MathsCBSE Class 12 · Continuity and Differentiability

Derivatives of Parametric Functions and Second Order Derivatives (Class 12)

Updated 15 October 2026 · Fact-checked

If x and y are both given in terms of a parameter t, then dy/dx = (dy/dt) ÷ (dx/dt). For the second derivative, differentiate dy/dx with respect to t and divide by dx/dt again: d²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt). It is not (d²y/dt²) ÷ (d²x/dt²), the most common mistake.

Find dy/dx, differentiate it with respect to t, then divide by dx/dt again; never divide d²y/dt² by d²x/dt².

Illustration drawn for QuestionBuddy.

Understand Parametric and Second Order Derivatives

Parametric form. Sometimes $x$ and $y$ are both expressed through a third variable, the parameter, such as $x = a (θ + sin θ)$ , $y = a (1 - cos θ)$ or the position of a ball $x = 20 t$ , $y = 203 t - 5 t^{2}$ . By the chain rule,
$\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ , provided $\frac{d x}{d t} \neq = 0$ .
You never need to eliminate $t$ . The answer is usually in terms of $t$ ; simplify it with identities (for example $\frac{s i n θ}{1 + c o s θ} = tan \frac{θ}{2}$ ).

Second order derivatives. For $y = f (x)$ , $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{d y}{d x})$ , also written $y^{''}$ or $y_{2}$ . For explicit functions just differentiate twice, using the product, quotient and chain rules. Standard results to know: $\frac{d ^{2}}{d x ^{2}} e^{a x} = a^{2} e^{a x}$ , $\frac{d ^{2}}{d x ^{2}} sin a x = - a^{2} sin a x$ .

Second derivative in parametric form. $\frac{d y}{d x}$ is a function of $t$ , so to differentiate it with respect to $x$ use the chain rule again:
$\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (\frac{d y}{d x}) \cdot \frac{d t}{d x} = \frac{\frac{d}{d t} ( \frac{d y}{d x} )}{\frac{d x}{d t}}$ .
The tempting shortcut $\frac{d ^{2} y / d t ^{2}}{d ^{2} x / d t ^{2}}$ is wrong: for $x = t^{2}$ , $y = t^{3}$ it gives $3 t$ , while the correct value is $\frac{3}{4 t}$ .

'Prove that' relations. Many board questions give $y$ and ask you to prove a relation such as $(1 + x^{2}) y^{''} + x y^{'} = 0$ . The tidy method is to find $y^{'}$ , rearrange it so there is no fraction or root (for example $1 + x^{2} y^{'} = 1$ ), and then differentiate that equation again. This avoids messy quotient rules. Functions like $e^{a x} (A cos b x + B sin b x)$ , $sin (m sin^{- 1} x)$ and $lo g (x + 1 + x^{2})$ lead to such relations.

Derivative of one function with respect to another. To differentiate $u (x)$ with respect to $v (x)$ , treat $x$ as the parameter: $\frac{d u}{d v} = \frac{d u / d x}{d v / d x}$ .

Meaning. The first derivative is a slope or a rate; the second derivative tells how that slope changes. For position $s (t)$ , $s^{'}$ is velocity and $s^{''}$ is acceleration. For a curve, a negative $\frac{d ^{2} y}{d x ^{2}}$ means it bends downward. These ideas return in maxima and minima.

NCERT covers parametric derivatives in Section 5.6 and second order derivatives in Section 5.7 (Chapter 5); both are frequent 3- and 5-mark board questions.

Parametric and Second Order Derivatives: formulas to remember

Parametric first derivative: $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$; Valid when $\frac{d x}{d t} \neq = 0$ .
Parametric second derivative: $\frac{d ^{2} y}{d x ^{2}} = \frac{\frac{d}{d t} ( \frac{d y}{d x} )}{\frac{d x}{d t}}$; Not $\frac{d ^{2} y / d t ^{2}}{d ^{2} x / d t ^{2}}$ .
Second derivative: $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{d y}{d x})$; Also written $y^{''}$ or $y_{2}$ .
Derivative with respect to another function: $\frac{d u}{d v} = \frac{d u / d x}{d v / d x}$; x acts as the parameter.
Useful half-angle forms: $\frac{s i n θ}{1 + c o s θ} = tan \frac{θ}{2}, 1 + cos θ = 2 cos^{2} \frac{θ}{2}$; Simplify parametric answers with these.
Standard second derivatives: $(e^{a x})^{''} = a^{2} e^{a x}, (sin a x)^{''} = - a^{2} sin a x, (lo g x)^{''} = - \frac{1}{x ^{2}}$; Handy in 'prove that' questions.

How to solve Parametric and Second Order Derivatives questions

Use this for parametric derivatives, second derivatives, and proofs of differential relations.

1Parametric: find $\frac{d x}{d t}$ and $\frac{d y}{d t}$ separately, then divide: $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ . Simplify with identities.
2Parametric second derivative: differentiate your simplified $\frac{d y}{d x}$ with respect to $t$ , then divide by $\frac{d x}{d t}$ (equivalently multiply by $\frac{d t}{d x}$ ).
3Explicit second derivative: differentiate once, simplify, then differentiate again with the product, quotient or chain rule.
4For 'prove that' relations: find $y^{'}$ , clear fractions and roots (for example multiply by $1 - x^{2}$ ), square if helpful, then differentiate the cleared equation.
5Substitute back $y$ or $y^{'}$ from earlier lines to reach the required relation.
6If a value is asked at a particular $t$ or $x$ , substitute only at the end.

Parametric and Second Order 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.

Parametric slope and bending

Suppose a Ferris wheel of radius 10 m has its centre 12 m above the ground. A cabin's position after turning through angle $t$ (radians) is $x = 10 cos t$ , $y = 12 + 10 sin t$ .

Find $\frac{d y}{d x}$ and $\frac{d ^{2} y}{d x ^{2}}$ in terms of $t$ . Evaluate the slope at $t = \frac{π}{4}$ and the second derivative at the top, $t = \frac{π}{2}$ .

Try it first, then show the solutionSolution

$\frac{d x}{d t} = - 10 sin t$ , $\frac{d y}{d t} = 10 cos t$ , so $\frac{d y}{d x} = - cot t$ .
$\frac{d}{d t} (- cot t) = cosec^{2} t$ , so $\frac{d ^{2} y}{d x ^{2}} = \frac{cosec ^{2} t}{- 10 s i n t} = - \frac{1}{10 s i n ^{3} t}$ .
At $t = \frac{π}{4}$ : $\frac{d y}{d x} = - cot \frac{π}{4} = - 1$ .
At $t = \frac{π}{2}$ : $\frac{d ^{2} y}{d x ^{2}} = - \frac{1}{10}$ ; negative, so the path bends downward at the top.

Answer: $\frac{d y}{d x} = - cot t$ , $\frac{d ^{2} y}{d x ^{2}} = - \frac{1}{10 s i n ^{3} t}$ ; slope $- 1$ at $t = \frac{π}{4}$ ; $- \frac{1}{10}$ at the top.

Memory hook: At the top of the wheel the path curves down: negative second derivative.

Second derivative as acceleration

Imagine a test car whose distance from the start line is $s (t) = 2 t^{3} - 3 t^{2}$ metres, $t$ seconds after the test begins (for $t \geq 1$ it moves forward).

Find the velocity and acceleration at $t = 2$ s, and the moment when the acceleration is zero.

Try it first, then show the solutionSolution

Velocity $v = \frac{d s}{d t} = 6 t^{2} - 6 t$ ; at $t = 2$ : $24 - 12 = 12$ m/s.
Acceleration $a = \frac{d ^{2} s}{d t ^{2}} = 12 t - 6$ ; at $t = 2$ : $18$ m/s².
$a = 0$ when $12 t - 6 = 0$ , i.e. $t = \frac{1}{2}$ s.

Answer: At $t = 2$ : velocity 12 m/s, acceleration 18 m/s². Acceleration is zero at $t = 0.5$ s.

Memory hook: First derivative = speedometer, second derivative = how hard you press the pedal.

Tricks to remember Parametric and Second Order Derivatives

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

Divide by dx/dt, every time
First derivative: $\frac{d y / d t}{d x / d t}$ . Second derivative: $\frac{d}{d t} (first)$ , again divided by $\frac{d x}{d t}$ .
Use it when: Parametric second derivatives.
Never divide second by second
$\frac{d ^{2} y}{d x ^{2}} \neq = \frac{d ^{2} y / d t ^{2}}{d ^{2} x / d t ^{2}}$ . Test with $x = t^{2}$ , $y = t^{3}$ : the shortcut gives $3 t$ , the truth is $\frac{3}{4 t}$ .
Use it when: Assertion-reason traps.
Clear, then differentiate
In 'prove that' questions, first make $y^{'}$ fraction-free (for example $1 + x^{2} y^{'} = 1$ ), then differentiate the whole line.
Use it when: Proofs like $(1 + x^{2}) y^{''} + x y^{'} = 0$ .
Half-angles tidy cycloids
$\frac{s i n θ}{1 \pm c o s θ}$ always simplifies with half-angle formulas to $tan \frac{θ}{2}$ or $cot \frac{θ}{2}$ .
Use it when: Cycloid-type parametric curves.

Parametric and Second Order Derivatives: solved examples

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

Example 1: A simple parametric pair · 2 marks · Very short answerBoard level

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$ , find $\frac{d y}{d x}$ .

Try it first, then show the solutionSolution

$\frac{d x}{d t} = 1 - \frac{1}{t ^{2}}$ , $\frac{d y}{d t} = 1 + \frac{1}{t ^{2}}$ . (1 mark)
$\frac{d y}{d x} = \frac{1 + \frac{1}{t ^{2}}}{1 - \frac{1}{t ^{2}}} = \frac{t ^{2} + 1}{t ^{2} - 1}$ , $t \neq = \pm 1$ . (1 mark)

Answer: $\frac{d y}{d x} = \frac{t ^{2} + 1}{t ^{2} - 1}$

Example 2: Second derivative of a product · 2 marks · Very short answerBoard level

Find $\frac{d ^{2} y}{d x ^{2}}$ if $y = x^{2} e^{- x}$ .

Try it first, then show the solutionSolution

$y^{'} = 2 x e^{- x} - x^{2} e^{- x} = e^{- x} (2 x - x^{2})$ . (1 mark)
$y^{''} = - e^{- x} (2 x - x^{2}) + e^{- x} (2 - 2 x) = e^{- x} (x^{2} - 4 x + 2)$ . (1 mark)

Answer: $\frac{d ^{2} y}{d x ^{2}} = e^{- x} (x^{2} - 4 x + 2)$

Example 3: Second derivative of a cycloid · 3 marks · Short answerHarder board level

If $x = a (θ - sin θ)$ and $y = a (1 - cos θ)$ , find $\frac{d y}{d x}$ and $\frac{d ^{2} y}{d x ^{2}}$ .

Try it first, then show the solutionSolution

$\frac{d x}{d θ} = a (1 - cos θ) = 2 a sin^{2} \frac{θ}{2}$ , $\frac{d y}{d θ} = a sin θ = 2 a sin \frac{θ}{2} cos \frac{θ}{2}$ . (1 mark)
$\frac{d y}{d x} = cot \frac{θ}{2}$ . (½ mark)
$\frac{d}{d θ} (cot \frac{θ}{2}) = - \frac{1}{2} cosec^{2} \frac{θ}{2}$ . (½ mark)
$\frac{d ^{2} y}{d x ^{2}} = \frac{- \frac{1}{2} cosec ^{2} \frac{θ}{2}}{2 a s i n ^{2} \frac{θ}{2}} = - \frac{1}{4 a} cosec^{4} \frac{θ}{2}$ . (1 mark)

Answer: $\frac{d y}{d x} = cot \frac{θ}{2}$ , $\frac{d ^{2} y}{d x ^{2}} = - \frac{1}{4 a} cosec^{4} \frac{θ}{2}$

Example 4: Proving a second-order relation · 3 marks · Short answerHarder board level

If $y = e^{2 x} (A cos x + B sin x)$ , where A and B are constants, prove that $\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 5 y = 0$ .

Try it first, then show the solutionSolution

$y^{'} = 2 e^{2 x} (A cos x + B sin x) + e^{2 x} (- A sin x + B cos x) = 2 y + e^{2 x} (B cos x - A sin x)$ . (1 mark)
So $y^{'} - 2 y = e^{2 x} (B cos x - A sin x)$ . Differentiate: $y^{''} - 2 y^{'} = 2 e^{2 x} (B cos x - A sin x) + e^{2 x} (- B sin x - A cos x)$ . (1 mark)
$= 2 (y^{'} - 2 y) - y$ . Hence $y^{''} - 2 y^{'} - 2 y^{'} + 4 y + y = 0$ , i.e. $y^{''} - 4 y^{'} + 5 y = 0$ . (1 mark)

Answer: $y^{''} - 4 y^{'} + 5 y = 0$

Example 5: Clear the root, then differentiate · 3 marks · Short answerHOTS / competency

If $y = lo g (x + 1 + x^{2})$ , prove that $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x} = 0$ .

Try it first, then show the solutionSolution

$y^{'} = \frac{1}{x + 1 + x ^{2}} (1 + \frac{x}{1 + x ^{2}}) = \frac{1}{x + 1 + x ^{2}} \cdot \frac{1 + x ^{2} + x}{1 + x ^{2}} = \frac{1}{1 + x ^{2}}$ . (1 mark)
So $1 + x^{2} y^{'} = 1$ . Differentiate: $1 + x^{2} y^{''} + \frac{x}{1 + x ^{2}} y^{'} = 0$ . (1 mark)
Multiply by $1 + x^{2}$ : $(1 + x^{2}) y^{''} + x y^{'} = 0$ . (1 mark)

Answer: $(1 + x^{2}) y^{''} + x y^{'} = 0$

Example 6: Case study: the path of a thrown ball · 4 marks · Case-basedHOTS / competency

Suppose a ball is thrown at 40 m/s at $60°$ to the ground. Taking $g = 10$ m/s², its position after $t$ seconds is $x = 20 t$ , $y = 203 t - 5 t^{2}$ (metres).
(i) Find $\frac{d y}{d x}$ in terms of $t$ . (1 mark)
(ii) Find the slope of the path at $t = 2$ . (1 mark)
(iii) Find the time and height at which $\frac{d y}{d x} = 0$ . (1 mark)
(iv) Find $\frac{d ^{2} y}{d x ^{2}}$ and say what its sign means. (1 mark)

Try it first, then show the solutionSolution

(i) $\frac{d x}{d t} = 20$ , $\frac{d y}{d t} = 203 - 10 t$ , so $\frac{d y}{d x} = 3 - \frac{t}{2}$ . (1 mark)
(ii) At $t = 2$ : $3 - 1 \approx 0.73$ . (1 mark)
(iii) $3 - \frac{t}{2} = 0 \Rightarrow t = 23$ s; height $= 203 (23) - 5 (12) = 120 - 60 = 60$ m. (1 mark)
(iv) $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (3 - \frac{t}{2}) \div 20 = - \frac{1}{40}$ ; negative and constant: the path is a downward-bending parabola. (1 mark)

Answer: (i) $3 - \frac{t}{2}$ (ii) $3 - 1$ (iii) $t = 23$ s, 60 m (iv) $- \frac{1}{40}$ , bends downward.

Where students lose marks in Parametric and Second Order Derivatives

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

Mistake: Dividing second derivatives
The parameter must be converted with $\frac{d t}{d x}$ , not by dividing $\frac{d ^{2} y}{d t ^{2}}$ by $\frac{d ^{2} x}{d t ^{2}}$ .
Wrong working
$x = t^{2}$ , $y = t^{3}$ : $\frac{d ^{2} y}{d x ^{2}} = \frac{6 t}{2} = 3 t$ .
Right working
$\frac{d y}{d x} = \frac{3 t}{2}$ , so $\frac{d ^{2} y}{d x ^{2}} = \frac{3/2}{2 t} = \frac{3}{4 t}$ .
Do this: Always divide by $\frac{d x}{d t}$ after differentiating $\frac{d y}{d x}$ with respect to $t$ .
Usually costs: 2 to 3 marks.
Mistake: Forgetting to divide by dx/dt in the second step
Differentiating $\frac{d y}{d x}$ with respect to $t$ gives $\frac{d}{d t}$ , not $\frac{d}{d x}$ .
Wrong working
$\frac{d y}{d x} = cot \frac{θ}{2} \Rightarrow \frac{d ^{2} y}{d x ^{2}} = - \frac{1}{2} cosec^{2} \frac{θ}{2}$ .
Right working
$\frac{d ^{2} y}{d x ^{2}} = \frac{- \frac{1}{2} cosec ^{2} \frac{θ}{2}}{a ( 1 - c o s θ )} = - \frac{1}{4 a} cosec^{4} \frac{θ}{2}$ .
Do this: Write the formula $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d θ} (\frac{d y}{d x}) \cdot \frac{d θ}{d x}$ before using it.
Usually costs: The final 1 to 2 marks.
Mistake: Messy quotient rule in proofs
Differentiating $\frac{1}{1 + x ^{2}}$ directly invites errors.
Wrong working
$y^{''} = \frac{- x}{( 1 + x ^{2} )}$ (power wrong).
Right working
From $1 + x^{2} y^{'} = 1$ : $1 + x^{2} y^{''} + \frac{x}{1 + x ^{2}} y^{'} = 0$ .
Do this: Clear fractions and roots before the second differentiation.
Usually costs: 1 to 2 marks.
Mistake: Not simplifying the first derivative
An unsimplified $\frac{d y}{d x}$ makes the second differentiation long and error-prone.
Wrong working
$\frac{d y}{d x} = \frac{a s i n θ}{a ( 1 - c o s θ )}$ differentiated by the quotient rule.
Right working
Simplify to $cot \frac{θ}{2}$ first, then differentiate.
Do this: Look for identities after every derivative.
Usually costs: Time, and often a sign.

Exam tips

A 5-mark question asking for a parametric second derivative appeared in the 2025 and 2026 board papers; show both $\frac{d x}{d t}$ and $\frac{d y}{d t}$ before dividing.
Write $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (\frac{d y}{d x}) \cdot \frac{d t}{d x}$ as a formula line; it earns the method mark.
For 'prove that' relations, the final line must match the required form exactly; rearrange signs if needed.
If a value at a point is asked, simplify the general expression first, then substitute.
Check that $\frac{d x}{d t} \neq = 0$ at the given point; otherwise $\frac{d y}{d x}$ is not defined there.

Practice questions on Parametric and Second Order 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
If $x = a t^{2}$ and $y = 2 a t$ , then $\frac{d y}{d x}$ is:
(a) $t$
(b) $\frac{1}{t}$
(c) $2 t$
(d) $\frac{2}{t}$
Hint
$\frac{d y / d t}{d x / d t} = \frac{2 a}{2 a t}$ .
Answer
(b) $\frac{1}{t}$ .
Q2 · 2 marks · Very short answerBoard level
If $x = e^{t} + e^{- t}$ and $y = e^{t} - e^{- t}$ , show that $\frac{d y}{d x} = \frac{x}{y}$ .
Hint
$\frac{d x}{d t} = e^{t} - e^{- t} = y$ .
Answer
$\frac{d x}{d t} = e^{t} - e^{- t} = y$ and $\frac{d y}{d t} = e^{t} + e^{- t} = x$ , so $\frac{d y}{d x} = \frac{x}{y}$ .
Q3 · 2 marks · Very short answerBoard level
Find $\frac{d ^{2} y}{d x ^{2}}$ if $y = tan x$ .
Hint
$\frac{d}{d x} sec^{2} x = 2 sec x \cdot sec x tan x$ .
Answer
$y^{'} = sec^{2} x$ , $y^{''} = 2 sec^{2} x tan x$ .
Q4 · 3 marks · Short answerHarder board level
If $x = 2 sec θ$ and $y = 3 tan θ$ , find $\frac{d ^{2} y}{d x ^{2}}$ at $θ = \frac{π}{4}$ .
Hint
First show $\frac{d y}{d x} = \frac{3}{2} cosec θ$ .
Answer
$\frac{d y}{d x} = \frac{3 s e c ^{2} θ}{2 s e c θ t a n θ} = \frac{3}{2} cosec θ$ . $\frac{d ^{2} y}{d x ^{2}} = \frac{- \frac{3}{2} cosec θ c o t θ}{2 s e c θ t a n θ} = - \frac{3}{4} cot^{3} θ$ . At $θ = \frac{π}{4}$ : $- \frac{3}{4}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $x = t^{2}$ and $y = t^{3}$ , then $\frac{d ^{2} y}{d x ^{2}} = \frac{d ^{2} y / d t ^{2}}{d ^{2} x / d t ^{2}} = 3 t$ .
Reason (R): $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (\frac{d y}{d x}) \cdot \frac{d t}{d x}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Use R: $\frac{d y}{d x} = \frac{3 t}{2}$ .
Answer
(d). By R, $\frac{d ^{2} y}{d x ^{2}} = \frac{3}{2} \cdot \frac{1}{2 t} = \frac{3}{4 t}$ , not $3 t$ ; A is false and R is true.
Q6 · 2 marks · Very short answerHarder board level
Differentiate $e^{x^{2}}$ with respect to $cos x$ .
Hint
Let $u = e^{x^{2}}$ , $v = cos x$ ; find $\frac{d u / d x}{d v / d x}$ .
Answer
$\frac{d u}{d x} = 2 x e^{x^{2}}$ , $\frac{d v}{d x} = - sin x$ , so $\frac{d u}{d v} = - \frac{2 x e ^{x^{2}}}{s i n x}$ .
Q7 · 3 marks · Short answerHOTS / competency
If $y = sin (m sin^{- 1} x)$ , prove that $(1 - x^{2}) \frac{d ^{2} y}{d x ^{2}} - x \frac{d y}{d x} + m^{2} y = 0$ .
Hint
Write $1 - x^{2} y^{'} = m cos (m sin^{- 1} x)$ and square.
Answer
$y^{'} = \frac{m c o s ( m s i n ^{- 1} x )}{1 - x ^{2}}$ , so $(1 - x^{2}) y^{'2} = m^{2} (1 - y^{2})$ . Differentiate: $- 2 x y^{'2} + 2 (1 - x^{2}) y^{'} y^{''} = - 2 m^{2} y y^{'}$ . Divide by $2 y^{'}$ : $(1 - x^{2}) y^{''} - x y^{'} + m^{2} y = 0$ .

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.State the parametric formula for $\frac{d y}{d x}$ .Show answer
$\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ .
2.State the correct formula for $\frac{d ^{2} y}{d x ^{2}}$ in parametric form.Show answer
$\frac{d}{d t} (\frac{d y}{d x}) \div \frac{d x}{d t}$ .
3.If $x = cos t$ , $y = sin t$ , what is $\frac{d y}{d x}$ ?Show answer
$- cot t$ .
4.What is $\frac{d ^{2}}{d x ^{2}} (e^{3 x})$ ?Show answer
$9 e^{3 x}$ .
5.What is $\frac{d ^{2}}{d x ^{2}} (lo g x)$ ?Show answer
$- \frac{1}{x ^{2}}$ .
6.How do you differentiate $u (x)$ with respect to $v (x)$ ?Show answer
$\frac{d u}{d v} = \frac{d u / d x}{d v / d x}$ .

Parametric and Second Order Derivatives: frequently asked questions

Why is d²y/dx² not equal to (d²y/dt²)/(d²x/dt²)?

Because dy/dx is a function of t, and differentiating it with respect to x needs the chain rule: d/dx = (d/dt)(dt/dx). So d²y/dx² equals d/dt(dy/dx) divided by dx/dt. Dividing the two second derivatives gives the wrong answer.

Do I need to eliminate the parameter?

No. Find dy/dt and dx/dt and divide. Eliminating the parameter is usually longer and is not required.

How should I approach 'prove that' second-derivative questions?

Find the first derivative, rearrange it to remove fractions or square roots (squaring if needed), then differentiate that equation. Substitute earlier results to reach the required relation.

Which NCERT sections cover this topic?

Section 5.6 (Derivatives of Functions in Parametric Forms) and Section 5.7 (Second Order Derivative) of Chapter 5 in NCERT Mathematics Part I for Class 12.

Are Rolle's theorem and the mean value theorem still in the syllabus?

No. They were dropped from the 2026-27 syllabus, so questions on this chapter end with second order derivatives.