Difficulty: Intermediate | Prerequisites: Basic differentiation rules (power rule, chain rule), function composition, domain and range.
This is the first half of Unit 7, which introduces inverse functions and shows you how to differentiate them. In earlier units you learned to differentiate standard functions directly. Here, the question flips: if you know f and its derivative, how do you find the derivative of f's inverse without ever writing down an explicit formula for that inverse? This matters because many functions you will integrate later (particularly inverse trig functions) rely on exactly this idea. You should be comfortable with the chain rule and with solving basic algebraic equations before starting.
If a function f has an inverse g, the derivative of g at a point is the reciprocal of the derivative of f at the corresponding point. The formula g'(x) = 1 / f'(g(x)) lets you compute inverse derivatives from a table or equation without ever solving for the inverse explicitly.
Inverse function
A function g such that f(g(x)) = x and g(f(x)) = x. Graphically, g is the reflection of f over the line y = x.
In simple terms, the inverse "undoes" whatever f does. If f turns 2 into 5, then g turns 5 back into 2.
One-to-one function
A function where each output corresponds to exactly one input; passes the horizontal line test. Only one-to-one functions have inverses that are themselves functions.
In simple terms, no two different x-values produce the same y-value.
Reciprocal of the derivative (inverse derivative rule)
The principle that g'(x) = 1 / f'(g(x)), where g is the inverse of f. Also written as dy/dx = 1 / (dx/dy).
In simple terms, the slope of the inverse at a point is "one over" the slope of the original function at the matching point.
Let f be a function whose domain is an interval (a, b). If f has an inverse function, two results follow:
If f is continuous on its domain, then f⁻¹ is continuous on its domain.
If f is differentiable on an interval containing c and f'(c) ≠ 0, then f⁻¹ is differentiable at the corresponding point.
The condition f'(c) ≠ 0 is critical. Where the original function has a horizontal tangent (slope zero), the inverse has a vertical tangent and is not differentiable there.
Start with y = f(x).
Swap x and y to get x = f(y).
Solve for y. The result is y = f⁻¹(x).
Example: y = x³ + 8. Swap to get x = y³ + 8. Solve: y = ∛(x – 8). So f⁻¹(x) = ∛(x – 8).
If g is the inverse of f, then:
$$g'(x) = \frac{1}{f'(g(x))}$$
Alternate notation: dy/dx = 1 / (dx/dy).
The big idea: the derivative of a function at (x, y) is the reciprocal of its inverse's derivative at (y, x).
Given f(x) = x³ – 1, find (f⁻¹)'(x) when a = 26.
Step 1: Find x such that f(x) = 26. Solve x³ – 1 = 26, so x³ = 27, so x = 3.
Step 2: Compute f'(x) = 3x². At x = 3, f'(3) = 3(9) = 27.
Step 3: Apply the formula. (f⁻¹)'(26) = 1 / f'(3) = 1/27.
This technique is common on exams: you are given a table or formula for f and asked to find the derivative of f⁻¹ at a particular output value.
Inverse derivative formula:
g'(x) = 1 / f'(g(x))
Or equivalently: dy/dx = 1 / (dx/dy)
Algebraic inverse procedure:
Write y = f(x)
Swap x and y
Solve for y
Inverse functions appear whenever you need to "reverse" a known relationship. For instance, if a temperature conversion formula gives Fahrenheit from Celsius, its inverse gives Celsius from Fahrenheit. The derivative of that inverse tells you how sensitive the reverse conversion is, which matters in engineering unit-conversion pipelines and sensor calibration.
Students often think (f⁻¹)'(a) = 1 / f'(a). It does not. You must evaluate f' at the point g(a), not at a itself. The input to f' is the x-value that f maps to a, not a directly.
Confusing "inverse" with "reciprocal." f⁻¹(x) is the inverse function, not 1/f(x). The notation is unfortunate, but they are completely different objects.
Forgetting to check that f'(c) ≠ 0 before applying the formula. If f' equals zero at the relevant point, the inverse is not differentiable there.
Assuming every function has an inverse. Only one-to-one functions (or functions restricted to a one-to-one interval) have inverses.
⚠️ The formula g'(x) = 1 / f'(g(x)) is tested frequently, usually with a table of values for f and f'. Make sure you can identify which row of the table to read.
⚠️ AP free-response questions often give you f and ask for (f⁻¹)' at a specific output. Practise the three-step process (find the matching x, compute f' there, take the reciprocal).
⚠️ Don't confuse the inverse derivative formula with the derivatives of inverse trig functions (covered in Part 2). They are related but distinct skills.
True or false: If f'(4) = 0, then f⁻¹ is differentiable at the point corresponding to x = 4.
Fill in the blank: To find (f⁻¹)'(a), you first need to find the value of x such that f(x) = ______.
True or false: The graph of an inverse function is a reflection of the original over the line y = x.
If f(2) = 7 and f'(2) = 5, what is (f⁻¹)'(7)?
True or false: f⁻¹(x) means the same thing as 1/f(x).
Q: If f(4) = 10 and f'(4) = 3, find (f⁻¹)'(10).
A: Since f(4) = 10, we need f'(4) = 3. So (f⁻¹)'(10) = 1/f'(4) = 1/3.
Q: Given f(x) = x³ – 1, find (f⁻¹)'(26).
A: Solve x³ – 1 = 26 to get x = 3. f'(x) = 3x², so f'(3) = 27. Therefore (f⁻¹)'(26) = 1/27.
Q: Why can't you use the inverse derivative formula when f'(c) = 0?
A: Because the formula requires dividing by f'(g(x)). If that derivative is zero, you would be dividing by zero. Geometrically, the original function has a horizontal tangent, so the inverse has a vertical tangent and is not differentiable at that point.
Q: Find the inverse of f(x) = 2x + 6 algebraically.
A: Swap x and y: x = 2y + 6. Solve for y: y = (x – 6)/2. So f⁻¹(x) = (x – 6)/2.
This material connects directly to the chain rule from Unit 3, since the inverse derivative formula is derived by differentiating f(g(x)) = x using the chain rule. It also sets up the derivatives of inverse trigonometric functions (covered in Part 2 of these notes), which are essential for a family of integration techniques tested on the AP exam.
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