). The slopes have opposite signs because the original function increases on the first interval and decreases on the second. The output \(\frac12\) alone does not identify the branch; the selected interval does.
A Reliable Branch-Selection Routine
Before computing an inverse derivative, make the branch choice explicit. This prevents solving for an input without checking whether that input belongs to the inverse being used.
Identify the interval or set on which the original function is being treated as one-to-one.
Find the outputs produced on that restriction. They form the inverse’s domain; the restricted domain forms its range.
Solve \(f(x_0)=y_0\), then keep only the solution in the selected domain.
Evaluate \(f'(x_0)\). If it is nonzero and the inverse is differentiable there, the inverse derivative is \(1/f'(x_0)\).
Common Mistake: Choosing a Root Before Choosing a Branch
Suppose the inverse is defined using \(f(x)=x^2\) restricted to \((-\infty,0]\), and we need its derivative at \(y=9\). A tempting but incorrect calculation is to choose \(x=3\), then write
This calculation uses the positive input, which is not in the selected domain. It finds the derivative for the nonnegative inverse branch. On the specified nonpositive branch, \(x_0=-3\), and the correct calculation is
The correction is not merely to choose the other sign at the end. The inverse branch determines the original input before the derivative is evaluated. Substituting an input from a different branch changes the function whose derivative is being calculated.
Points Where the Reciprocal Calculation Fails
Even after the branch is identified, the reciprocal formula has a condition: the corresponding derivative must be nonzero. For \(f(x)=x^2\), both inverse branches meet at \(y=0\), corresponding to \(x=0\). But \(f'(0)=0\), so the formula does not produce a finite inverse derivative there. Indeed, the square-root branches do not have a finite derivative at \(0\).
Endpoints also deserve care. A restricted interval can give an inverse whose domain includes an endpoint, while a usual two-sided derivative at that endpoint may not be defined. Check whether the question asks for an interior derivative or an endpoint one-sided rate, and do not apply an interior reciprocal formula without checking its conditions.
Check Your Understanding
For each question, identify the branch before using any inverse-derivative calculation.
- For \(f(x)=x^2\) restricted to \([0,\infty)\), what are the domain and range of the inverse?
- For the same function restricted to \((-\infty,0]\), what input corresponds to inverse input \(y=16\)?
- For that nonpositive branch, what is the inverse derivative at \(y=16\)?
- Why can two inverse branches have different derivative values at the same output?
- What condition on \(f'(x_0)\) is required for the usual reciprocal formula to give a finite inverse derivative?
