01 · Corresponding inputs
You know f(2) = 7 and f′(2) = 5. If g = f⁻¹ on a suitable branch, find g(7) and g′(7).
Hint
The inverse returns the original input.
Worked solution
g(7) = 2 and g′(7) = 1/5. These facts do not give g′(2).
Understand · explore · practise
Derivatives of inverse functions
Differentiate inverse functions using reciprocal gradients at corresponding points. Learn when dy/dx = 1/(dx/dy) is valid, keep the correct branch and handle zero-gradient exceptions.
Before you startInverse functions and domains; chain rule; tangents and normals
01 / Reflect the point and tangent
Swap the coordinates before comparing gradients.
If f(a) = b, the original curve contains (a, b) and its inverse contains (b, a). Reflection in y = x swaps a small horizontal change with a small vertical change. A non-zero gradient therefore becomes its reciprocal.
(f⁻¹)′(b) = 1/f′(a), where b = f(a) and f′(a) ≠ 0
Work on an invertible branch with the differentiability conditions needed for the inverse rule.
The derivative is taken at b on the inverse, not at a. Use the explorer to compare the corresponding points, including decreasing functions and zero original gradients.
Reflect the point and its tangentExplore
f(x) = x³; inverse f⁻¹(x) = ∛x.
Original point (1, 1); inverse point (1, 1). Original gradient 3; inverse gradient 0.3333.
The gradients are reciprocals, with the same sign. They are not negative reciprocals.
Both axes use the same scale. Changing to x² restricts a to non-negative inputs so its inverse is a function. At a zero original gradient, inspect the vertical inverse tangent rather than dividing by zero.
02 / Derive the inverse rule
Differentiate the identity f(g(x)) = x.
Let g = f⁻¹ on a valid branchWorked example
f(g(x)) = x
Following the inverse by the original returns the input.
f′(g(x))g′(x) = 1
Apply the chain rule where both derivatives exist.
g′(x) = 1/f′(g(x))
Divide only when the original derivative is non-zero.
For a continuously differentiable f with f′(a) ≠ 0, a local differentiable inverse exists around a. A globally one-to-one function may still have a zero derivative at some point, so global invertibility alone is not enough for a finite inverse derivative everywhere.
Watch a tangent reflect into its reciprocal slope
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Find an inverse slope without solving for the inverse
A known point can be enough.
f(x) = x³ + 2x + 1; find (f⁻¹)′(4)Worked example
f(1) = 1 + 2 + 1 = 4
The corresponding original input is 1.
f′(x) = 3x² + 2 > 0
f is strictly increasing and has a non-zero derivative everywhere.
(f⁻¹)′(4) = 1/f′(1) = 1/5
No cubic inversion is needed.
02 · Another cubic
For f(x) = x³ + 4x − 2, find the derivative of its inverse at x = 3.
Hint
Check f(1), then f′(1).
Worked solution
f(1) = 3 and f′(x) = 3x² + 4 > 0. The inverse derivative at 3 is 1/7.
03 · A decreasing function
For f(x) = 6 − 3x, find the inverse derivative at x = 0.
Hint
The original slope is negative.
Worked solution
The inverse is (6 − x)/3 with derivative −1/3. Equivalently, f(2) = 0 and 1/f′(2) = −1/3.
04 / Choose an invertible branch
The branch decides the sign of an inverse derivative.
f(x) = x² − 4 on two different domainsWorked example
For x ≥ 0: f⁻¹(x) = √(x + 4)
The inverse has domain x ≥ −4.
(f⁻¹)′(x) = 1/[2√(x + 4)], x > −4
The positive branch is increasing.
For original x ≤ 0: f⁻¹(x) = −√(x + 4)
The inverse derivative is −1/[2√(x + 4)] for x > −4.
Without a restriction, x² − 4 is not one-to-one and has no single inverse function. At the endpoint x = −4 the inverse exists but has no finite derivative.
04 · Positive branch
Let f(x) = x² + 5 for x ≥ 0. Find the inverse derivative at x = 14.
Hint
The corresponding original input is 3.
Worked solution
f′(3) = 6, so (f⁻¹)′(14) = 1/6.
05 · Negative branch
Repeat question 04 when the original domain is x ≤ 0.
Hint
The corresponding original input is now −3.
Worked solution
f′(−3) = −6, so the inverse derivative is −1/6.
05 / Differentiate when x is expressed in terms of y
Take the easy derivative first, then invert it.
dy/dx = 1/(dx/dy), provided dx/dy ≠ 0
This applies locally where y is a differentiable function of x.
x = y³ + y; find dy/dx when y = 2Worked example
dx/dy = 3y² + 1
Differentiate x with respect to y.
At y = 2, dx/dy = 13
This is non-zero.
dy/dx = 1/13, at the point (10, 2)
The order of the coordinates is still (x, y).
If the question asks for an answer in terms of x, eliminate y where practical or use the given point. Otherwise a correctly stated derivative in terms of y may be appropriate.
06 · Exponential relation
If x = 2eʸ + 3, find dy/dx in terms of x and state the domain.
Hint
dx/dy = 2eʸ = x − 3.
Worked solution
dy/dx = 1/(x − 3), for x > 3. The inverse is ln[(x − 3)/2].
07 · Power relation
For x = y⁵ + 2y, find the tangent gradient dy/dx at y = −1.
Hint
The point is (−3, −1).
Worked solution
dx/dy = 5y⁴ + 2 = 7 at y = −1. The tangent gradient is 1/7.
06 / Recover the logarithm derivative
Use the inverse of the exponential.
y = ln x is equivalent to x = eʸWorked example
dx/dy = eʸ = x
The original exponential derivative is positive.
dy/dx = 1/x
The inverse exists for x > 0.
For y = logₐx: x = aʸ
Here a > 0 and a ≠ 1.
dy/dx = 1/(x ln a), x > 0
The sign is negative when 0 < a < 1.
08 · A base below one
Find the derivative of y = log₁⁄₂x at x = 4.
Hint
ln(1/2) = −ln 2.
Worked solution
The derivative is 1/[x ln(1/2)], so at x = 4 it is −1/(4ln 2). The logarithm is decreasing.
07 / Build tangents and normals at the reflected point
Reciprocal gradients and perpendicular gradients are different.
f(2) = 9 and f′(2) = 4; let g = f⁻¹Worked example
The inverse point is (9, 2)
Swap the coordinates.
Inverse tangent: y − 2 = (x − 9)/4
Use reciprocal slope 1/4.
Inverse normal: y − 2 = −4(x − 9)
Now take the negative reciprocal of the inverse tangent slope.
The original tangent and inverse tangent generally are not perpendicular. Their non-zero slopes multiply to +1, whereas perpendicular finite slopes multiply to −1.
09 · A negative tangent
Suppose f(−1) = 3 and f′(−1) = −2. Find the tangent and normal to the inverse at x = 3.
Hint
The inverse point is (3, −1).
Worked solution
The tangent is y + 1 = −(x − 3)/2. Its normal has slope 2, so y + 1 = 2(x − 3).
10 · Use the implicit point
For x = y³ + 2y, find the normal at y = 1.
Hint
Compute x and dx/dy at y = 1.
Worked solution
The point is (3, 1), with dx/dy = 5 and dy/dx = 1/5. The normal is y − 1 = −5(x − 3).
08 / Handle a zero original derivative
Do not write 1/0 as a finite slope.
The cubic f(x) = x³ is one-to-one on all real numbers. At zero f′(0) = 0. Its inverse g(x) = ∛x is continuous there, but its difference quotient g(h)/h = 1/|h|²ᐟ³ grows without bound as h tends to zero. The inverse has a vertical tangent, not a finite derivative.
A square root at its domain endpoint also has a vertical tangent in the one-sided sense. This does not make the endpoint an ordinary two-sided differentiable point.
11 · An inverse with a vertical tangent
For f(x) = x⁵, describe its inverse at the origin and its derivative away from zero.
Hint
Write x = y⁵ and differentiate with respect to y.
Worked solution
The inverse is the real fifth root. For x ≠ 0 its derivative is 1/[5|x|⁴ᐟ⁵], which is positive. It has a vertical tangent and no finite derivative at zero.
09 / Test the direction and the evaluation point
An inverse derivative is not the reciprocal of the function value.
12 · Compare three expressions
For f(x) = e²ˣ, compare (f⁻¹)′(e²), 1/f′(e²) and 1/f(e²).
Hint
For the first expression, solve f(a) = e².
Worked solution
a = 1, so (f⁻¹)′(e²) = 1/(2e²). The other two are 1/[2exp(2e²)] and exp(−2e²). They evaluate different quantities and are not interchangeable.
Write a small correspondence table before differentiating: original input a, original output b, original gradient f′(a), inverse gradient at b. This prevents evaluating a correct formula at the wrong input.
10 / Check branch, point and non-zero slope
A reciprocal rule needs a matching point.
- Choose a one-to-one branch before using an inverse function.
- If f(a) = b, use 1/f′(a) for the inverse derivative at b.
- Keep the sign: reflection uses a reciprocal, not a negative reciprocal.
- If dx/dy is easier, calculate it first and invert only where non-zero.
- Analyse zero-gradient points and domain endpoints separately.
Section 1 of 10 · Reflect the point and tangent