Continuity and Differentiability Class 12 NCERT Solutions Chapter 5 Ex 5.4

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.4 cover all 10 questions on derivatives of exponential and logarithmic functions. Collegedunia's subject experts solve each question with clear, exam-ready steps that match the 2026-27 NCERT print. The free PDF of these solutions is available to download on this page.

These solutions are written by experienced Class 12 Maths teachers and checked against the official NCERT answer key. Every question shows the formula recall, the substitution, and the boxed final derivative on separate lines.

• CBSE Weightage: 8-10 marks (Continuity and Differentiability plus Applications of Derivatives, Calculus unit total 35 marks)
• JEE Main Weightage: 6-8% of the Mathematics section, with 2-3 questions on exponential and logarithmic differentiation
• Exercise 5.4 Question Count: 10 questions covering direct e^x and log x derivatives, then their composite extensions

Each solution starts with the relevant identity, then layers the chain rule or quotient rule as needed, so you can transfer the same structure to a CBSE board answer script.

Continuity and Differentiability Ex 5.4 Solved Step by Step (Video)

Source: NCERT Wallah on YouTube

How will Collegedunia's NCERT Solutions for Exercise 5.4 Help You?

Exercise 5.4 is the foundation for Exercise 5.5's logarithmic differentiation, so a weak grasp here costs marks later too. Collegedunia's solution PDF is built to fix exactly this.

• Every differentiation rule named on its own line, with the formula and the substituted variable clearly separated.
• Formula recall printed at the top of any question that uses the general exponential rule d/dx a^x = a^x log a or a change-of-base logarithm.
• Expert Solution block showing the alternate route via taking logarithms on both sides for the quotient-rule questions.
• Tip callouts at common mistake points: writing log a ambiguously, forgetting the inner chain-rule factor, and sign errors on log of a negative argument.

Important Concepts Covered in Class 12 Maths Exercise 5.4

Exercise 5.4 stays inside the four concept boxes below. Reviewing them before attempting the questions cuts solving time by roughly half.

The exponential and logarithmic identities drive most calculus-based growth and decay problems in Class 12, so practice here pays off well beyond this chapter.

Exam Relevance of Continuity and Differentiability Exercise 5.4

Continuity and Differentiability is part of the Calculus unit, which carries 35 marks in the CBSE Class 12 Maths paper. Exercise 5.4 contributes most of the 2-mark and 3-mark exponential and logarithmic differentiation questions. A quick look at recent board and JEE appearances:

The exponential function reappears in Application of Derivatives, Integrals, and Differential Equations, so Ex 5.4 has the longest reach of any exercise in this chapter.

Common Mistakes Students Make in Exercise 5.4

These solutions follow the same formal notation as the official NCERT print, line by line. Three error patterns cause most of the marks lost on this exercise in CBSE board scripts.

• Writing log a ambiguously. In NCERT, log without a base means natural log. Students who treat it as base 10 get the wrong constant in the general exponential rule.
• Forgetting the inner chain factor on e^f(x) and log(f(x)). The derivative needs the extra f'(x) factor, not just the outer function.
• Quotient-rule simplification errors: students often forget to square the denominator or invert the numerator subtraction order.

Other Resources for Class 12 Maths Chapter 5 Continuity and Differentiability

Pair the Exercise 5.4 solutions with the rest of the Chapter 5 resource library for complete preparation.

Exercise-wise Breakdown of the Continuity and Differentiability Chapter

The Continuity and Differentiability chapter splits into 7 numbered exercises plus a Miscellaneous Exercise. The table below maps every exercise to the concept it tests.

NCERT Solutions for Class 12 Maths: All Chapters

Chapter-by-chapter NCERT Solutions for the rest of Class 12 Maths, aligned to the 2026-27 print.

The Exercise 5.4 solutions PDF is available above as a free download, aligned to the 2026-27 NCERT Class 12 Maths syllabus.

All NCERT Solutions for Continuity and Differentiability Ex 5.4 with Step-by-Step Working

Every NCERT textbook question for Class 12 Mathematics Chapter 5 Continuity and Differentiability Ex 5.4 is listed below with its full Solution and Expert Solution hidden inside collapsible tabs. Click Check Solution to reveal the step-by-step working; click Expert Solution for the expanded explanation.

Questions

Q 5.1

Differentiate e^xsin x with respect to x.

Check SolutionExpert Solution

Concept used. Quotient rule: (u/v)' = (u'v - uv')/v². Here u = e^x, u' = e^x, v = sin x, v' = cos x.

1. Apply the quotient rule: ddxe^xsin x = e^x sin x - e^x cos xsin² x.
2. Factor e^x: = e^x (sin x - cos x)sin² x.

ddx(e^xsin x) = e^x(sin x - cos x)sin² x, x ≠ nπ.

AS

Aarav Sharma

M.Sc Mathematics, IIT Bombay

Verified Expert

Quick reading. Quotient rule with e^x on top and sin x on bottom.

1. Numerator after rule: e^x sin x - e^x cos x.
2. Factor e^x: e^x(sin x - cos x).
3. Divide by sin² x.

e^x(sin x - cos x)/sin² x.

Q 5.2

Differentiate e^{sin-1 x} with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule. ddx e^f(x) = e^f(x) f'(x). Here f(x) = sin^-1 x with derivative 1/√1 - x².

1. Let y = e^{sin-1 x}. Then dydx = e^{sin-1 x} · ddx(sin^-1 x) = e^{sin-1 x} · 1√1 - x².

dydx = e^{sin-1 x}√1 - x², -1 < x < 1.

SI

Sneha Iyer

Ph.D Mathematics, IIT Delhi

Verified Expert

Quick reading. Standard e^u chain rule with u = sin^-1 x.

1. Outer derivative: e^{sin-1 x}.
2. Inner derivative: 1/√1-x².

e^{sin-1 x}/√1 - x².

Q 5.3

Differentiate e^x3 with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule. ddx e^f(x) = e^f(x) f'(x) with f(x) = x³, f'(x) = 3x².

1. Differentiate: ddx e^x3 = e^x3 · 3 x² = 3 x² e^x3.

3 x² e^x3.

VG

Vivaan Gupta

M.Tech CS, IIT Madras

Verified Expert

Quick reading. e^x3 times derivative of exponent.

1. Derivative of x³ is 3x².
2. Multiply: 3 x² e^x3.

3 x² e^x3.

Q 5.4

Differentiate sin(tan^-1 e^-x) with respect to x.

Check SolutionExpert Solution

Concept used. Three-layer chain: outer sin, middle tan^-1, inner e^-x.

1. Let u = e^-x, v = tan^-1 u, y = sin v.
2. Derivatives layer by layer: aligned dydv &= cos v = cos(tan^-1 e^-x),
   dvdu &= 11 + u² = 11 + e^-2x,
   dudx &= -e^-x. aligned
3. Chain rule (three layers): dydx = cos(tan^-1 e^-x) · 11 + e^-2x · (-e^-x) = -e^-x cos(tan^-1 e^-x)1 + e^-2x.

dydx = -e^-x cos(tan^-1 e^-x)1 + e^-2x.

AM

Aanya Mehta

Ph.D Pure Mathematics, IISc Bangalore

Verified Expert

Structural observation. Triple chain; differentiate sin, then tan^-1, then e^-x.

1. sin' = cos(tan^-1e^-x).
2. tan^-1' = 1/(1 + e^-2x).
3. (e^-x)' = -e^-x.
4. Multiply.

-e^-xcos(tan^-1 e^-x)/(1 + e^-2x).

Q 5.5

Differentiate log(cos e^x) with respect to x.

Check SolutionExpert Solution

Concept used. Three-layer chain: outer log, middle cos, inner e^x.

1. Let u = e^x, v = cos u, y = log v.
2. Layer-wise derivatives: aligned dydv &= 1v = 1cos e^x,
   dvdu &= -sin u = -sin e^x,
   dudx &= e^x. aligned
3. Multiply: dydx = 1cos e^x · (-sin e^x) · e^x = -e^x tan(e^x).

dydx = -e^x tan(e^x) where cos e^x ≠ 0.

PS

Priya Singh

M.Sc Applied Mathematics, IIT Kanpur

Verified Expert

Quick reading. Three layers; sin/cos = tan produces a clean answer.

1. Layer 1: 1/cos e^x.
2. Layer 2: -sin e^x.
3. Layer 3: e^x.
4. Product: -e^x tan e^x.

-e^x tan e^x.

Q 5.6

Differentiate e^x + e^x2 + e^x3 + e^x4 + e^x5 with respect to x.

Check SolutionExpert Solution

Concept used. Sum rule, then chain rule on each e^xk: ddx e^xk = e^xk · k x^{k - 1}.

1. Differentiate term-by-term: aligned ddx e^x &= e^x,
   ddx e^x2 &= 2x e^x2,
   ddx e^x3 &= 3x² e^x3,
   ddx e^x4 &= 4x³ e^x4,
   ddx e^x5 &= 5x⁴ e^x5. aligned
2. Add: dydx = e^x + 2x e^x2 + 3x² e^x3 + 4 x³ e^x4 + 5 x⁴ e^x5.

dydx = e^x + 2x e^x2 + 3x² e^x3 + 4x³ e^x4 + 5 x⁴ e^x5.

KJ

Karan Joshi

M.Sc Mathematics, IIT Bombay

Verified Expert

Quick reading. Five separate e^xk derivatives summed.

1. Each e^xk produces k x^k-1 e^xk.
2. Sum five contributions.

_k=1⁵ k x^k-1 e^xk.

Q 5.7

Differentiate √e^√x with respect to x, x > 0.

Check SolutionExpert Solution

Concept used. Simplify first using exponent laws: √e^√x = e^√x/2. Then apply chain rule.

1. Rewrite: y = √e^√x = (e^√x)^1/2 = e^√x/2.
2. Differentiate using ddx e^u = e^u u' with u = √x/2, u' = 12 · 12√x = 14 √x: dydx = e^√x/2 · 14 √x = e^√x/24 √x = √e^√x4 √x.

dydx = √e^√x4√x (x > 0).

PK

Pranav Kapoor

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. Exponent simplification before differentiation.

1. √e^√x = e^√x/2.
2. Derivative: e^√x/2 · 14√x.

e^√x/2/(4√x).

Q 5.8

Differentiate log(log x) with respect to x, x > 1.

Check SolutionExpert Solution

Concept used. Chain rule with ddx log u = 1u · u'.

1. Let u = log x, so y = log u, with u' = 1/x.
2. Chain rule: dydx = 1u · u' = 1log x · 1x = 1x log x.

dydx = 1x log x (x > 1).

TP

Tara Pillai

Ph.D Mathematics, IIT Delhi

Verified Expert

Quick reading. Two layers of log derivatives.

1. Outer: 1/log x.
2. Inner: 1/x.
3. Product: 1/(x log x).

1/(x log x).

Q 5.9

Differentiate cos xlog x with respect to x, x > 0.

Check SolutionExpert Solution

Concept used. Quotient rule with u = cos x, u' = -sin x, v = log x, v' = 1/x.

1. Apply quotient rule: dydx = u' v - u v'v² = (-sin x)(log x) - (cos x)(1/x)(log x)².
2. Simplify the numerator: = -(sin x)(log x) - cos xx(log x)² = -x (sin x)(log x) + cos xx (log x)².

dydx = -x (sin x)(log x) + cos xx (log x)².

KR

Krishna Rao

M.Sc Mathematics, IIT Bombay

Verified Expert

Structural observation. Quotient with cos/log; bring the 1/x inside by multiplying numerator and denominator by x.

1. Raw quotient rule output: ((-sin x) log x - cos x/x)/(log x)².
2. Multiply num and den by x to clear the inner fraction.

-(x sin x log x + cos x)/(x (log x)²).

Q 5.10

Differentiate cos(log x + e^x) with respect to x, x > 0.

Check SolutionExpert Solution

Concept used. Chain rule with outer cos and inner log x + e^x. Derivative of the inner: 1/x + e^x.

1. Let u = log x + e^x, y = cos u.
2. dudx = 1x + e^x.
3. Chain rule: dydx = -sin u · dudx = -sin(log x + e^x) (1x + e^x).

dydx = -(1x + e^x) sin(log x + e^x) (x > 0).

MC

Meera Chatterjee

M.Sc Mathematics, ISI Kolkata

Verified Expert

Quick reading. Standard chain rule with sum-derivative inside.

1. Outer: -sin(log x + e^x).
2. Inner derivative: 1/x + e^x.
3. Multiply.

-(1/x + e^x)sin(log x + e^x).