AP Syllabus focus:
‘Use the definition of an inverse function together with the chain rule to derive the formula (f⁻¹)'(x) = 1 / f'(f⁻¹(x)), when f'(x) exists and is nonzero.’
This section develops how the derivative of an inverse function arises from composing a function with its inverse, emphasizing the chain rule and conditions guaranteeing differentiability and validity.
Deriving the Derivative Formula for an Inverse Function
Understanding the Relationship Between a Function and Its Inverse
A central idea in calculus is that a function and its inverse undo each other’s operations. For a one-to-one differentiable function f, its inverse f⁻¹ satisfies the identity f(f⁻¹(x)) = x for every x in the domain of the inverse.
This diagram shows the graphs of a function and its inverse reflected across the line , illustrating that each point corresponds to on the inverse. Although the example uses a specific square and square-root pair, the reflection property it demonstrates is general and directly relevant for understanding inverse functions. The detail exceeds syllabus requirements slightly but remains conceptually aligned. Source.
Inverse Function: A function f⁻¹ such that f(f⁻¹(x)) = x for all x in the domain of f⁻¹.
This identity suggests that when we differentiate the composition using the chain rule, we can isolate the derivative of the inverse and express it in terms of values from the original function. Understanding this structural relationship is essential before applying differentiation techniques.
Applying the Chain Rule to the Composition
To derive the derivative of an inverse, we begin with the fundamental composition equation. Differentiating both sides with respect to x requires careful use of the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function.
Chain Rule: The derivative of f(g(x)) is f′(g(x))·g′(x).
Because the right-hand side of the identity f(f⁻¹(x)) = x differentiates to 1, the chain rule allows us to express the derivative of the inverse entirely in terms of f and f′, avoiding the need for an explicit formula for the inverse function.
Derivation Structure
Differentiating the identity produces a form that can be algebraically rearranged to isolate the derivative of the inverse. While no explicit computational example is included here, it is important to recognize the logical sequence required in AP-level reasoning. Each step stems from properties of inverse functions, composition, and differentiation.
The chain rule applied to f(f⁻¹(x)) gives f′(f⁻¹(x))·(f⁻¹)'(x). Since the derivative of x is 1, this expression must equal 1, allowing us to solve for the derivative of the inverse. This leads to the formula emphasized in the AP specification.
= Derivative of the inverse function
= Derivative of f evaluated at the corresponding input to f^{-1}
This equation is central in AP Calculus AB because it enables students to compute derivatives of inverse functions without explicitly determining the algebraic form of the inverse.
Why the Formula Works
The formula reflects the reciprocal relationship between slopes of inverse functions. If f is increasing at a certain rate, its inverse increases at the reciprocal of that rate because the two functions exchange the roles of x and y on their graphs. The chain rule formalizes this geometric intuition.
This figure displays tangent lines to a function and its inverse at corresponding points, highlighting that their slopes are reciprocals. The labeled slopes and visually encode the relationship . The symbolic notation slightly exceeds syllabus detail but remains directly supportive of the chain-rule-based reasoning. Source.
When f′(x) is large, small changes in x create large changes in f(x); therefore, the inverse must change slowly, resulting in a small derivative. Conversely, if f′(x) is small but nonzero, the slope of the inverse becomes large.
Conditions Required for Validity
In the AP syllabus, students must not only use the formula but also understand when it applies. These conditions ensure the formula’s validity:
The function must be one-to-one. Without one-to-one behavior, an inverse function does not exist over the interval.
The original function must be differentiable. Differentiability guarantees the existence of f′(x), which is necessary for applying the chain rule.
The derivative must be nonzero at the relevant point. If f′(x) = 0, the reciprocal is undefined, and the inverse fails to be differentiable there.
The point of evaluation must correspond to a point in the domain of f⁻¹. This ensures that f′ is evaluated at a valid input.
Each of these criteria reinforces the interplay between algebraic definitions and graphical behavior, preparing students to justify use of the formula in rigorous AP-style reasoning.
Conceptual Interpretation
The derivative of an inverse encapsulates how inverse functions transform rates of change. Because inverse functions swap inputs and outputs, the derivative must reflect this interchange. The reciprocal form of the derivative formula emerges naturally from this conceptual symmetry: slopes along f become reciprocal slopes along f⁻¹. This symmetry, grounded in the chain rule and inverse definition, provides a unifying perspective for mastering advanced differentiation topics.
FAQ
If the derivative of the original function is zero at a point, the function has a horizontal tangent there. A horizontal tangent corresponds to an infinite or undefined reciprocal slope for the inverse, meaning the inverse cannot have a well-defined derivative.
A non-zero derivative ensures a locally one-to-one mapping with a consistent slope, allowing the reciprocal slope of the inverse to exist.
Local linearity means that a differentiable function behaves like a straight line when viewed extremely close-up. If f is locally linear near a point, then f inverse behaves like the inverse of that line.
Because the slope of a line and the slope of its inverse are reciprocals, this linear approximation naturally leads to the inverse derivative formula for smooth functions.
Yes, provided you can read corresponding x- and y-values and estimate the slope of f at the required point.
To apply the formula:
• Identify the point on the graph where f(x) equals the given output.
• Estimate the slope of f at that point.
• Take the reciprocal of that slope to approximate the inverse’s derivative.
Concavity affects how rapidly the derivative of f changes but does not alter the reciprocal relationship itself. The formula still holds as long as f remains one-to-one and differentiable with a non-zero derivative.
However, if concavity causes f ’ to approach zero, the derivative of the inverse becomes large, reflecting steep changes in the inverse function’s graph.
The formula depends only on evaluating the original function and its derivative at the appropriate corresponding point.
This means you can compute the derivative of the inverse even when:
• The algebraic inverse is extremely difficult to find.
• The inverse cannot be written in closed form.
• You only have tables or graphs of f.
The method relies purely on the structure of the composition f(f inverse(x)) = x.
Practice Questions
Question 1 (1–3 marks)
A function f is differentiable and one-to-one. It is known that f(2) = 5 and f ’(2) = 4.
Find the value of (f inverse)’(5).
Give a reason for your answer.
Question 1
• Correct use of the inverse derivative relationship earns 1 mark:
(f inverse)’(5) = 1 / f ’(2)
• Substitution of f ’(2) = 4 earns 1 mark:
1 / 4
• Correct justification earns 1 mark:
Because 5 = f(2), the derivative of the inverse at 5 corresponds to the reciprocal of f ’(2).
Question 2 (4–6 marks)
Let f be a differentiable, one-to-one function. A table of values for f and f ’ is shown below:
x: 1 2 3 4
f(x): 4 7 9 11
f ’(x): 3 5 2 4
(a) Determine the value of f inverse(9).
(b) Use the inverse derivative formula to find (f inverse)’(9).
(c) Explain why the inverse derivative formula is valid for this function on the given interval.
Question 2
(a) 1 mark
• Correct identification: f inverse(9) = 3, since f(3) = 9.
(b) 3 marks
• Statement of inverse derivative formula earns 1 mark:
(f inverse)’(9) = 1 / f ’(3)
• Correct substitution earns 1 mark:
= 1 / 2
• Correct final value earns 1 mark:
= 0.5
(c) 2 marks
• Correct explanation that f is one-to-one on the interval (using strictly increasing values in the table) earns 1 mark.
• Explanation that f ’ is non-zero at each listed point, ensuring differentiability of the inverse, earns 1 mark.
