IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~10 min read

Differentiating Inverse Functions

Some inverses are ugly to write explicitly — what’s the inverse of x³ + 2x + 1? The trick: don’t bother inverting. If y = f⁻¹(x), just rewrite as x = f(y), differentiate that with respect to y, then flip: dy/dx = 1/(dx/dy). The same idea proves d/dx[³√x], d/dx[ln x], and d/dx[arcsin x] — three results you’ll meet again before this chapter is done.

📘 What you need to know

The setup: if y = f⁻¹(x), then by the cancellation property f(y) = x. Write it as x = f(y) — that’s the form you’ll differentiate.
The key formula: dy/dx = 1/(dx/dy), provided dx/dy ≠ 0. Comes from the chain rule: dx/dx = (dx/dy)(dy/dx) = 1.
The result is in terms of y, not x. That’s expected — leave it in y unless the question demands x.
To get the gradient at a specific x-value: find the corresponding y first by solving f(y) = x, then substitute that y into dy/dx.
Geometric meaning: f and f⁻¹ are reflections of each other across the line y = x. At corresponding points (a, b) on f and (b, a) on f⁻¹, the tangent slopes are RECIPROCALS.
Where it breaks: dx/dy = 0 means f has a horizontal tangent — and f⁻¹ has a VERTICAL tangent there, so dy/dx is undefined at that point.
Use the formula booklet for the standard inverse-derivative results (d/dx[ln x] = 1/x, d/dx[arcsin x] = 1/√(1−x²), etc.). Each one PROVED via this technique.
For tangent/normal problems at a point on y = f⁻¹(x), find dy/dx as the tangent slope; normal slope is −1/(dy/dx).

The technique — rewrite, differentiate, reciprocate

Inverse function derivative If y = f⁻¹(x), then x = f(y) and dydx = 1dx/dy

The formula comes from the chain rule applied to the identity x = x: differentiating both sides with respect to x gives 1 = (dx/dy) · (dy/dx), so the two derivatives multiply to 1 — they’re reciprocals.

Why it works

dxdx = dxdy · dydx = 1

chain rule + the fact that dx/dx = 1 ⟹ the derivatives are reciprocals

When it fails

dx/dy = 0

if f has a horizontal tangent at y, then f⁻¹ has a vertical tangent at x — dy/dx undefined there

Geometric meaning — reflection across y = x

The graph of y = f⁻¹(x) is the graph of y = f(x) reflected across the line y = x. Reflection swaps coordinates: a point (a, b) on f becomes (b, a) on f⁻¹. And the slope of the reflected tangent is the RECIPROCAL of the original — which is exactly what the formula says.

The blue curve y = x² and the orange curve y = √x are reflections across the dashed purple line y = x. At point P = (1.5, 2.25) on f, the tangent has slope 3. At the reflected point P’ = (2.25, 1.5) on f⁻¹, the tangent has slope 1/3 — exactly the reciprocal.

The intuition: reflection swaps “rise” and “run”. If at P the tangent rises 3 for every 1 across, then after reflecting both axes are swapped, so at P’ it rises 1 for every 3 across — slope 1/3.

🧭 Recipe — differentiate y = f⁻¹(x)

Rewrite y = f⁻¹(x) as x = f(y) — apply f to both sides. You do NOT need to invert f explicitly.
Differentiate x = f(y) with respect to y to get dx/dy (an expression in y).
Reciprocate: dy/dx = 1/(dx/dy). The result is still in terms of y.
For a specific x-value: solve f(y) = x to find the corresponding y, then substitute into dy/dx. For an expression in x, substitute y = f⁻¹(x) back in (often via the relation f(y) = x).
Use the gradient for tangent or normal lines: tangent slope = dy/dx; normal slope = −1/(dy/dx) = −(dx/dy).

Worked examples

WE 1

Express the derivative of an inverse in terms of y

The function f is defined by f(x) = 2x³ + 5x − 1. Find an expression in terms of y for the derivative dydx of y = f⁻¹(x).

Step 1 — rewrite y = f⁻¹(x) as x = f(y) y = f⁻¹(x) ⟹ x = f(y) = 2y³ + 5y − 1 Step 2 — differentiate with respect to y dx/dy = 6y² + 5 Step 3 — reciprocate dy/dx = 1/(dx/dy) = 1/(6y² + 5) dy/dx = 1/(6y² + 5) notice you never have to invert f explicitly — solving 2y³ + 5y − 1 = x for y is a cubic equation. The inverse-function technique sidesteps it completely.

WE 2

Gradient of the inverse at a specific point

The function f is defined by f(x) = x² + 4x for x ≥ −2. Find the gradient of the curve y = f⁻¹(x) at the point where y = 1.

Step 1 — rewrite as x = f(y) x = y² + 4y Step 2 — differentiate dx/dy = 2y + 4 Step 3 — reciprocate dy/dx = 1/(2y + 4) Step 4 — substitute y = 1 dy/dx|_y=1 = 1/(2(1) + 4) = 1/6 gradient = 1/6 since the y-coordinate was given, no need to solve for it from the original equation. If only x had been given (say, “at x = 5”), you’d first solve y² + 4y = 5 to find y = 1.

WE 3

Derive d/dx[³√x] using the inverse-function technique

By considering y = ³√x as the inverse of y = x³, show that ddx [³√x] = 13x^2/3.

Step 1 — y = ³√x = x^(1/3); the inverse relation gives x = y³ y = x^(1/3) ⟹ x = y³ Step 2 — differentiate x = y³ with respect to y dx/dy = 3y² Step 3 — reciprocate dy/dx = 1/(3y²) Step 4 — convert back to x using y = x^(1/3) y² = (x^(1/3))² = x^(2/3) dy/dx = 1/(3 · x^(2/3)) = (1/3) · x^(−2/3) ✓ d/dx[³√x] = 1/(3 x^(2/3)) = (1/3) x^(−2/3) this also matches the power rule: d/dx[x^n] = n · x^(n−1), with n = 1/3 giving (1/3) · x^(−2/3). The inverse-function technique PROVES the power rule for fractional exponents.

WE 4

Derive d/dx[arcsin x] using the inverse-function technique

By considering y = arcsin x as the inverse of y = sin x on its principal domain, show that ddx [arcsin x] = 1√(1 − x²).

Step 1 — y = arcsin x ⟹ x = sin y, where −π/2 ≤ y ≤ π/2 x = sin y Step 2 — differentiate with respect to y dx/dy = cos y Step 3 — reciprocate dy/dx = 1/cos y Step 4 — express cos y in terms of x using cos²y + sin²y = 1 cos²y = 1 − sin²y = 1 − x² on y ∈ [−π/2, π/2], cos y ≥ 0, so cos y = +√(1 − x²) ⟹ dy/dx = 1/√(1 − x²) ✓ d/dx[arcsin x] = 1/√(1 − x²) the “+” sign on the square root matters: it’s forced by arcsin’s principal range. If we’d allowed any y with sin y = x, cos y could be negative and the formula wouldn’t be single-valued.

WE 5

Equation of the tangent to y = f⁻¹(x) at a given x-value

The function f is defined by f(x) = x³ + 2x + 1. Find the equation of the tangent to the curve y = f⁻¹(x) at the point where x = 4.

Step 1 — find the y-coordinate by solving f(y) = 4 y³ + 2y + 1 = 4 ⟹ y³ + 2y − 3 = 0 try y = 1: 1 + 2 − 3 = 0 ✓ point of tangency is (4, 1) Step 2 — x = y³ + 2y + 1 ⟹ differentiate dx/dy = 3y² + 2 Step 3 — reciprocate and evaluate at y = 1 dy/dx = 1/(3y² + 2) at y = 1: dy/dx = 1/(3 + 2) = 1/5 Step 4 — tangent through (4, 1) with slope 1/5 y − 1 = (1/5)(x − 4) 5y − 5 = x − 4 ⟹ x − 5y + 1 = 0 (or y = (x + 1)/5) tangent: y = (x + 1)/5 the trial-and-error guess for y³ + 2y − 3 = 0 worked because the question was designed with integer y. On a calculator paper, you’d just use the GDC’s equation solver.

WE 6

Equation of the normal to y = f⁻¹(x) — function involving e^x

The function f is defined by f(x) = e^x + x. Find the equation of the normal to the curve y = f⁻¹(x) at the point where x = 1.

Step 1 — find the y-coordinate by solving f(y) = 1 e^y + y = 1 try y = 0: e^0 + 0 = 1 + 0 = 1 ✓ point is (1, 0) Step 2 — x = e^y + y ⟹ differentiate dx/dy = e^y + 1 Step 3 — reciprocate and evaluate at y = 0 dy/dx = 1/(e^y + 1) at y = 0: dy/dx = 1/(1 + 1) = 1/2 Step 4 — slope of normal = −1/(slope of tangent) slope of normal = −1/(1/2) = −2 Step 5 — normal through (1, 0) with slope −2 y − 0 = −2(x − 1) ⟹ y = −2x + 2 (or 2x + y = 2) normal: y = −2x + 2 because dy/dx is in terms of y, you couldn’t have plugged x = 1 directly — you HAD to find y = 0 first. This is the standard order of operations for inverse-function gradients.

💡 Top tips

Set out the working clearly: write the original f(x), the inverse y = f⁻¹(x), the rewrite x = f(y), then dx/dy and dy/dx. Each on its own line.
Never try to invert f explicitly if it’s ugly. The point of the technique is to skip that step entirely.
The answer in terms of y is fine for most questions. Only convert to x if explicitly asked.
For specific-point questions: identify whether x or y is given. If x, solve for y first; if y, substitute directly.
Tangent slope vs normal slope: tangent = dy/dx; normal = −(dx/dy) — the negative reciprocal. Easy to confuse on Paper 1 under time pressure.

⚠ Common mistakes

Substituting an x-value into dy/dx(y) — the variable in the derivative is y, not x. Find the y first.
Inverting wrong: dy/dx = 1/(dx/dy), NOT dy/dx = dx/dy. Reciprocals, not “swap and keep”.
Forgetting the principal-range argument in WE 4 — without it, cos y could be ±√(1−x²), making d/dx[arcsin x] ambiguous. The range pins the sign.
Chain-rule slip when differentiating x = f(y) with composite f — the variable is y, so terms like d/dy[(2y+3)⁴] = 4(2y+3)³ · 2, NOT 4(2y+3)³.
Confusing tangent and normal: slope of normal is −1/(tangent slope). Drop the minus sign and you’ve found the wrong line.

Up next: Implicit Differentiation. When the equation can’t be solved for y at all — like x² + y² = 25 (a circle) — you differentiate both sides with respect to x, treating y as an implicit function of x. The chain rule gives d/dx[g(y)] = g′(y) · dy/dx, and you rearrange to isolate dy/dx. Same chain-rule idea as this note, applied to a much wider class of curves.

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