# Parametric Differentiation

> Parametric differentiation for JEE Mathematics: dy/dx = (dy/dt)/(dx/dt), the second-derivative trap, ellipse, parabola and cycloid parametrisations worked through.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/parametric-differentiation
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Parametric Differentiation", PrepElephant, https://prepelephant.com/topics/jee/mathematics/parametric-differentiation

## Direct answer

When x and y are both driven by a parameter t, the derivative is a ratio, not a chain: dy/dx = (dy/dt)/(dx/dt), valid wherever dx/dt ≠ 0. The second derivative is where marks are won and lost — it is d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt), obtained by treating dy/dx as a new function of t and differentiating once more with respect to t, then dividing by dx/dt. It is emphatically not the ratio of second derivatives (d²y/dt²)/(d²x/dt²). Standard parametrisations to recognise on sight: the parabola (at^2, 2at) giving dy/dx = 1/t, the ellipse (a cos θ, b sin θ) giving dy/dx = -(b/a) cot θ, and the cycloid x = a(θ - sin θ), y = a(1 - cos θ) giving dy/dx = cot(θ/2).

## What you must remember

- **First derivative:** dy/dx = (dy/dt)/(dx/dt) — chain rule rearranged; it fails only where dx/dt = 0 (a vertical tangent, like θ = 0 on the cycloid).
- **Second derivative:** d²y/dx² = [d/dt(dy/dx)] / (dx/dt); never compute it as (y'')/(x'') of the separate second derivatives.
- **Parabola (at^2, 2at):** dy/dx = 2a/(2at) = 1/t; the tangent ty = x + at^2 follows in one step.
- **Ellipse (a cos θ, b sin θ):** dy/dx = (b cos θ)/(-a sin θ) = -(b/a) cot θ; horizontal tangent at θ = π/2, vertical at θ = 0.
- **Cycloid:** x = a(θ - sin θ), y = a(1 - cos θ); dy/dx = sin θ/(1 - cos θ) = cot(θ/2), so the cusp at θ = 0 has a vertical tangent.
- **Tangent and normal equations:** at parameter t, the tangent is y - y(t) = (dy/dx)(x - x(t)) — parametric questions usually end here.
- **Speed interpretation:** √((dx/dt)^2 + (dy/dt)^2) is the speed along the curve; physics-flavoured questions test the same derivatives in disguise.

## The second derivative done correctly

Differentiate the cycloid twice and the trap becomes visible. First, dy/dx = a sin θ / (a(1 - cos θ)) = sin θ/(1 - cos θ); the half-angle identity simplifies this to cot(θ/2). For the second derivative, differentiate cot(θ/2) with respect to θ: -(1/2) cosec^2(θ/2). Then divide by dx/dθ = a(1 - cos θ) = 2a sin^2(θ/2): d²y/dx² = -(1/2) cosec^2(θ/2) ÷ (2a sin^2(θ/2)) = -(1/4a) cosec^4(θ/2). Check what the wrong route would have produced: (d²y/dθ²)/(d²x/dθ²) = (a cos θ)/(a sin θ) = cot θ — a completely different function, and instant zero marks. The logic is simple: d²y/dx² means d/dx(dy/dx), and since dy/dx is known as a function of θ, the x-derivative must travel through the θ-derivative and divide by dx/dθ. Write that sentence in your solution and the error cannot recur.

## How the exam frames it

JEE Main keeps this mechanical: given x = f(t), y = g(t), find dy/dx or d²y/dx² at a stated parameter — the ellipse and parabola parametrisations dominate, with sin/cos or t and t^2 pairings. Advanced prefers the cycloid and astroid (x = a cos^3 θ, y = a sin^3 θ, where dy/dx = -tan θ), asks where the tangent is horizontal or vertical, or requests tangent/normal equations that feed into length-of-tangent or area questions. The near-certain distractor in any second-derivative multiple choice is the ratio of second derivatives — often planted as the first option. A second, quieter trap: evaluating at a point where dx/dt = 0 and reporting "derivative does not exist" when the true answer is a vertical tangent (infinite slope); check dx/dt before concluding. Parametric differentiation sits inside the calculus unit (application of derivatives) for both papers and reliably yields at least one question.

## Frequently asked questions

### What is the formula for dy/dx when x and y are functions of t?

dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0; where dx/dt = 0 the tangent is vertical.

### Why is d²y/dx² not equal to (d²y/dt²)/(d²x/dt²)?

Because d²y/dx² means differentiating dy/dx (a function of t) with respect to x, so you must first differentiate with respect to t and then divide by dx/dt: d²y/dx² = [d/dt(dy/dx)]/(dx/dt).

### What is dy/dx for the ellipse x = a cos θ, y = b sin θ?

-(b/a) cot θ, zero (horizontal tangent) at θ = π/2 and infinite (vertical tangent) at θ = 0.

### What is the slope of the cycloid at a general point?

cot(θ/2), from sin θ/(1 - cos θ); near the cusp at θ = 0 the slope blows up, a vertical tangent.

### How do you write the tangent at a parametric point?

y - y(t) = (dy/dx)|_t × (x - x(t)), substituting the ratio (dy/dt)/(dx/dt) evaluated at that parameter value.