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"Differentiate sin²x with respect to e^(cos x)" looks like a new kind of question, but it uses a rule you already know. This post gives the rule, four worked examples with a check for each, and the second-derivative mistake that costs the most marks.
Where is this in NCERT and the JEE Main syllabus?
The NCERT Class 12 chapter Continuity and Differentiability (2026-27 reprint) has section 5.6, "Derivatives of Functions in Parametric Forms". If x = f(t) and y = g(t), then dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0. Differentiating u with respect to v is the same rule with x playing the part of t. The current book has no separate exercise with this heading; Exercise 5.6 and Miscellaneous questions 12 and 17 train the same skill.
The JEE (Main) 2026 syllabus lists differentiation of "composite and implicit functions" and "derivatives of order upto two". It does not name parametric forms, but the chain rule behind them is in it.
What is the rule?
If u and v are both functions of x, then
du/dv = (du/dx) ÷ (dv/dx), wherever dv/dx ≠ 0.
Think of x as a hidden parameter. You never need to write u as a function of v.
Example 1: a trigonometric pair
Differentiate u = sin²x with respect to v = e^(cos x).
du/dx = 2 sin x cos x.
dv/dx = e^(cos x) × (−sin x).
du/dv = 2 sin x cos x ÷ (−sin x e^(cos x)) = −2 cos x e^(−cos x), for sin x ≠ 0.
At x = π/3: −2 × (1/2) × e^(−1/2) = −e^(−1/2) ≈ −0.607.
Example 2: check by rewriting
Differentiate sec²x with respect to tan x. By the rule: (2 sec²x tan x) ÷ sec²x = 2 tan x.
Check: sec²x = 1 + tan²x = 1 + v², and the derivative of 1 + v² with respect to v is 2v = 2 tan x. When you can rewrite u in terms of v, do it once as a check.
Example 3: an inverse-trigonometric pair
Differentiate u = tan⁻¹[(√(1 + x²) − 1)/x] with respect to v = tan⁻¹x, for x ≠ 0.
Put x = tan θ with −π/2 < θ < π/2, so √(1 + x²) = sec θ (positive in this interval). Then (sec θ − 1)/tan θ = (1 − cos θ)/sin θ = tan(θ/2). Since θ/2 lies between −π/4 and π/4, inside the principal range, u = θ/2 = v/2. So du/dv = 1/2.
We checked this numerically at x = 0.3, 1.7 and −2.2, and the ratio of derivatives came out as 0.5 each time. Our post on substitutions in inverse-trig differentiation covers cases where the range forces a sign change.
Example 4: the second-derivative trap
Find d²u/dv² for u = x² and v = x³.
First derivative: du/dv = 2x ÷ 3x² = 2/(3x).
Second derivative: differentiate 2/(3x) with respect to x, then divide by dv/dx again:
d²u/dv² = (−2/(3x²)) ÷ 3x² = −2/(9x⁴).
Check: u = v^(2/3), so du/dv = (2/3)v^(−1/3) and d²u/dv² = −(2/9)v^(−4/3) = −2/(9x⁴). It matches.
The tempting shortcut, u″ ÷ v″ = 2 ÷ 6x = 1/(3x), is wrong. It does not even have the right sign. NCERT's Miscellaneous question 17 asks for a second derivative in parametric form, where the same trap applies.
Which mistakes cost marks?
Dividing the wrong way round: du/dv needs du/dx on top.
Forgetting the chain-rule factor, such as the −sin x in Example 1.
Ignoring points where dv/dx = 0; the rule fails there.
Using u″/v″ for the second derivative.
In inverse-trig substitutions, not checking that θ/2 stays inside the principal range.
Why explain each step aloud?
In a classic study, Chi and colleagues (1989) found that students who learned well from worked examples produced many self-explanations linking each step to a principle, while weaker students produced few and relied on copying examples later. With this rule, the explanation is short: "x is the hidden parameter, so I divide the two x-derivatives."
A practice routine
Do NCERT Exercise 5.6 questions 1 to 6 and say the rule aloud for each.
Write three "u with respect to v" pairs of your own and check one by rewriting, as in Example 2.
Find d²u/dv² for two of them, then check with a direct method.
Mix in two implicit-differentiation questions from our logarithmic and implicit differentiation guide, so you practise choosing the method.
If a chain-rule factor keeps disappearing, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can see the exact step where it dropped out.
Try this now: differentiate x³ with respect to x², then find the second derivative, and check by writing x³ as a power of x².
