UG Mathematics Booster Test 3 - Advanced Differentiation -

🚀

UG Mathematics Booster Test 3 - Advanced Differentiation

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

UG Mathematics Booster Test 3 - Advanced Differentiation

1 / 20

QUESTION 1 OF 20

A region is bounded by
\(y=sin⁡(x^{2})\).
Applying the composite chain rule, what defines the slope of the curve \(\frac{dy}{dx}\)?

Answer Review

1 A region is bounded by
\(y=sin⁡(x^{2})\).
Applying the composite chain rule, what defines the slope of the curve \(\frac{dy}{dx}\)?

A \(cos⁡(x^{2})\)

B \(2xcos⁡(x^{2})\)

C \(2xsin⁡(x^{2})\)

D \(sin⁡(2x)\)

💡 Short Explanation

Differentiate outer function first Multiply by derivative of inner function \(x^{2}\) Chain rule gives \(2xcos⁡(x^{2})\)

✅ Correct Answer Explanation

The function is composite because \(sin⁡\)acts on \(x^{2}\). By chain rule: \(\frac{d}{dx}[sin⁡(x^{2})]=cos⁡(x^{2})⋅2x\) Option B correctly applies outer derivative and inner derivative. Option A ignores inner differentiation. Option C incorrectly keeps sine unchanged. Option D confuses with trigonometric identity differentiation.

❓ Why Other Options Are Incorrect

Option A → \(cos⁡(x^{2})\)omits derivative of inner function \(x^{2}\), violating chain rule.
Option C → \(2xsin⁡(x^{2})\)incorrectly differentiates sine as sine instead of cosine.
Option D → \(sin⁡(2x)\)is unrelated to differentiation of composite trigonometric functions.

🎯 Strategy

Used: Elimination

Application:

Check whether both outer and inner derivatives appear together after differentiation.

Final Logic:

Composite functions always require multiplication by derivative of inner term.

🧠 Memory Trick

"Outer derivative × Inner derivative"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Chain Rule

2 A parametric probability curve uses
\(x=t^{2}\)and \(y=sin⁡t\).
The derivative evaluates to:

A \(\frac{\cos\,t}{2t}\)

B \(\frac{\cos\,t}{2t}\)

C \(2tcos⁡t\)

D \(\sin\,t\)

💡 Short Explanation

Use parametric differentiation formula Compute \(dy/dt\) and \(dx/dt\) separately Divide both derivatives

✅ Correct Answer Explanation

For parametric equations: \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\) Here, \(\frac{dy}{dt}=cos⁡t,\frac{dx}{dt}=2t\) Thus, \(\frac{dy}{dx}=\frac{\cos\,t}{2t}\) Option B is correct. Option C multiplies instead of divides. Option D is original function. Option A duplicates correct expression.

❓ Why Other Options Are Incorrect

Option A → Same as correct answer but duplicated; officially accepted option remains B.
Option C → Incorrectly multiplies derivatives instead of taking quotient.
Option D → Represents \(y\), not derivative.

🎯 Strategy

Used: Substitution

Application:

Substitute derivatives into parametric derivative formula directly.

Final Logic:

Parametric differentiation always uses quotient of derivatives.

🧠 Memory Trick

"Parametric = Divide the rates"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Parametric Differentiation

3 Assertion (A): The second order derivative of
\(y=(x^{2}+1)^{3}\)
is
\(\frac{d^{2}y}{dx^{2}}=6(x^{2}+1)(5x^{2}+1)\)
Reason (R): The first derivative is
\(\frac{dy}{dx}=3(x^{2}+1)^{2}⋅2x\)
and the chain rule differentiates further into the second derivative.

A Both true, R explains A

B Both true, R does not explain A

C A true, R false

D A false, R true

💡 Short Explanation

First derivative is correct Second derivative calculation given is incorrect Assertion formula mismatch occurs

✅ Correct Answer Explanation

Differentiating: \(y=(x^{2}+1)^{3}\) First derivative: \(\frac{dy}{dx}=6x(x^{2}+1)^{2}\) Second derivative becomes: \(\frac{d^{2}y}{dx^{2}}=6(x^{2}+1)(5x^{2}+1)\) The assertion incorrectly states \(3x^{2}+1\). Hence Assertion is false. Reason correctly explains chain-rule-based differentiation.

❓ Why Other Options Are Incorrect

Option A → Incorrect because assertion formula is mathematically wrong.
Option B → Both statements are not true since assertion contains wrong coefficient.
Option C → Reason is actually correct because first derivative and chain-rule process are valid.

🎯 Strategy

Used: Elimination

Application:

Verify differentiation carefully before checking logical relation.

Final Logic:

A single algebraic mismatch makes assertion false.

🧠 Memory Trick

"Differentiate twice—check coefficients carefully"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Higher Order Derivatives

4 Identify the INCORRECT relation regarding higher order trigonometric derivatives:

A If \(y=sin⁡x\), then \(\frac{d^{2}y}{dx^{2}}=-sin⁡x\).

B The second derivative of \(\cos\,x\) is \(-cos⁡x\).

C If \(y=sin⁡(x^{2})\), the chain rule is unnecessary for the second derivative.

D Second order derivatives are denoted by \(\frac{d^{2}y}{dx^{2}}\).

💡 Short Explanation

Composite functions always require chain rule \(sin⁡(x^{2})\)is composite Second derivative still uses chain rule

✅ Correct Answer Explanation

The function \(sin⁡(x^{2})\)is composite, so differentiation requires chain rule repeatedly. \(\frac{d}{dx}[sin⁡(x^{2})]=2xcos⁡(x^{2})\) Hence Option C is incorrect. Options A and B are standard second derivatives. Option D correctly denotes second-order differentiation.

❓ Why Other Options Are Incorrect

Option A → Correct because differentiating sine twice returns \(-sin⁡x\).
Option B → Correct since second derivative of cosine is negative cosine.
Option D → Standard notation for second derivative is correctly written.

🎯 Strategy

Used: Odd One Out

Application:

Identify statement contradicting basic chain rule principle.

Final Logic:

Composite trigonometric functions always need chain rule.

🧠 Memory Trick

"Composite ⇒ Chain rule compulsory"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Higher Order Derivatives

5 If the derivative of a continuous function is
\(\frac{dy}{dx}=2xcos⁡(x^{2}),\)
the original three-chain function integrates back to:

A \(\sin\,x\)

B \(cos⁡(x^{2})\)

C \(sin⁡(x)\)

D \(sin⁡(x^{2})\)

💡 Short Explanation

Recognize chain rule pattern Derivative matches differentiation of \(sin⁡(x^{2})\) Reverse differentiation through integration

✅ Correct Answer Explanation

Differentiating \(sin⁡(x^{2})\)gives: \(\frac{d}{dx}[sin⁡(x^{2})]=2xcos⁡(x^{2})\) Hence integrating \(2xcos⁡(x^{2})\)restores \(sin⁡(x^{2})\). Option B differentiates differently. Options A and C involve simple sine functions without composite structure.

❓ Why Other Options Are Incorrect

Option A → Derivative of \(\sin\,x\) is simply \(\cos\,x\).
Option B → Derivative becomes \(-2xsin⁡(x^{2})\), not given expression.
Option C → Same issue as Option A; lacks composite term.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Match derivative pattern with known chain-rule derivative forms.

Final Logic:

Expression exactly matches derivative of \(sin⁡(x^{2})\).

🧠 Memory Trick

"\(2xcos⁡(x^{2})\) comes from \(sin⁡(x^{2})\)"

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Reverse Chain Rule

6 Arrange the extended logarithmic differentiation steps for
\(y=x^{x}\):
1. Express as \(ln⁡y=xln⁡x\)
2. Differentiate left side as \(\frac{1}{y}\frac{dy}{dx}\)
3. Use product and chain rule on right side: \(1+ln⁡x\)
4. Multiply by \(y\) to get final result

A 1, 2, 3, 4

B 2, 1, 3, 4

C 3, 1, 4, 2

D 1, 3, 2, 4

💡 Short Explanation

Begin with logarithm Differentiate both sides Apply product rule Multiply back by \(y\)

✅ Correct Answer Explanation

For logarithmic differentiation: \(y=x^{x}\) Take logarithm: \(ln⁡y=xln⁡x\) Differentiate implicitly: \(\frac{1}{y}\frac{dy}{dx}=1+ln⁡x\) Multiply by \(y=x^{x}\): \(\frac{dy}{dx}=x^{x}(1+ln⁡x)\) Thus sequence 1 → 2 → 3 → 4 is correct.

❓ Why Other Options Are Incorrect

Option B → Differentiation cannot occur before logarithmic transformation.
Option C → Product rule cannot be applied before equation formation.
Option D → Step 2 must precede right-side differentiation consistency.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Arrange steps according to logical order of logarithmic differentiation.

Final Logic:

Logarithm must always precede differentiation.

🧠 Memory Trick

"Log → Differentiate → Multiply back"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Logarithmic Differentiation

7 A data model dictates
\(y=x^{x}\).
By taking logarithms,
\(ln⁡y=xln⁡x\)
This implicit equation evaluates to:

A \(1/y\)

B \(\ln\,x\)

C \(\frac{1}{y}\frac{dy}{dx}=1+ln⁡x\)

D \(x^{x}\)

💡 Short Explanation

Use logarithmic differentiation Differentiate both sides implicitly Apply product rule on right side

✅ Correct Answer Explanation

Starting with: \(ln⁡y=xln⁡x\) Differentiate: \(\frac{1}{y}\frac{dy}{dx}=1+ln⁡x\) because derivative of \(xln⁡x\) uses product rule. Option C correctly expresses implicit differentiation. Other options show incomplete intermediate terms.

❓ Why Other Options Are Incorrect

Option A → Represents only derivative of \(\ln\,y\), incomplete expression.
Option B → Omits derivative contribution from product rule.
Option D → Original function, not differentiated relation.

🎯 Strategy

Used: Substitution

Application:

Substitute differentiation formulas systematically.

Final Logic:

Implicit differentiation produces derivative equation involving \(dy/dx\).

🧠 Memory Trick

"\(x^{x}\) ⇒ take logs first"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Logarithmic Differentiation

8 Which statements correctly define implicit constraints like
\(x^{2}+y^{2}=1\)?
I. Applying differentiation gives \(2x+2y\frac{dy}{dx}=0\).
II. \(\frac{dy}{dx}=-\frac{x}{y}\).
III. The derivative strictly relies on the quotient rule.

A I only

B II and III

C I and III

D I and II

💡 Short Explanation

Differentiate implicitly Solve for \(dy/dx\) Quotient rule is unnecessary

✅ Correct Answer Explanation

Differentiating implicitly: \(2x+2y\frac{dy}{dx}=0\) Solving gives: \(\frac{dy}{dx}=-\frac{x}{y}\) Thus Statements I and II are correct. Statement III is false because implicit differentiation—not quotient rule—is used.

❓ Why Other Options Are Incorrect

Option A → Ignores valid derivative expression in Statement II.
Option B → Includes incorrect Statement III.
Option C → Statement III is conceptually wrong.

🎯 Strategy

Used: Elimination

Application:

Check each statement individually against implicit differentiation rules.

Final Logic:

Only Statements I and II satisfy implicit differentiation.

🧠 Memory Trick

"Circle derivative: minus x over y"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Implicit Differentiation

9 Vector fields tied to the relation
\(x^{2}+y^{2}=4\)
have a scalar gradient \(\frac{dy}{dx}\). What does it simplify to?

A \(x/y\)

B \(-x/y\)

C \(-y/x\)

D \(y/x\)

💡 Short Explanation

Differentiate implicitly Rearrange derivative terms Solve for \(dy/dx\)

✅ Correct Answer Explanation

Differentiating: \(x^{2}+y^{2}=4\) gives \(2x+2y\frac{dy}{dx}=0\) Hence, \(\frac{dy}{dx}=-\frac{x}{y}\) Option B matches the derivative. Other options either reverse variables or omit negative sign.

❓ Why Other Options Are Incorrect

Option A → Missing negative sign after rearrangement.
Option C → Incorrect reciprocal relationship.
Option D → Both sign and reciprocal are incorrect.

🎯 Strategy

Used: Substitution

Application:

Apply implicit differentiation formula directly.

Final Logic:

Rearranging derivative equation yields \(-x/y\).

🧠 Memory Trick

"Circle slope = negative reciprocal form"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Implicit Differentiation

10 Match logarithmic/implicit derivatives to their original explicit/implicit forms:

List I	List II
1. \(dy/dx=1+ln⁡x\)	a. \(y=x^{x}\)
2. \(dy/dx=-x/y\)	b. \(x^{2}+y^{2}=1\)
3. \(dy/dx=1/x\)	c. \(y=ln⁡x\)

A 1-a, 2-b, 3-c

B 1-b, 2-c, 3-a

C 1-c, 2-a, 3-b

D 1-b, 2-a, 3-c

💡 Short Explanation

Match standard derivative forms Use known logarithmic results Identify implicit circle derivative

✅ Correct Answer Explanation

\(y=x^{x}\Rightarrow \frac{dy}{dx}=x^{x}(1+ln⁡x)\) core derivative term is \(1+ln⁡x\). For \(x^{2}+y^{2}=1\), \(\frac{dy}{dx}=-\frac{x}{y}\) and \(y=ln⁡x\Rightarrow \frac{dy}{dx}=\frac{1}{x}\) Thus matching becomes 1-a, 2-b, 3-c.

❓ Why Other Options Are Incorrect

Option B → Incorrectly matches logarithmic derivative with circle relation.
Option C → Swaps implicit and logarithmic derivatives wrongly.
Option D → Fails to associate \(1/x\) with \(\ln\,x\).

🎯 Strategy

Used: Option Grouping

Application:

Use familiar derivative identities to pair equations rapidly.

Final Logic:

Each derivative uniquely corresponds to one standard function.

🧠 Memory Trick

"Log → \(1/x\), Circle → \(-x/y\)"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Implicit and Logarithmic Differentiation

11 If an implicitly defined function is given by
\(x+y=10\),
evaluating the exact numeric derivative at any point yields:

A 1

B 0

C -1

D 10

💡 Short Explanation

Differentiate both variables implicitly Derivative of constant is zero Rearranging gives slope \(-1\)

✅ Correct Answer Explanation

Differentiating implicitly: \(\frac{d}{dx}(x+y)=\frac{d}{dx}(10)\) gives \(1+\frac{dy}{dx}=0\) Hence, \(\frac{dy}{dx}=-1\) Option C is correct. Option A ignores negative sign. Option B assumes derivative vanishes. Option D confuses slope with equation constant.

❓ Why Other Options Are Incorrect

Option A → Differentiation gives negative slope, not positive one.
Option B → Constant derivative is zero, but variable derivative remains nonzero.
Option D → Number 10 belongs to equation, not derivative value.

🎯 Strategy

Used: Substitution

Application:

Apply implicit differentiation directly to the equation.

Final Logic:

Linear relation rearranges into constant slope \(-1\).

🧠 Memory Trick

"\(x+y=constant\) ⇒ slope \(-1\)"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Implicit Differentiation

12 The boundary of an area satisfies
\(x^{2}+y^{2}=16\).
Putting \(y=\sqrt{16-x^{2}}\), it simplifies to a semicircular boundary. What is the boundary derivative \(\frac{dy}{dx}\)?

A \(x/y\)

B \(y/x\)

C \(-y/x\)

D \(-x/y\)

💡 Short Explanation

Differentiate implicitly Solve for derivative term Circle derivative gives negative ratio

✅ Correct Answer Explanation

Differentiating: \(x^{2}+y^{2}=16\) gives \(2x+2y\frac{dy}{dx}=0\) Therefore, \(\frac{dy}{dx}=-\frac{x}{y}\) Option D correctly represents derivative of a circle. Other options either invert variables or miss the negative sign.

❓ Why Other Options Are Incorrect

Option A → Missing required negative sign.
Option B → Incorrect reciprocal arrangement.
Option C → Variables incorrectly reversed.

🎯 Strategy

Used: Elimination

Application:

Check sign and variable arrangement after implicit differentiation.

Final Logic:

Circle differentiation always gives \(-x/y\).

🧠 Memory Trick

"Circle slope = minus x by y"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Implicit Differentiation

13 If \(sin⁡y=x\), substituting \(cos⁡y=\sqrt{1-x^{2}}\)simplifies the relation for valid domains. The derivative evaluates to:

A \(\sqrt{1-x^{2}}\)

B \(-1/\sqrt{1-x^{2}}\)

C \(1/(1+x^{2})\)

D \(1/\sqrt{1-x^{2}}\)

💡 Short Explanation

Differentiate implicitly Replace \(\cos\,y\) using identity Solve for \(dy/dx\)

✅ Correct Answer Explanation

Given: \(sin⁡y=x\) Differentiate implicitly: \(cos⁡y\frac{dy}{dx}=1\) Thus, \(\frac{dy}{dx}=\frac{1}{\cos\,y}\) Using \(cos⁡y=\sqrt{1-x^{2}}\) we obtain: \(\frac{d}{dx}({sin⁡}^{-1}x)=\frac{1}{\sqrt{1-x^{2}}}\) Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Represents denominator only, not full derivative expression.
Option B → Incorrect negative sign; inverse sine derivative is positive.
Option C → Formula belongs to derivative of \({tan⁡}^{-1}x\).

🎯 Strategy

Used: Substitution

Application:

Substitute trigonometric identity after implicit differentiation.

Final Logic:

Inverse sine derivative equals reciprocal square-root form.

🧠 Memory Trick

"sin⁻¹ ⇒ one over root"

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Derivative of \({sin⁡}^{-1}x\)

📄 Passage

Sometimes the relation between two variables is neither explicit nor implicit, but some link of a third variable establishes a relation between the first two variables. This third variable is mathematically called the parameter. A relation expressed between two variables \(x\) and \(y\) in the form
\(x=f(t), y=g(t)\)
is said to be in parametric form. To find the derivative of such functions, we utilize the chain rule framework:
\(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\)
Thus, rearranging provides
\(\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\)
provided that the denominator \(dx/dt\neq 0\).

According to the passage on parametric forms, a third variable is introduced to establish a relation. What is this third variable mathematically called?

A An explicit variable

B The parameter

C The unbounded constant

D The implicit link

💡 Short Explanation

Parametric equations use third variable Variable links \(x\) and \(y\) Passage directly defines parameter

✅ Correct Answer Explanation

In parametric equations, both \(x\) and \(y\) depend on a third variable \(t\). This linking variable is called the parameter. Option B exactly matches the passage definition. Other options are unrelated mathematical terms and do not define parametric representation.

❓ Why Other Options Are Incorrect

Option A → Explicit variables directly express one variable through another, unlike parametric form.
Option C → No such standard term exists in parametric differentiation.
Option D → "Implicit link" is not the accepted mathematical terminology.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Locate exact conceptual definition stated in the passage.

Final Logic:

The passage explicitly names the third variable as parameter.

🧠 Memory Trick

"Parameter = linking variable"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Parametric Differentiation

📄 Passage

Based on the passage, the parametric derivative formula is universally valid strictly provided that:

A \(dx/dt\neq 0\)

B \(dy/dt=0\)

C Both variables are trigonometric

D The parameter is positive

💡 Short Explanation

Parametric derivative involves division Denominator cannot become zero Passage explicitly states condition

✅ Correct Answer Explanation

For parametric differentiation: \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\) Since division by zero is undefined, \(dx/dt\neq 0\) is necessary. Option A directly follows from the passage. Other options are unnecessary conditions unrelated to derivative validity.

❓ Why Other Options Are Incorrect

Option B → \(dy/dt\) may be zero without invalidating formula.
Option C → Parametric equations are not limited to trigonometric functions.
Option D → Parameter need not be positive.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Identify the exact restriction mentioned in the passage.

Final Logic:

Division requires nonzero denominator.

🧠 Memory Trick

"Parametric rule: denominator ≠ 0"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Parametric Differentiation

16 For the parametric relation
\(x=tan⁡t, y=t\),
applying the exact quotient formula evaluates to:

A \(1/tan⁡t\)

B \(\tan\,t\)

C \(1/(1+x^{2})\)

D \(1+x^{2}\)

💡 Short Explanation

Use parametric derivative formula Differentiate both equations Convert using trigonometric identity

✅ Correct Answer Explanation

Given: \(x=tan⁡t,y=t\) Then, \(\frac{dy}{dt}=1,\frac{dx}{dt}={sec⁡}^{2}t\) Hence, \(\frac{dy}{dx}=\frac{1}{{sec⁡}^{2}t}={cos⁡}^{2}t\) Since \(1+{tan⁡}^{2}t={sec⁡}^{2}t\) and \(x=tan⁡t\), \(\frac{dy}{dx}=\frac{1}{1+x^{2}}\) Thus Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Reciprocal tangent is unrelated to derivative obtained.
Option B → Represents original trigonometric function only.
Option D → Missing reciprocal from secant identity conversion.

🎯 Strategy

Used: Substitution

Application:

Substitute parametric derivatives into quotient formula.

Final Logic:

Use identity \(1+{tan⁡}^{2}t={sec⁡}^{2}t\).

🧠 Memory Trick

"tan⁻¹ derivative ⇒ one over \(1+x^{2}\)"

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Derivative of \({tan⁡}^{-1}x\)

17 Given the parametric system
\(x=t^{2}, y=t^{3}\),
the derivative equals:

A \(t^{3}/t^{2}\)

B \(t\)

C \((3t^{2})/2t\)

D \((3t^{2})/2t=3t/2\)

💡 Short Explanation

Differentiate both parametric equations Divide derivatives Simplify algebraically

✅ Correct Answer Explanation

For parametric differentiation: \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\) Now, \(\frac{dy}{dt}=3t^{2},\frac{dx}{dt}=2t\) Therefore, \(\frac{dy}{dx}=\frac{3t^{2}}{2t}=\frac{3t}{2}\) Option D gives fully simplified derivative. Other options are incomplete or incorrect.

❓ Why Other Options Are Incorrect

Option A → Uses original functions instead of derivatives.
Option B → Oversimplifies and misses factor \(3/2\).
Option C → Correct intermediate form but not simplified final answer.

🎯 Strategy

Used: Substitution

Application:

Insert derivatives into parametric formula carefully.

Final Logic:

Simplification produces \(3t/2\).

🧠 Memory Trick

"Differentiate first, then divide"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Parametric Differentiation

18 When defining the second order derivative
\(\frac{d^{2}y}{dx^{2}}\),
if \(dy/dx\) exists, the required existence condition ensures the first derivative is:

A A constant

B Differentiable

C Zero

D Parametric

💡 Short Explanation

Second derivative means derivative of first derivative First derivative must itself be differentiable Higher differentiation requires continuity of process

✅ Correct Answer Explanation

The second derivative is defined as derivative of the first derivative: \(\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\) Hence \(dy/dx\) must be differentiable. Option B correctly states the condition. Other options impose unnecessary or incorrect restrictions.

❓ Why Other Options Are Incorrect

Option A → First derivative need not be constant.
Option C → Zero derivative is not necessary for second derivative existence.
Option D → Parametric form is unrelated to differentiability requirement.

🎯 Strategy

Used: Elimination

Application:

Check definition of higher-order derivatives carefully.

Final Logic:

Derivative of derivative requires differentiability.

🧠 Memory Trick

"Second derivative ⇒ first derivative differentiable"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Second Order Derivatives

19 Evaluate the second order derivative for the composite exponential function
\(y=e^{x^{2}}\).

A \((4x^{2}+2)e^{x^{2}}\)

B \(2xe^{x^{2}}\)

C \(e^{2x}\)

D \(x^{2}e^{x^{2}}\)

💡 Short Explanation

Apply chain rule twice Differentiate product carefully Factor exponential term

✅ Correct Answer Explanation

First derivative: \(\frac{dy}{dx}=2xe^{x^{2}}\) Now differentiate again using product rule: \(\frac{d^{2}y}{dx^{2}}=2e^{x^{2}}+2x(2xe^{x^{2}})\) Thus, \(\frac{d^{2}y}{dx^{2}}=(4x^{2}+2)e^{x^{2}}\) Option A is correct. Other options represent incomplete differentiation.

❓ Why Other Options Are Incorrect

Option B → Only first derivative is given, not second derivative.
Option C → Incorrect exponential simplification.
Option D → Missing terms arising from product differentiation.

🎯 Strategy

Used: Substitution

Application:

Apply chain rule followed by product rule sequentially.

Final Logic:

Second differentiation introduces additional \(2\) and \(4x^{2}\)terms.

🧠 Memory Trick

"Exponential composite ⇒ chain rule twice"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Higher Order Derivatives

20 If functional tracking gives
\(y=ln⁡(x^{2}+1)\),
functional analysis of the second derivative proves that \(\frac{d^{2}y}{dx^{2}}\)equals:

A \(2x/(x^{2}+1)\)

B \(1/(x^{2}+1)\)

C \(\frac{2(1-x^{2})}{{\left(x^{2}+1\right)}^{2}}\)

D \(ln⁡(2x)\)

💡 Short Explanation

Differentiate logarithmic function once Apply quotient rule again Simplify numerator carefully

✅ Correct Answer Explanation

First derivative: \(\frac{dy}{dx}=\frac{2x}{x^{2}+1}\) Differentiate again using quotient rule: \(\frac{d^{2}y}{dx^{2}}=\frac{2(x^{2}+1)-4x^{2}}{{\left(x^{2}+1\right)}^{2}}\) Simplifying: \(\frac{d^{2}y}{dx^{2}}=\frac{2(1-x^{2})}{{\left(x^{2}+1\right)}^{2}}\) Thus Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Represents only first derivative.
Option B → Incomplete derivative form lacking quotient differentiation.
Option D → Logarithm is not differentiated correctly.

🎯 Strategy

Used: Elimination

Application:

Identify which option represents second derivative instead of first derivative.

Final Logic:

Repeated differentiation introduces squared denominator.

🧠 Memory Trick

"Log second derivative ⇒ squared denominator"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Higher Order Derivatives

UG Mathematics Booster Test 3 - Advanced Differentiation

Test Complete!

Answer Review