Parametric Differentiation
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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.