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Parametric differentiation looks like a free mark: divide dy/dt by dx/dt and you are done. The first derivative is that easy. The second derivative is where marks go, because the obvious shortcut is wrong. This post sets out both formulas, checks them against a curve you can also write as y = f(x), and works three examples, including one from projectile motion.
Where is this in NCERT and the syllabus?
The NCERT Class 12 chapter Continuity and Differentiability (Reprint 2026-27) covers "Derivatives of Functions in Parametric Forms" in section 5.6, with a relation x = f(t), y = g(t) and t as the parameter, and "Second Order Derivative" in section 5.7. The CBSE Mathematics curriculum for 2026-27 lists "derivative of functions expressed in parametric forms" and "Second order derivatives" in the same unit.
The JEE (Main) 2026 syllabus lists differentiation of "composite and implicit functions" and "derivatives of order upto two", but does not name parametric forms. The JEE (Advanced) 2026 brochure lists "derivatives of order two" and parametric equations of the circle, parabola, ellipse and hyperbola. So parametric differentiation is a direct CBSE topic and, for JEE, an application of the chain rule to curves the syllabus does name. Check the documents for your exam year.
What are the two formulas?
First derivative, from the chain rule (dy/dt = dy/dx × dx/dt):
dy/dx = (dy/dt)/(dx/dt), wherever dx/dt ≠ 0.
Second derivative. dy/dx is itself a function of t. To differentiate it with respect to x, differentiate with respect to t and divide by dx/dt again:
d²y/dx² = [d/dt (dy/dx)]/(dx/dt).
OpenStax Calculus Volume 2, section 7.2 derives the same formula from the definition d²y/dx² = d/dx(dy/dx).
Worked example 1: a curve you can check
Take x = t², y = t³ with t > 0.
dx/dt = 2t, dy/dt = 3t², so dy/dx = 3t²/2t = 3t/2.
d/dt(3t/2) = 3/2, so d²y/dx² = (3/2)/(2t) = 3/(4t).
Check by removing t: y = x^(3/2), so dy/dx = (3/2)x^(1/2) = 3t/2 and d²y/dx² = (3/4)x^(−1/2) = 3/(4t). Both match. The wrong shortcut gives (6t)/(2) = 3t, which fails the check.
Worked example 2: an ellipse
Take x = 3 cos t, y = 2 sin t, the ellipse x²/9 + y²/4 = 1.
dx/dt = −3 sin t, dy/dt = 2 cos t, so dy/dx = −(2/3) cot t.
d/dt[−(2/3) cot t] = (2/3) cosec²t. Divide by dx/dt: d²y/dx² = (2/3) cosec²t/(−3 sin t) = −2/(9 sin³t).
At t = π/4, the point is (3/√2, √2) ≈ (2.12, 1.41).
Slope: dy/dx = −2/3.
Second derivative: −2/(9 × (1/√2)³) = −4√2/9 ≈ −0.63. Negative, as expected on the upper half of an ellipse, which curves downward.
Tangent: y − √2 = −(2/3)(x − 3/√2), which simplifies to 2x + 3y = 6√2.
The wrong shortcut gives (−2 sin t)/(−3 cos t) = (2/3) tan t = +2/3 at π/4, the wrong sign as well as the wrong size.
Vertical tangents appear where dx/dt = 0 but dy/dt ≠ 0. Here that is t = 0 and t = π, the points (±3, 0).
Worked example 3: a projectile
A ball is thrown at speed u and angle α: x = ut cos α, y = ut sin α − ½gt².
dy/dx = (u sin α − gt)/(u cos α) = tan α − gt/(u cos α).
d/dt(dy/dx) = −g/(u cos α). Divide by dx/dt = u cos α: d²y/dx² = −g/(u² cos²α).
The second derivative is constant, which is exactly what a parabola needs. It matches the trajectory equation y = x tan α − gx²/(2u² cos²α). With u = 20 m/s, α = 30° and g = 9.8 m/s² (sample numbers), d²y/dx² = −9.8/(400 × 0.75) ≈ −0.0327 per metre. Here the wrong shortcut is not even defined, because d²x/dt² = 0.
Which mistakes cost marks?
Dividing second derivatives. Use the formula above every time.
Forgetting the second division by dx/dt. Differentiating dy/dx with respect to t gives d/dt(dy/dx), not d²y/dx².
Ignoring points where dx/dt = 0. The slope is undefined there; check for vertical tangents.
Sign slips with trigonometric derivatives, especially d/dt(cot t) = −cosec²t.
Not checking. When you can remove the parameter easily, do it once to confirm.
For curves given implicitly rather than parametrically, see our post on implicit and logarithmic differentiation. For tangent and normal questions, see tangents and normals.
A practice routine
Day 1: NCERT Exercise 5.6, questions 1 to 6, first derivatives only.
Day 2: find d²y/dx² for three of those, and check one by removing the parameter.
Day 3: redo Example 2 for x = a cos t, y = b sin t and show d²y/dx² = −b/(a² sin³t).
Day 4: for the cycloid x = t − sin t, y = 1 − cos t, find d²y/dx² at t = π/2. (Answer: −1.)
If you understand the formula but keep making algebra slips, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, take any parametric curve from your textbook and compute d²y/dx² both the right way and the wrong way, so you can see the difference.
