Second order derivative
NCERT Class 12 Mathematics Chapter 5 Section 5.7
Quick answer
The second order derivative is the derivative of dy/dx, written dy/dx, f″(x), y″ or y; it measures how fast the slope changes. To prove a relation between y and its derivatives, find y = dy/dx and y and substitute, clearing square roots first. For example, y = A sin x + B cos x gives y = −y, so y + y = 0.
With your textbook
NCERT Mathematics Part I (Class 12), Chapter 5 “Continuity and Differentiability”, Section 5.7 “Second Order Derivative”, pages 137–140.
Open the official NCERT chapter (PDF, 3.2 MB, ncert.nic.in) (opens in a new tab)
Checked against the 2026-27 reprint on .
What is a second order derivative?
Second order derivative
- If y = f(x) and its derivative f′(x) is differentiable, the derivative of dy/dx is the second order derivative, d/dx (dy/dx), written d2y/dx2 (NCERT Section 5.7, page 137).
- Other notations: f″(x), D2y, y″ or y2 (NCERT page 137); NCERT writes the first derivative as y1 (page 139).
We remark that higher order derivatives may be defined similarly.
NCERT Mathematics Part I (Class 12), Section 5.7, page 137
In plain words: differentiate once to get the slope, then differentiate the slope to see how fast the slope itself changes. For a moving object, if x is time and y is position, dy/dx is the velocity and d2y/dx2 the acceleration, the idea behind the velocity at an instant in Class 11.
NCERT Example 35 (page 138): for y = x3 + tan x, dy/dx = 3x2 + sec2 x. Differentiating again, the derivative of sec2 x is 2 sec x × sec x tan x, by the chain rule, so
d2y/dx2 = 6x + 2 sec2 x tan x.
Proving a relation between y and its derivatives
Many questions give y and ask you to show an equation linking y, y1 and y2. Find y1 and y2, then substitute.
NCERT Example 36 (page 138): for y = A sin x + B cos x, y1 = A cos x − B sin x and y2 = −A sin x − B cos x = −y. So
A relation that recurs
dy/dx + y = 0, for y = A sin x + B cos x
NCERT Example 37 (page 138): for y = 3e2x + 2e3x, y1 = 6e2x + 6e3x and y2 = 12e2x + 18e3x. Then y2 − 5y1 + 6y: the e2x terms give 12 − 30 + 18 = 0 and the e3x terms give 18 − 30 + 12 = 0, so y2 − 5y1 + 6y = 0.
Clear the roots first. NCERT Example 38 (page 139): for y = sin x, y1 = 1/√(1 − x2). Squaring and clearing the fraction, (1 − x2)y12 = 1. Differentiating with the product rule, (1 − x2) × 2y1y2 − 2x y12 = 0, and dividing by 2y1 (which is never 0):
(1 − x2)y2 − xy1 = 0.
Implicit and parametric second derivatives
NCERT sets both kinds in its Miscellaneous Exercise on Chapter 5 (Q15 and Q17, page 145).
Implicit. On a branch of x2 + y2 = 25 with y ≠ 0, differentiating gives 2x + 2yy1 = 0, so y1 = −x/y. Differentiate again by the quotient rule, then put in y1 = −x/y:
y2 = −(y − xy1)/y2
= −(x2 + y2)/y3 = −25/y3.
Parametric. If x and y are given in terms of t, find dy/dx in terms of t, differentiate that with respect to t, and divide by dx/dt, provided it is non-zero:
d2y/dx2 = [d/dt (dy/dx)] ÷ (dx/dt).
It is not (d2y/dt2) ÷ (d2x/dt2). For x = a(cos t + t sin t) and y = a(sin t − t cos t), the curve of NCERT Exercise 5.6, Q10 with t for θ, dx/dt = at cos t and dy/dx = tan t. So, wherever at cos t ≠ 0,
d2y/dx2 = sec2 t ÷ (at cos t)
= sec3 t/(at).
Solved examples
Example 1: Second derivatives (NCERT Exercise 5.7, Q3, Q5 and Q8, page 139)
Find d2y/dx2 for (i) y = x cos x; (ii) y = x3 log x; (iii) y = tan x.
(i) y1 = cos x − x sin x, so y2 = −sin x − sin x − x cos x. (ii) y1 = 3x2 log x + x2, so y2 = 6x log x + 3x + 2x. (iii) y1 = (1 + x2)−1, so y2 = −(1 + x2)−2 × 2x.
Answer: (i) −x cos x − 2 sin x; (ii) x(5 + 6 log x); (iii) −2x/(1 + x2)2.
Example 2: An exponential times a cosine (NCERT Exercise 5.7, Q7, page 139)
Find d2y/dx2 for y = e6x cos 3x.
y1 = 6e6x cos 3x − 3e6x sin 3x = 3e6x(2 cos 3x − sin 3x). Differentiating again, y2 = 18e6x(2 cos 3x − sin 3x) + 3e6x(−6 sin 3x − 3 cos 3x) = e6x(27 cos 3x − 36 sin 3x).
Answer: 9e6x(3 cos 3x − 4 sin 3x).
Example 3: In terms of y alone (NCERT Exercise 5.7, Q12, page 140)
If y = cos x, find d2y/dx2 in terms of y alone.
Here x = cos y. y1 = −1/√(1 − x2) = −1/sin y, for −1 < x < 1, since then sin y = √(1 − x2) > 0 for 0 < y < π. The endpoints have no finite first derivative. Then y2 = d/dx (−cosec y) = cosec y cot y × y1 = cosec y cot y × (−cosec y).
Answer: d2y/dx2 = −cot y cosec2 y.
Example 4: An equation with logarithms (NCERT Exercise 5.7, Q13, page 140)
If y = 3 cos (log x) + 4 sin (log x), show that x2y2 + xy1 + y = 0.
Put u = log x, so du/dx = 1/x and y = 3 cos u + 4 sin u. Then xy1 = −3 sin u + 4 cos u. Differentiating this product, xy2 + y1 = (−3 cos u − 4 sin u)/x, which is −y/x. Multiply by x.
Answer: x2y2 + xy1 = −y, that is x2y2 + xy1 + y = 0.
Example 5: Two exponentials (NCERT Exercise 5.7, Q14, page 140)
If y = Aemx + Benx, show that y2 − (m + n)y1 + mny = 0.
y1 = Amemx + Bnenx and y2 = Am2emx + Bn2enx. The coefficient of Aemx in the combination is m2 − (m + n)m + mn = 0, and likewise for Benx.
Answer: the combination is 0, as required. (NCERT Example 37 is the case A = 3, m = 2, B = 2, n = 3.)
Common mistakes to avoid
Common mistake
Squaring the first derivative instead of differentiating again
d2y/dx2 is not (dy/dx)2. For y = x3, y2 = 6x while y12 = 9x4.
Common mistake
Dropping the chain rule on the second step
In NCERT Example 35, the derivative of sec2 x is 2 sec x × (sec x tan x), not 2 sec x.
Common mistake
Fighting a square root
With y1 = 1/√(1 − x2), differentiating directly is messy. Square and clear the fraction first, as in NCERT Example 38, then differentiate.
Quick revision summary
- The second order derivative is d2y/dx2 = d/dx (dy/dx), also written f″(x), y″ or y.
- To prove a relation, find y = dy/dx and y and substitute; clear square roots and fractions before the second differentiation.
- Implicit and parametric curves: differentiate y again with the chain rule; in parametric form, d2y/dx2 = [d/dt (dy/dx)] ÷ (dx/dt).
- y = A sin x + B cos x gives y = −y, so y + y = 0; y = Aemx + Benx satisfies y − (m + n)y + mny = 0.
Formula sheet
- dy/dx = d/dx (dy/dx)
- y = A sin x + B cos x ⇒ y″ + y = 0
- y = sin x ⇒ (1 − x)y″ − xy′ = 0 (|x| < 1)
Questions students ask
What is the difference between the first and the second derivative?
The first derivative dy/dx is the slope of a graph, while the second derivative d2y/dx2 is the rate at which that slope changes, so it shows how the graph bends. Where f″(x) > 0 on an interval the slope increases and the graph bends upwards, like a cup, and where f″(x) < 0 it bends downwards (beyond NCERT, concave up and concave down); NCERT’s Chapter 6 uses the sign of f″ in the second derivative test. The signs of f′ and f″ are independent: sin x falls from π/2 to 3π/2, yet bends downwards before π, where y″ = −sin x < 0, and upwards after it.
Why is the second derivative written d2y/dx2?
The second derivative is written d2y/dx2 because it is d/dx applied twice to y: the two d’s collect on top and the two dx’s at the bottom, so the notation means two differentiations with respect to x, not differentiation with respect to x. This is operator notation, not an ordinary quotient that can be cancelled algebraically. Its units, as for the dimensions of any derivative, are those of y over the square of those of x: for a position in metres against time in seconds it is an acceleration, in m s−2. For a square’s area A = s2 against its side s, d2A/ds2 = 2, a pure number.
Is d2y/dx2 equal to 1/(d2x/dy2)?
No, d2y/dx2 is not the reciprocal of d2x/dy2, although dy/dx = 1/(dx/dy) wherever dx/dy ≠ 0: differentiating 1/(dx/dy) with respect to x by the chain rule gives d2y/dx2 as −d2x/dy2 divided by (dx/dy)3. For y = ex, that is x = log y, dx/dy = 1/y and d2x/dy2 = −1/y2, so the formula gives (1/y2) × y3 = y = ex, the right answer, while the reciprocal, −y2, is not even positive.
Is every differentiable function twice differentiable?
No, a function can be differentiable everywhere and still have no second derivative at some point. Take f(x) = 0 for x ≤ 0 and x2 for x > 0: its derivative is 0 for x < 0 and 2x for x > 0, and also 0 at x = 0, where both one-sided difference quotients tend to 0 (the right-hand one is h2/h = h). But f′ then has one-sided derivatives 0 and 2 at 0, a corner like that of |x| at 0, so f″(0) does not exist, which is why NCERT’s definition first asks that f′ be differentiable.
What are higher order derivatives, and are they in the JEE Main syllabus?
Higher order derivatives are the third, fourth and later derivatives, each the derivative of the one before: d3y/dx3 is the derivative of d2y/dx2, and the derivatives of sin x run cos x, −sin x, −cos x, then sin x again at the fourth. NCERT’s Chapter 5 only remarks that they may be defined similarly (page 137), and the JEE Main 2027 syllabus, Unit 7, names “derivatives of order upto two”. They return in its Unit 9, “Ordinary differential equations, their order and degree”, since the order of a differential equation is that of its highest derivative and can be 3 or more.
Practice questions
Try each question first, then open the answer to check your working.
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Q1. Find d2y/dx2 for (i) x20; (ii) log x; (iii) ex sin 5x (NCERT Exercise 5.7, Q2, Q4 and Q6, page 139).
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(i) 20 × 19 x18 = 380x18. (ii) y1 = 1/x, y2 = −1/x2. (iii) y1 = ex(sin 5x + 5 cos 5x), y2 = ex(sin 5x + 5 cos 5x + 5 cos 5x − 25 sin 5x) = 2ex(5 cos 5x − 12 sin 5x).
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Q2. If y = 500e7x + 600e−7x, show that d2y/dx2 = 49y (NCERT Exercise 5.7, Q15, page 140).
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y1 = 3500e7x − 4200e−7x and y2 = 24 500e7x + 29 400e−7x = 49(500e7x + 600e−7x) = 49y.
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Q3. If ey(x + 1) = 1, show that d2y/dx2 = (dy/dx)2 (NCERT Exercise 5.7, Q16, page 140).
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The equation requires x > −1. Take logarithms: y + log (x + 1) = 0, so y = −log (x + 1). Then y1 = −1/(x + 1) and y2 = 1/(x + 1)2 = y12.
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Q4. If y = 5 cos x − 3 sin x, prove that d2y/dx2 + y = 0 (NCERT Exercise 5.7, Q11, page 139).
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y1 = −5 sin x − 3 cos x and y2 = −5 cos x + 3 sin x = −y. (It is NCERT Example 36 with A = −3, B = 5.)