MathsISC Class 12 · Continuity, Differentiability and Differentiation

Second-Order Derivatives and Parametric Differentiation

Updated 15 October 2026 · Fact-checked

The second derivative d²y/dx² is the derivative of dy/dx. For parametric curves x = f(t), y = g(t), dy/dx = (dy/dt)/(dx/dt), and d²y/dx² is the t-derivative of dy/dx divided by dx/dt. In "show that" proofs, clear fractions after the first derivative, then differentiate again.

Understand Second-Order and Parametric Derivatives

Second derivatives. If $y = f (x)$ , its derivative $\frac{d y}{d x}$ (also written $y_{1}$ or $y^{'}$ ) is again a function of $x$ . Differentiating once more gives the second-order derivative $\frac{d ^{2} y}{d x ^{2}}$ (also $y_{2}$ or $y^{''}$ ). The CISCE syllabus covers successive differentiation up to the second order. The second derivative measures how fast the slope is changing: in motion, if $s$ is position, $\frac{d s}{d t}$ is velocity and $\frac{d ^{2} s}{d t ^{2}}$ is acceleration; on a graph, it tells you whether the curve bends upwards or downwards.

Proof questions. The standard ISC question gives $y$ in terms of $x$ and asks you to show that some combination such as $(1 - x^{2}) y_{2} - x y_{1} = 2$ holds. Two techniques make these short:
1. Rearrange after the first derivative. Before differentiating again, multiply out any fractions or square roots so that the equation has no denominators. For $y = (sin^{- 1} x)^{2}$ : $y_{1} = \frac{2 s i n ^{- 1} x}{1 - x ^{2}}$ , so $1 - x^{2} y_{1} = 2 sin^{- 1} x$ . Squaring gives $(1 - x^{2}) y_{1}^{2} = 4 y$ , which is free of $sin^{- 1}$ and roots. Differentiating that is easy. The ISC 2025 examiners gave exactly this advice after candidates struggled with the second differentiation.
2. **Replace the original function by $y$ .** Wherever the original expression reappears, write $y$ . The target relation is usually in $x$ , $y$ , $y_{1}$ , $y_{2}$ only.
When differentiating a product like $(1 - x^{2}) y_{1}^{2}$ , use the product rule and the chain rule: $\frac{d}{d x} (y_{1}^{2}) = 2 y_{1} y_{2}$ . Then divide through by a common factor (often $2 y_{1}$ ) at the end.

Parametric functions. A curve can be described by $x = f (t)$ , $y = g (t)$ , with $t$ a parameter. Then
$\frac{d y}{d x} = \frac{d y / d t}{d x / d t}, \frac{d x}{d t} \neq = 0.$
For the second derivative, remember that $\frac{d y}{d x}$ is a function of $t$ , not of $x$ . So
$\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{\frac{d}{d t} ( \frac{d y}{d x} )}{d x / d t} .$
The commonest mistake is to write $\frac{d ^{2} y}{d x ^{2}} = \frac{d ^{2} y / d t ^{2}}{d ^{2} x / d t ^{2}}$ , which is wrong. For the parabola $x = a t^{2}$ , $y = 2 a t$ : $\frac{d y}{d x} = \frac{2 a}{2 a t} = \frac{1}{t}$ , and $\frac{d ^{2} y}{d x ^{2}} = \frac{- 1/ t ^{2}}{2 a t} = - \frac{1}{2 a t ^{3}}$ .

Standard second-derivative facts.

$y = A cos k x + B sin k x \Rightarrow y_{2} = - k^{2} y$ .
$y = A e^{m x} + B e^{n x}$ satisfies $y_{2} - (m + n) y_{1} + mn y = 0$ .
$y = sin (lo g x)$ or $cos (lo g x)$ satisfies $x^{2} y_{2} + x y_{1} + y = 0$ .

These patterns often show what the final relation should look like.

Link with applications. The second derivative test for maxima and minima, on the applications-of-derivatives pages, uses $\frac{d ^{2} y}{d x ^{2}}$ : negative means a local maximum, positive a local minimum.

Second-Order and Parametric Derivatives: formulas to remember

Second derivative: $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{d y}{d x})$; Also written y₂ or y″.
Parametric first derivative: $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$; dx/dt ≠ 0.
Parametric second derivative: $\frac{d ^{2} y}{d x ^{2}} = \frac{\frac{d}{d t} ( \frac{d y}{d x} )}{d x / d t}$; Not (d²y/dt²) ÷ (d²x/dt²).
Derivative of y₁²: $\frac{d}{d x} (y_{1}^{2}) = 2 y_{1} y_{2}$; Chain rule, used after squaring in proofs.
Trigonometric solutions: $y = A cos k x + B sin k x \Rightarrow y_{2} + k^{2} y = 0$; Differentiating sin or cos twice brings back the function with −k².
Exponential solutions: $y = A e^{m x} + B e^{n x} \Rightarrow y_{2} - (m + n) y_{1} + mn y = 0$; m and n are the roots of r² − (m + n)r + mn = 0.
Motion: $v = \frac{d s}{d t}, a = \frac{d ^{2} s}{d t ^{2}}$; Acceleration is the second derivative of position.

How to solve Second-Order and Parametric Derivatives questions

For "show that" questions on second derivatives, and for parametric second derivatives, use these steps.

1Find $y_{1}$ carefully, using the chain, product or quotient rule as needed.
2Rearrange $y_{1}$ to remove fractions and roots: multiply by the denominator, and square if a square root remains. Replace the original function by $y$ where possible.
3Differentiate the rearranged equation with respect to $x$ , using the product rule on every product and $\frac{d}{d x} (y_{1}^{2}) = 2 y_{1} y_{2}$ .
4Divide by any common factor (often $2 y_{1}$ ) and rearrange into the required form.
5For parametric curves, find $\frac{d x}{d t}$ and $\frac{d y}{d t}$ , then $\frac{d y}{d x}$ as their quotient, simplified in terms of $t$ .
6Differentiate $\frac{d y}{d x}$ with respect to $t$ , then divide by $\frac{d x}{d t}$ to get $\frac{d ^{2} y}{d x ^{2}}$ .
7If a value is asked at a point, substitute the value of $t$ only after simplifying.

Second-Order and Parametric Derivatives in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Second derivative as acceleration

Suppose a test car on a straight track is $s (t) = t^{3} - 6 t^{2} + 15 t$ metres from the start after $t$ seconds.

Find its velocity and acceleration. When is the acceleration zero, and what is the velocity then? Is the car ever at rest?

Try it first, then show the solutionSolution

Velocity $v = \frac{d s}{d t} = 3 t^{2} - 12 t + 15$ ; acceleration $a = \frac{d ^{2} s}{d t ^{2}} = 6 t - 12$ .
$a = 0$ when $t = 2$ s. Then $v = 12 - 24 + 15 = 3$ m/s.
Before $t = 2$ the car slows down ( $a < 0$ ), after it speeds up ( $a > 0$ ), so 3 m/s is its smallest velocity.
$v = 3 (t^{2} - 4 t + 5) = 3 [(t - 2)^{2} + 1] > 0$ , so the car is never at rest.

Answer: $v = 3 t^{2} - 12 t + 15$ , $a = 6 t - 12$ ; at $t = 2$ s the acceleration is zero and the velocity is 3 m/s, its minimum. The car never stops.

Memory hook: Position, velocity, acceleration: differentiate once for speed, twice for the push.

Parametric second derivative

Imagine a Ferris wheel of radius 10 m. Measured from the lowest point, a seat is at $x = 10 sin t$ , $y = 10 - 10 cos t$ after turning through angle $t$ .

Find $\frac{d y}{d x}$ and $\frac{d ^{2} y}{d x ^{2}}$ in terms of $t$ , and their values at $t = \frac{π}{3}$ .

Try it first, then show the solutionSolution

$\frac{d x}{d t} = 10 cos t$ , $\frac{d y}{d t} = 10 sin t$ , so $\frac{d y}{d x} = tan t$ .
$\frac{d ^{2} y}{d x ^{2}} = \frac{s e c ^{2} t}{10 c o s t} = \frac{1}{10 c o s ^{3} t}$ .
At $t = \frac{π}{3}$ : $\frac{d y}{d x} = 3$ ; $cos^{3} \frac{π}{3} = \frac{1}{8}$ , so $\frac{d ^{2} y}{d x ^{2}} = \frac{8}{10} = 0.8$ .

Answer: $\frac{d y}{d x} = tan t$ , $\frac{d ^{2} y}{d x ^{2}} = \frac{1}{10 c o s ^{3} t}$ ; at $t = \frac{π}{3}$ these are $3$ and $0.8$ .

Memory hook: For a parametric curve, the second derivative is "d/dt of the slope, over dx/dt".

Tricks to remember Second-Order and Parametric Derivatives

Short memory aids. Say each one aloud once, then use it on a question the same day.

Clear, then differentiate
After y₁, clear every fraction and root before differentiating again.
Use it when: All "show that" second-derivative proofs.
Slope over speed
Parametric d²y/dx² = (d/dt of the slope) ÷ (dx/dt).
Use it when: Avoiding the wrong ratio of second derivatives.
Name it y
Whenever the original expression reappears, write y.
Use it when: Reaching a relation in x, y, y₁, y₂ only.
Square then halve
Squaring y₁ leads to 2y₁y₂ on differentiating; divide by 2y₁ at the end.
Use it when: Proofs with sin⁻¹x or square roots.

Second-Order and Parametric Derivatives: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A direct second derivative · 1 mark · MCQBoard level

If $y = x^{3} - 2 x$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $x = 1$ is:
(a) 1
(b) 3
(c) 6
(d) 0

Try it first, then show the solutionSolution

$y_{1} = 3 x^{2} - 2$ , $y_{2} = 6 x$ , so $y_{2} (1) = 6$ .

Answer: (c) 6.

Example 2: Parametric parabola · 2 marks · Short answerBoard level

If $x = a t^{2}$ and $y = 2 a t$ , find $\frac{d y}{d x}$ and $\frac{d ^{2} y}{d x ^{2}}$ .

Try it first, then show the solutionSolution

$\frac{d x}{d t} = 2 a t$ , $\frac{d y}{d t} = 2 a$ , so $\frac{d y}{d x} = \frac{1}{t}$ . (1 mark)
$\frac{d ^{2} y}{d x ^{2}} = \frac{\frac{d}{d t} ( 1/ t )}{d x / d t} = \frac{- 1/ t ^{2}}{2 a t} = - \frac{1}{2 a t ^{3}}$ . (1 mark)

Answer: $\frac{d y}{d x} = \frac{1}{t}$ , $\frac{d ^{2} y}{d x ^{2}} = - \frac{1}{2 a t ^{3}}$ .

Example 3: An exponential combination · 3 marks · Short answerHarder board level

If $y = A e^{2 x} + B e^{- 3 x}$ , show that $\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 6 y = 0$ .

Try it first, then show the solutionSolution

$y_{1} = 2 A e^{2 x} - 3 B e^{- 3 x}$ . (1 mark)
$y_{2} = 4 A e^{2 x} + 9 B e^{- 3 x}$ . (1 mark)
$y_{2} + y_{1} - 6 y = (4 + 2 - 6) A e^{2 x} + (9 - 3 - 6) B e^{- 3 x} = 0$ . (1 mark)

Answer: Proved.

Example 4: x sin(log x) · 3 marks · Short answerHarder board level

If $y = x sin (lo g x)$ , prove that $x^{2} \frac{d ^{2} y}{d x ^{2}} - x \frac{d y}{d x} + 2 y = 0$ .

Try it first, then show the solutionSolution

$y_{1} = sin (lo g x) + x cos (lo g x) \cdot \frac{1}{x} = sin (lo g x) + cos (lo g x)$ . (1 mark)
$y_{2} = \frac{c o s ( l o g x ) - s i n ( l o g x )}{x}$ , so $x^{2} y_{2} = x cos (lo g x) - x sin (lo g x)$ . (1 mark)
$x^{2} y_{2} - x y_{1} + 2 y = x cos (lo g x) - x sin (lo g x) - x sin (lo g x) - x cos (lo g x) + 2 x sin (lo g x) = 0$ . (1 mark)

Answer: Proved.

Example 5: Squaring to remove sin⁻¹ · 3 marks · Short answerHarder board level

If $y = (sin^{- 1} x)^{2}$ , show that $(1 - x^{2}) \frac{d ^{2} y}{d x ^{2}} - x \frac{d y}{d x} = 2$ .

Try it first, then show the solutionSolution

$y_{1} = \frac{2 s i n ^{- 1} x}{1 - x ^{2}}$ , so $1 - x^{2} y_{1} = 2 sin^{- 1} x$ . Squaring: $(1 - x^{2}) y_{1}^{2} = 4 y$ . (1 mark)
Differentiate: $(1 - x^{2}) 2 y_{1} y_{2} - 2 x y_{1}^{2} = 4 y_{1}$ . (1 mark)
Divide by $2 y_{1}$ : $(1 - x^{2}) y_{2} - x y_{1} = 2$ . (1 mark)

Answer: Proved.

Example 6: An exponential of sin⁻¹x · 5 marks · Long answerHOTS / competency

If $y = e^{m s i n^{- 1} x}$ , prove that $(1 - x^{2}) \frac{d ^{2} y}{d x ^{2}} - x \frac{d y}{d x} - m^{2} y = 0$ .

Try it first, then show the solutionSolution

$y_{1} = e^{m s i n^{- 1} x} \cdot \frac{m}{1 - x ^{2}} = \frac{m y}{1 - x ^{2}}$ . (1 mark)
Rearrange: $1 - x^{2} y_{1} = m y$ . Square: $(1 - x^{2}) y_{1}^{2} = m^{2} y^{2}$ . (1 mark)
Differentiate: $(1 - x^{2}) 2 y_{1} y_{2} + y_{1}^{2} (- 2 x) = 2 m^{2} y y_{1}$ . (2 marks)
Divide by $2 y_{1}$ (non-zero, since $y > 0$ and $m \neq = 0$ ): $(1 - x^{2}) y_{2} - x y_{1} = m^{2} y$ , that is $(1 - x^{2}) y_{2} - x y_{1} - m^{2} y = 0$ . (1 mark)

Answer: Proved. Rearranging before the second differentiation keeps the work to four lines.

Where students lose marks in Second-Order and Parametric Derivatives

The slips examiners see most often, and what to write instead.

Mistake: Dividing second derivatives in parametric form
d²y/dx² is not (d²y/dt²)/(d²x/dt²); dy/dx is a function of t and must be differentiated with respect to t and then divided by dx/dt.
Wrong working
For $x = a t^{2}$ , $y = 2 a t$ : $\frac{d ^{2} y}{d x ^{2}} = \frac{0}{2 a} = 0$ .
Right working
$\frac{d ^{2} y}{d x ^{2}} = \frac{- 1/ t ^{2}}{2 a t} = - \frac{1}{2 a t ^{3}}$ .
Do this: Always write $\frac{d}{d t} (\frac{d y}{d x}) \div \frac{d x}{d t}$ .
Usually costs: The whole second part.
Mistake: Differentiating before rearranging
The ISC 2025 examiners noted errors in the second differentiation and advised rearranging after the first derivative.
Do this: Clear fractions and roots from y₁, then differentiate the cleaner equation.
Usually costs: 2 to 3 marks in a 4- or 5-mark proof.
Mistake: Forgetting the chain rule on y₁²
y₁ is a function of x, so its square differentiates to 2y₁y₂, not 2y₁.
Do this: Write d/dx(y₁²) = 2y₁ · y₂ explicitly.
Usually costs: 1 to 2 marks.
Mistake: Leaving the original function in the answer
The target relation uses y; leaving e^(m sin⁻¹x) or sin⁻¹x unreplaced hides the result.
Do this: Replace every occurrence of the original expression by y before the final step.
Usually costs: 1 mark.

Exam tips

Name your derivatives $y_{1}$ and $y_{2}$ at the start; it keeps long proofs readable.
After finding $y_{1}$ , write the rearranged form on its own line; it is often a marked step.
In a parametric question, simplify $\frac{d y}{d x}$ fully in terms of $t$ before differentiating it again.
State when you divide by $y_{1}$ or $2 y_{1}$ , and why it is non-zero, if the question is a proof.
For motion questions, give units: m/s for velocity and m/s² for acceleration.

Practice questions on Second-Order and Parametric Derivatives

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$\frac{d ^{2}}{d x ^{2}} (e^{3 x})$ equals:
(a) $3 e^{3 x}$
(b) $9 e^{3 x}$
(c) $e^{3 x}$
(d) $6 e^{3 x}$
Hint
Differentiate twice.
Answer
(b) $9 e^{3 x}$ .
Q2 · 2 marks · Short answerBoard level
If $x = 2 cos t$ and $y = 3 sin t$ , find $\frac{d y}{d x}$ at $t = \frac{π}{4}$ .
Hint
$\frac{d y}{d x} = \frac{3 c o s t}{- 2 s i n t}$ .
Answer
$\frac{d y}{d x} = - \frac{3}{2} cot t$ ; at $t = \frac{π}{4}$ it is $- \frac{3}{2}$ .
Q3 · 3 marks · Short answerHarder board level
If $x = a (θ - sin θ)$ and $y = a (1 - cos θ)$ , find $\frac{d ^{2} y}{d x ^{2}}$ .
Hint
First show $\frac{d y}{d x} = cot \frac{θ}{2}$ .
Answer
$\frac{d y}{d x} = \frac{s i n θ}{1 - c o s θ} = cot \frac{θ}{2}$ . Then $\frac{d ^{2} y}{d x ^{2}} = \frac{- \frac{1}{2} cosec ^{2} \frac{θ}{2}}{a ( 1 - c o s θ )} = \frac{- \frac{1}{2} cosec ^{2} \frac{θ}{2}}{2 a s i n ^{2} \frac{θ}{2}} = - \frac{1}{4 a s i n ^{4} ( θ /2 )}$ .
Q4 · 2 marks · Short answerHarder board level
If $y = tan x$ , show that $\frac{d ^{2} y}{d x ^{2}} = 2 y (1 + y^{2})$ .
Hint
Write $y_{1}$ in terms of $y$ first.
Answer
$y_{1} = sec^{2} x = 1 + y^{2}$ . Then $y_{2} = 2 y y_{1} = 2 y (1 + y^{2})$ .
Q5 · 3 marks · Short answerHarder board level
If $y = e^{x} sin x$ , prove that $\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 2 y = 0$ .
Hint
Use the product rule twice.
Answer
$y_{1} = e^{x} (sin x + cos x)$ , $y_{2} = 2 e^{x} cos x$ . Then $y_{2} - 2 y_{1} + 2 y = 2 e^{x} cos x - 2 e^{x} sin x - 2 e^{x} cos x + 2 e^{x} sin x = 0$ .
Q6 · 3 marks · Short answerHOTS / competency
If $y = (x + 1 + x^{2})^{m}$ , prove that $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x} - m^{2} y = 0$ .
Hint
Show that $1 + x^{2} y_{1} = m y$ first.
Answer
$y_{1} = m (x + 1 + x^{2})^{m - 1} (1 + \frac{x}{1 + x ^{2}}) = \frac{m y}{1 + x ^{2}}$ . Square: $(1 + x^{2}) y_{1}^{2} = m^{2} y^{2}$ . Differentiate: $2 (1 + x^{2}) y_{1} y_{2} + 2 x y_{1}^{2} = 2 m^{2} y y_{1}$ . Divide by $2 y_{1}$ : $(1 + x^{2}) y_{2} + x y_{1} - m^{2} y = 0$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a ball is thrown so that after $t$ seconds it is at $x = 20 t$ m horizontally and $y = 15 t - 5 t^{2}$ m vertically.
(i) Find $\frac{d y}{d x}$ in terms of $t$ . (1 mark)
(ii) Find the highest point of the path. (1 mark)
(iii) Show that $\frac{d ^{2} y}{d x ^{2}}$ is constant and say what this tells you about the path. (1 mark)
(iv) Find the slope of the path when the ball lands. (1 mark)
Hint
The ball lands when $y = 0$ again.
Answer
(i) $\frac{15 - 10 t}{20}$ . (ii) Slope 0 at $t = 1.5$ : $(30, 11.25)$ . (iii) $\frac{d ^{2} y}{d x ^{2}} = \frac{- 10/20}{20} = - \frac{1}{40}$ , constant and negative: the path is a parabola opening downwards. (iv) Lands at $t = 3$ : slope $\frac{15 - 30}{20} = - \frac{3}{4}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the parametric formula for $\frac{d ^{2} y}{d x ^{2}}$ .Show answer
$\frac{d}{d t} (\frac{d y}{d x}) \div \frac{d x}{d t}$ .
2.If $y = cos 3 x$ , what is $y_{2} + 9 y$ ?Show answer
0.
3.What is $\frac{d}{d x} (y_{1}^{2})$ ?Show answer
$2 y_{1} y_{2}$ .
4.If $x = t^{2}$ , $y = t^{3}$ , find $\frac{d y}{d x}$ .Show answer
$\frac{3 t}{2}$ .
5.If $s = 5 t^{2} + 2 t$ , find the acceleration.Show answer
10 (units of distance per second squared).
6.Which order of derivatives is in the ISC syllabus?Show answer
Successive differentiation up to the second order.

Second-Order and Parametric Derivatives: frequently asked questions

How do I find the second derivative of a parametric function?

First find dy/dx = (dy/dt)/(dx/dt) and simplify it in terms of t. Then differentiate that expression with respect to t and divide by dx/dt. Do not divide d²y/dt² by d²x/dt².

Why do textbooks square y₁ in proofs?

Squaring removes square roots such as √(1 − x²) and often lets you replace the original expression by y. The squared equation is then easy to differentiate, and dividing by 2y₁ at the end gives the required relation.

What does the second derivative tell us?

It is the rate of change of the slope. In motion it is acceleration; on a graph, a positive value means the curve bends upwards (like a cup) and a negative value means it bends downwards. It is used in the second derivative test for maxima and minima.

Are third derivatives asked in ISC?

No. The CISCE 2027 syllabus limits successive differentiation to the second order.

How many marks do second-derivative proofs carry?

They usually appear as 3- to 5-mark questions. In ISC 2025, a 4-mark question asked candidates to prove a relation between y, dy/dx and d²y/dx² for a function defined through tan and log.