Real Analysis 39 | Derivatives of Inverse Functions [dark version]

TL;DR

If f: I → J is bijective and differentiable at x0 with f′(x0) ≠ 0, then the inverse’s differentiability depends on behavior near y0 = f(x0).

Briefing Cornell Notes

Briefing

A general rule links the derivative of a function to the derivative of its inverse: when an inverse exists and behaves continuously at the corresponding point, the inverse’s slope is the reciprocal of the original function’s slope. The payoff is immediate—this framework produces the classic derivative formula for the natural logarithm without treating it as a special case.

The discussion starts by recalling the geometric relationship between exponential and logarithm. Reflecting the graph of the exponential function across the line y = x swaps x- and y-values, turning exponential into logarithm. Since differentiability corresponds to having a well-defined tangent slope at each point, the reflection suggests that differentiability should transfer between a function and its inverse—provided the tangent slope doesn’t collapse into a “flat” case. That leads to a key restriction: at the point x0, the derivative f′(x0) must be nonzero. If f′(x0) were zero, the inverse would develop an infinite slope at the corresponding point, breaking differentiability.

To make the idea precise, the argument considers intervals I and J and a bijection f: I → J with inverse f⁻¹: J → I. Fix a point x0 in I where f is differentiable and f′(x0) ≠ 0, and let y0 = f(x0). The derivative of the inverse at y0 is defined through a limit of difference quotients using sequences yn → y0 (with yn ≠ y0). The key algebraic step rewrites the inverse difference quotient in terms of f by using xn = f⁻¹(yn), so that f(xn) = yn. After substitution, the inverse’s difference quotient becomes the reciprocal of the original function’s difference quotient. Taking limits then yields the relationship between derivatives—assuming the inverse is continuous at y0 so that xn → x0.

The resulting theorem is: if f is differentiable at x0 with nonzero derivative, and f⁻¹ is continuous at y0 = f(x0), then f⁻¹ is differentiable at y0, with (f⁻¹)′(y0) = 1 / f′(x0).

The minus-one notation is clarified: it refers to the inverse function on the left, while on the right it means taking the reciprocal of the number f′(x0).

Finally, the rule is applied to the natural logarithm. Since ln is the inverse of the exponential function exp, and exp′(y) = exp(y), the reciprocal formula collapses cleanly. With y = ln(x) (or equivalently matching variables so that exp and ln cancel), the derivative of ln becomes 1/x. The conclusion is a reusable identity: d/dx [ln x] = 1/x, which the discussion flags as a tool for later work such as integration.

Cornell Notes

The derivative of an inverse function can be computed from the derivative of the original function. For a bijection f: I → J with inverse f⁻¹, pick x0 in I and set y0 = f(x0). If f is differentiable at x0 and f′(x0) ≠ 0, and if f⁻¹ is continuous at y0, then f⁻¹ is differentiable at y0. The inverse’s derivative equals the reciprocal of the original derivative: (f⁻¹)′(y0) = 1 / f′(x0). Using this with ln as the inverse of exp yields (ln x)′ = 1/x, because exp′(t) = exp(t) and exp and ln cancel appropriately.

Why must f′(x0) be nonzero when differentiating an inverse?

If f′(x0) = 0, the tangent to f at x0 is horizontal. After reflecting across y = x to form the inverse, that horizontal tangent becomes a vertical tangent, which corresponds to an infinite slope. A vertical tangent means the inverse cannot have a finite derivative at the corresponding point y0 = f(x0), so differentiability fails.

How does the inverse difference quotient get rewritten using f?

Starting from the definition (f⁻¹)′(y0) = lim_{n→∞} (f⁻¹(yn) − f⁻¹(y0)) / (yn − y0), set xn = f⁻¹(yn). Then f(xn) = yn and f(x0) = y0. The quotient becomes (xn − x0) / (f(xn) − f(x0)), which is the reciprocal of the usual difference quotient for f at x0.

Why is continuity of f⁻¹ at y0 needed for the limit?

The limit for the inverse derivative uses yn → y0. Continuity of f⁻¹ at y0 ensures xn = f⁻¹(yn) → f⁻¹(y0) = x0. Without xn → x0, the difference quotient for f might not approach f′(x0), so the reciprocal-limit step would not be justified.

What exactly does the formula (f⁻¹)′(y0) = 1 / f′(x0) mean?

The “−1” on the left refers to the inverse function f⁻¹. On the right, the “1/” indicates taking the reciprocal of the number f′(x0). So the inverse’s slope at y0 is the reciprocal of the original function’s slope at x0.

How does this produce the derivative of ln x?

ln x is the inverse of exp. Apply (f⁻¹)′(y0) = 1 / f′(x0) with f = exp. Since exp′(t) = exp(t), the reciprocal becomes 1 / exp(t). Matching variables so that y0 = exp(t) = x gives (ln x)′ = 1/x.

Review Questions

State the conditions under which the inverse function f⁻¹ is differentiable at y0 = f(x0).
Derive the relationship between (f⁻¹)′(y0) and f′(x0) using difference quotients.
Use the inverse-derivative rule to compute d/dx[ln x].

Key Points

1
If f: I → J is bijective and differentiable at x0 with f′(x0) ≠ 0, then the inverse’s differentiability depends on behavior near y0 = f(x0).
2
Continuity of f⁻¹ at y0 is required to ensure that xn = f⁻¹(yn) converges to x0 when yn → y0.
3
The inverse derivative is computed from the original derivative via a reciprocal relationship: (f⁻¹)′(y0) = 1 / f′(x0).
4
A zero derivative f′(x0) would create an infinite slope for the inverse, so differentiability of the inverse fails in that case.
5
The derivative of the natural logarithm follows immediately because ln is the inverse of exp and exp′(t) = exp(t).
6
The final identity to remember is (ln x)′ = 1/x, derived from the inverse-derivative rule rather than treated as a standalone fact.

Highlights

The inverse function’s slope equals the reciprocal of the original function’s slope, provided the inverse is continuous at the matching point.

The nonzero-slope condition f′(x0) ≠ 0 prevents the inverse from developing a vertical tangent.

Continuity of f⁻¹ at y0 guarantees the sequence xn = f⁻¹(yn) actually approaches x0, making the limit work.

Applying the rule to exp and ln yields the classic result (ln x)′ = 1/x.

Topics

Inverse Derivatives
Natural Logarithm
Reciprocal Slopes
Continuity Conditions
Difference Quotients