The Derivative of an Inverse Function
If f is one-to-one and differentiable, its inverse f⁻¹ is differentiable too, and the two derivatives are reciprocals at mirrored points:
(f⁻¹)′(b) = 1 / f′(a), where a = f⁻¹(b), i.e. f(a) = b
The remarkable part is what the formula does not require: a formula for f⁻¹. Many important inverses, ln, arctan, arcsin, have derivatives most easily obtained this way, and for functions like x³ + x the inverse has no elementary formula at all, yet its derivative at any point is two steps of arithmetic. This calculator performs those steps: it solves f(a) = b numerically, warns when several solutions reveal a failure of one-to-oneness, evaluates f′(a) symbolically, and returns the reciprocal, with the mirror-image geometry drawn across the line y = x.
How to Use the Inverse Function Derivative Calculator
Enter f, the value b at which you want the inverse's derivative, and an interval in which to search for the preimage a. The steps show the rule, the solved a with its verification f(a) = b, the symbolic f′ and its value at a, and the final reciprocal. If the search finds multiple preimages, the tool flags the function as not one-to-one on that interval and proceeds with the first branch, mirroring what "restrict the domain" means in practice, the x² chip demonstrates. The plot shows f, the dashed mirror line y = x, and the two reflected points (a, b) and (b, a): the graph of f⁻¹ passes through the second one with exactly the reciprocal slope.
Worked Example
Let f(x) = x³ + x and compute (f⁻¹)′(2). First solve f(a) = 2: numerically a = 1 fits, since 1 + 1 = 2. Then differentiate:
f′(x) = 3x² + 1, f′(1) = 4 ⇒ (f⁻¹)′(2) = 1/4
No formula for f⁻¹ was ever needed, and none exists in elementary terms. The exp chip reproduces a classical derivation: for f = eᵀ at b = 5, the preimage is a = ln 5 and f′(a) = e^(ln 5) = 5, so (ln)′(5) = 1/5, the general fact (ln x)′ = 1/x obtained in one line. The tan chip similarly delivers (arctan)′(1) = 1/sec²(π/4) = 1/2.
Why Reciprocal Slopes: the Mirror Argument
The graph of f⁻¹ is the graph of f reflected across the line y = x, because inverting a function swaps the roles of input and output. Reflection across that line swaps rise with run: a segment climbing 4 units over 1 becomes a segment climbing 1 unit over 4. Slopes therefore invert: m becomes 1/m. That is the whole theorem, made rigorous by the chain rule: differentiating the identity f(f⁻¹(x)) = x gives f′(f⁻¹(x)) · (f⁻¹)′(x) = 1, and solving for (f⁻¹)′ yields the reciprocal formula. The chain-rule derivation also exposes the failure mode: where f′(a) = 0, the formula divides by zero, and genuinely, a horizontal tangent on f reflects into a vertical tangent on f⁻¹, where no finite derivative exists. That is why √x, the inverse of x², has an infinite slope at 0.
Common Mistakes to Avoid
- Evaluating f′ at b instead of at a. The derivative of f is taken at the preimage a = f⁻¹(b), not at b itself. This is the error the notation invites and the step display guards against.
- Applying the formula where f is not one-to-one. x² at b = 4 has preimages ±2 with different reciprocal slopes (1/4 and −1/4); "the" inverse only exists after restricting the domain to one branch.
- Ignoring f′(a) = 0. The reciprocal of zero is not a number: the inverse has a vertical tangent there and is not differentiable, exactly where invertibility itself is marginal.
- Confusing f⁻¹ with 1/f. The inverse function undoes f; the reciprocal 1/f divides one by it. Their derivatives are unrelated, and the notation f⁻¹ regrettably resembles a power.
- Forgetting that domains and ranges swap. (f⁻¹)′ lives on the range of f. Asking for its value at a b outside that range is asking about a point where the inverse is undefined; the search step makes this visible.
Real-World Applications
Inverse-function derivatives appear whenever a measurement chain runs backward. Instrument calibration: a sensor maps temperature to voltage via f; reading temperature from voltage uses f⁻¹, and the sensitivity of the reading to voltage noise is exactly (f⁻¹)′ = 1/f′, so flat regions of the sensor curve (small f′) are precisely where readings become unstable. Economics inverts demand curves to get price as a function of quantity, with reciprocal marginal interpretations. Statistics differentiates quantile functions, inverses of distribution functions, by this rule: the density of a transformed variable carries the 1/f′ factor, the change-of-variables formula's core. In numerical computing, the rule prices the sensitivity of root-finding: how much a root moves per unit change in the target level is 1/f′ at the root, the conditioning number of the root-finding problem. Related pages: the inverse function calculator for the algebra and the chain rule calculator for the machinery behind the proof.
Frequently Asked Questions
Why is the derivative of the inverse the reciprocal?
Reflecting a graph across y = x swaps rise and run, so every slope m becomes 1/m. Analytically, differentiating f(f⁻¹(x)) = x by the chain rule forces f′·(f⁻¹)′ = 1. Both arguments give the same formula, one geometric, one algebraic.
Where exactly do I evaluate f′?
At the point a on f's side of the mirror: a = f⁻¹(b), the input whose output is b. The calculator finds a by solving f(a) = b before touching the derivative, and displays the check f(a) = b so the branch is unambiguous.
What if f(a) = b has several solutions?
Then f is not one-to-one on the search interval and has no single inverse there. Each solution belongs to a different branch with its own slope; restricting the domain (as arcsin does for sine) chooses a branch. The tool lists all preimages and proceeds with the first, stating exactly this caveat.
What happens when f′(a) = 0?
The formula divides by zero because the inverse genuinely fails to be differentiable: a horizontal tangent on f reflects to a vertical tangent on f⁻¹. The cube root at 0, inverse of x³, is the standard picture, continuous, invertible, infinite slope.
How does this rule give the derivative of ln x?
ln is the inverse of eᵀ. At any b > 0 the preimage is a = ln b, and f′(a) = e^(ln b) = b, so (ln)′(b) = 1/b. One line, no limit computations, and the same template yields the derivatives of arcsin, arccos, and arctan.
Do I ever need a formula for the inverse function itself?
Not for its derivative at a point. The rule runs entirely on f's side: solve one equation, evaluate one derivative, take a reciprocal. This is essential for functions like x³ + x or x + eᵀ whose inverses provably have no elementary formula, yet whose inverse derivatives are simple numbers.
What is the second derivative of an inverse function?
Differentiating the reciprocal rule again gives (f⁻¹)″(b) = −f″(a)/[f′(a)]³. The sign flip means an increasing f that is concave up has a concave-down inverse, visible in the exp/ln pair, and the cube in the denominator makes near-flat regions of f violently curved on the inverse.
How is this formula used in probability?
Quantile functions are inverses of CDFs, and simulating by inverse transform differentiates them: the density of the output is 1/f′ evaluated at the quantile, the Jacobian of the change of variables. Heavy tails correspond to flat CDF stretches, tiny f′, huge quantile sensitivity.
Is (f⁻¹)′ always positive?
It has the same sign as f′(a): inverses of increasing functions increase, of decreasing functions decrease. What is impossible is a sign change without passing through a non-invertible flat spot, since one-to-one differentiable functions on an interval have f′ of constant sign (or isolated zeros with matching one-sided behavior).
Why does the graph show the line y = x?
It is the mirror: the graphs of f and f⁻¹ are reflections across it, and the two marked points (a, b) and (b, a) are a mirrored pair. Seeing the pair makes the rule concrete, the tangent at one point and the tangent at its mirror image have reciprocal slopes.