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"If y = (sin⁻¹x)², show that (1 − x²)y″ − xy′ = 2." Many students differentiate twice, get a page of fractions and square roots, and cannot see how the relation appears. The relation appears easily if you rearrange before the second differentiation. This post gives that method, works three examples, and covers the parametric case where the usual instinct gives the wrong answer.
Where is this in the syllabus?
Section 5.7, Second Order Derivative, of the NCERT Class 12 chapter Continuity and Differentiability (2026-27 reprint) defines d²y/dx², also written y″ or y2, and its examples include y = A sin x + B cos x giving y″ + y = 0, and y = sin⁻¹x giving (1 − x²)y″ − xy′ = 0. Its exercises ask for relations such as x²y2 + xy1 + y = 0. The JEE (Main) 2026 syllabus lists "derivatives of order upto two", and the JEE (Advanced) 2026 brochure lists "derivatives of order two" and "derivatives up to order two of implicit functions". Check your year's syllabus.
What is the method?
Find y′.
Rearrange so that y′ is multiplied by whatever was in the denominator, and square if there is a square root. You want a clean equation linking x, y and y′.
Differentiate that equation, using the product rule.
Divide by any common factor (often 2y′) and tidy up.
NCERT shows both routes for y = sin⁻¹x: differentiating the fraction directly, and the "alternative" of writing (1 − x²)y1² = 1 first. The second is shorter, and on harder functions it is much shorter.
Worked example 1: (sin⁻¹x)²
Let y = (sin⁻¹x)². Show (1 − x²)y″ − xy′ = 2.
y′ = 2 sin⁻¹x / √(1 − x²).
Multiply across: √(1 − x²) y′ = 2 sin⁻¹x.
Square: (1 − x²)(y′)² = 4(sin⁻¹x)² = 4y. Now there are no roots and no inverse functions.
Differentiate: (1 − x²) · 2y′y″ − 2x(y′)² = 4y′.
Divide by 2y′ (it is not zero for x ≠ 0): (1 − x²)y″ − xy′ = 2.
Notice the step that made it work: replacing (sin⁻¹x)² by y. Use the original function to simplify whenever you can.
Worked example 2: a relation with constant coefficients
Let y = e^(2x)(A cos x + B sin x). Show y″ − 4y′ + 5y = 0.
y′ = 2e^(2x)(A cos x + B sin x) + e^(2x)(−A sin x + B cos x) = 2y + u, where u = e^(2x)(−A sin x + B cos x).
u′ = 2u + e^(2x)(−A cos x − B sin x) = 2u − y.
y″ = 2y′ + u′ = 2y′ + 2u − y. From the first line, u = y′ − 2y, so y″ = 2y′ + 2(y′ − 2y) − y = 4y′ − 5y.
Hence y″ − 4y′ + 5y = 0.
Naming the "other" bracket u keeps the work short. Without it you expand eight terms and collect them. If you meet linear differential equations later, you will recognise 2 ± i as the roots of m² − 4m + 5 = 0. Our post on differential equations for JEE covers where such equations come from.
Worked example 3: a parametric curve
x = t², y = t³. Find d²y/dx².
dy/dx = (dy/dt)/(dx/dt) = 3t²/(2t) = 3t/2.
d²y/dx² = d/dx (3t/2) = [d/dt (3t/2)] ÷ (dx/dt) = (3/2)/(2t) = 3/(4t).
The tempting wrong answer divides the second derivatives: (d²y/dt²)/(d²x/dt²) = 6t/2 = 3t. That is not d²y/dx². Check with the Cartesian form: y = x^(3/2) for t > 0, so y″ = (3/4)x^(−1/2) = 3/(4t). The two agree.
A second check on a curve you know: for the cycloid x = a(t − sin t), y = a(1 − cos t), the same method gives d²y/dx² = −1/[a(1 − cos t)²], which is −1/a at t = π/2.
Common mistakes and how to avoid them
Differentiating the fraction form directly. It works, but it is long and error-prone. Rearrange first.
Dividing by y′ without comment. Note that y′ ≠ 0, or state the values of x excluded.
Forgetting the chain rule on (y′)². Its derivative is 2y′y″, not 2y′.
Treating d²y/dx² as a ratio in parametric problems. Differentiate dy/dx with respect to t, then divide by dx/dt once more.
If implicit functions are the weak point, our post on implicit and logarithmic differentiation covers second derivatives of curves like x² + y² = r².
A practice routine
Day 1: Redo NCERT Examples 36 to 38 without looking, then compare.
Day 2: Show that y = log(x + √(x² + 1)) satisfies (1 + x²)y″ + xy′ = 0, and that y = e^(a sin⁻¹x) satisfies (1 − x²)y″ − xy′ − a²y = 0.
Day 3: Three parametric second derivatives, each checked against a Cartesian form where one exists.
Day 4: Five mixed relations, timed at four minutes each.
If you know the rules but freeze on the first rearrangement, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, prove the relation for y = e^(a sin⁻¹x) by squaring the y′ equation first.
