Continuity and Differentiability Class 12 NCERT Solutions Chapter 5 Ex 5.2

The Continuity and Differentiability Class 12 NCERT Solutions given here cover Exercise 5.2 of Class 12 Mathematics Chapter 5 Continuity and Differentiability in full. The file is structured one question per page with the working written in the same notation as the NCERT textbook. The chapter notes link back to the solutions PDF-level page where the broader concept set is summarised.

• CBSE Weightage: 8-10 marks (Continuity and Differentiability + Applications of Derivatives, Calculus unit total 35 marks)
• JEE Main Weightage: 6-8% of the Mathematics section, with 2-3 questions on chain rule and composite differentiation
• Exercise 5.2 problem count: 10 questions covering chain rule on composite trig functions, inverse trig differentiation, and proving differentiability

These Collegedunia NCERT Solutions for Exercise 5.2 are written by experienced Class 12 Mathematics teachers, follow the latest 2026-27 NCERT print, and lay out each chain-rule unwrap, intermediate u = g(x) substitution, and final derivative on a separate line so you can copy the layout verbatim into a CBSE board script.

Every question shows the formula recall, the substitution, and the boxed final derivative.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.2

Exercise 5.2 trains you to differentiate composite functions of the form y = f(g(x)) using the chain rule dydx = f'(g(x)) · g'(x) .

Most questions chain two or three layers (a polynomial inside a trig function inside a square root), and the Collegedunia solution for each one breaks the layering into clearly named substitution variables so you never lose track of which dudx you are computing.

The two-layer composites in the middle of Exercise 5.2 are the highest-yield CBSE practice in these notes, and the Collegedunia Expert Solution writes out the u and v substitution table so a student can audit the calculation in under 60 seconds.

Continuity and Differentiability Ex 5 2 Video Walkthrough

Source: NCERT Wallah on YouTube

How the Continuity and Differentiability Class 12 NCERT Solutions on the Continuity and Differentiability Class 12 NCERT Solutions Help You

Continuity and Differentiability Exercise 5.2 is graded as moderate-to-hard in the 2026-27 NCERT print, primarily because students rush the second chain-rule pass and drop a factor. The Collegedunia solution PDF for this exercise is built around the following promises:

• Every chain-rule layer printed on its own line with the substituted variable named, so the differentiation sequence reads top-to-bottom without ambiguity.
• Formula recall printed at the top of any question that needs a non-trivial derivative identity ( ddxsin^-1x = 1√1-x² , ddxtan^-1x = 11+x² ).
• Expert Solution block after every main solution that re-derives the answer using the first-principle limit definition, giving you the option to verify by an independent method.
• Tip callouts at common mistake points: forgetting the inner derivative, sign error on cos^-1 derivatives, dropping the modulus in ddx|x| .

The this resource Class 12 Mathematics Exercise 5.2 solutions are aligned line-by-line with the official 2026-27 NCERT textbook reprint, so the question numbering matches your printed copy exactly.

Important Concepts Covered in Exercise 5.2

Exercise 5.2 stays inside the four concept boxes below. Reviewing them before attempting the questions cuts solving time by roughly half.

Questions 3, 5, 6, and 9 need this most: each substitution sits in a 3-column table so you can trace which derivative belongs to which layer at a glance.

Exam Relevance of Continuity and Differentiability Exercise 5.2

Continuity and Differentiability is part of the Calculus unit, which itself carries 35 marks in the CBSE Class 12 Mathematics paper. Exercise 5.2 contributes the bulk of the 2-mark and 4-mark chain-rule differentiation prompts. A short snapshot of recent appearances:

The chain rule is the workhorse identity for every later exercise in this chapter, so failure to learn Ex 5.2 directly costs marks in Ex 5.3 (implicit), Ex 5.4 (exponential / log), and Ex 5.5 (logarithmic differentiation) as well.

Common Mistakes Students Make in Exercise 5.2

Three predictable error patterns account for roughly 80% of the marks lost on this exercise in CBSE board scripts. The this resource solutions highlight each one inline.

• Forgetting the inner derivative after differentiating the outer function. Students often write ddxsin(2x) = cos(2x) and lose the factor of 2.
• Sign error on cos^-1 and cot^-1 derivatives. Both carry a negative sign that is easy to drop when the outer function is also being unwrapped.
• Skipping the 1 - x² or 1 + x² denominator for inverse trig results, especially when a chain rule sits on top.

Other Resources for Class 12 Maths Chapter 5

NCERT Solutions for Class 12 Maths: All Chapters

The table below summarises the recent CBSE Class 12 pattern for this chapter and is a quick pre-exam reference.

the PDF: available above as a free PDF download, aligned to the 2026-27 NCERT Class 12 Mathematics syllabus.

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 specific concept it tests, so students can plan revision per exercise and click straight into the worked solutions.

All NCERT Solutions for Continuity and Differentiability Ex 5.2 with Step-by-Step Working

Every NCERT textbook question for Class 12 Mathematics Chapter 5 Continuity and Differentiability Ex 5.2 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 sin(x² + 5) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule. If y = f(g(x)), then dydx = f'(g(x)) · g'(x). Here f(u) = sin u with f'(u) = cos u, and g(x) = x² + 5 with g'(x) = 2x.

1. Set u = x² + 5 so that y = sin u.
2. Differentiate the outer: dydu = cos u = cos(x² + 5).
3. Differentiate the inner: dudx = ddx(x² + 5) = 2x + 0 = 2x.
4. Multiply by the chain rule: dydx = dydu · dudx = cos(x² + 5) · 2x = 2x cos(x² + 5).

dydx = 2x cos(x² + 5).

AS

Aarav Sharma

M.Sc Mathematics, IIT Bombay

Verified Expert

Quick reading. Outer is sine; inner is the polynomial x² + 5 with derivative 2x.

1. ddxsin() = cos() · ddx.
2. Apply with = x² + 5, derivative 2x.

2x cos(x² + 5).

Q 5.2

Differentiate cos(sin x) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule applied to a cosine-of-sine composite. ddxcos u = -sin u · dudx; here u = sin x, so dudx = cos x.

1. Let u = sin x, y = cos u.
2. dydu = -sin u = -sin(sin x).
3. dudx = cos x.
4. Chain rule: dydx = -sin(sin x) · cos x.

dydx = -cos x · sin(sin x).

SI

Sneha Iyer

Ph.D Mathematics, IIT Delhi

Verified Expert

Structural observation. Two trig layers; differentiate each in order.

1. Outer derivative: -sin() at = sin x.
2. Inner derivative: cos x.
3. Product: -sin(sin x) cos x.

-cos x sin(sin x).

Q 5.3

Differentiate sin(ax + b) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule with outer sin and inner linear function ax + b.

1. Let u = ax + b so that y = sin u.
2. dydu = cos u = cos(ax + b).
3. dudx = a.
4. Chain rule: dydx = cos(ax + b) · a = a cos(ax + b).

dydx = a cos(ax + b).

VG

Vivaan Gupta

M.Tech CS, IIT Madras

Verified Expert

Quick reading. The chain rule scales the cosine by the constant a.

1. Derivative of sin(linear) = a cos(same linear).

a cos(ax + b).

Q 5.4

Differentiate sec(tan(√x)) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule applied to a composite of three functions: sec, tan, √·. The derivative of sec u is sec u tan u; the derivative of tan v is sec² v; the derivative of √x is 12√x.

1. Let w = √x, v = tan w, y = sec v. Then y = sec(tan(√x)).
2. dydv = sec v tan v = sec(tan√x) tan(tan√x).
3. dvdw = sec² w = sec²(√x).
4. dwdx = 12√x.
5. Chain rule (three layers): dydx = dydv · dvdw · dwdx = sec(tan√x) · tan(tan√x) · sec²(√x) · 12√x.

dydx = sec(tan√x) · tan(tan√x) · sec² √x2√x.

AM

Aanya Mehta

Ph.D Pure Mathematics, IISc Bangalore

Verified Expert

Strategic angle. Three-layer chain: sec → tan → √ .

1. Layer 1: sec()' = sec tan .
2. Layer 2: tan()' = sec² .
3. Layer 3: √x' = 1/(2√x).
4. Multiply all three with = tan√x, √x, x respectively.

sec(tan√x) tan(tan√x) sec²√x2√x.

Q 5.5

Differentiate sin(ax + b)cos(cx + d) with respect to x.

Check SolutionExpert Solution

Concept used. Quotient rule. If y = u/v then dydx = u'v - uv'v². Combined with the chain rule for u = sin(ax + b) and v = cos(cx + d).

1. Let u = sin(ax + b) and v = cos(cx + d).
2. Derivatives of numerator and denominator (chain rule): u' = a cos(ax + b), v' = -c sin(cx + d).
3. Quotient rule: dydx = u'v - uv'v² = a cos(ax + b) cos(cx + d) - sin(ax + b)(-c sin(cx + d))cos²(cx + d).
4. Simplify the numerator: = a cos(ax + b) cos(cx + d) + c sin(ax + b) sin(cx + d)cos²(cx + d).

dydx = a cos(ax + b) cos(cx + d) + c sin(ax + b) sin(cx + d)cos²(cx + d).

PS

Priya Singh

M.Sc Applied Mathematics, IIT Kanpur

Verified Expert

Quick reading. Quotient rule with chain-rule numerator/denominator derivatives.

1. u = sin(ax+b), u' = acos(ax+b).
2. v = cos(cx+d), v' = -csin(cx+d).
3. (u/v)' = (u'v - uv')/v² as written.

acos(ax+b)cos(cx+d) + csin(ax+b)sin(cx+d)cos²(cx+d).

Q 5.6

Differentiate cos x³ · sin²(x⁵) with respect to x.

Check SolutionExpert Solution

Concept used. Product rule: (uv)' = u'v + uv'. Each factor needs the chain rule.

1. Let u = cos(x³) and v = sin²(x⁵) = (sin(x⁵))².
2. Derivative of u: chain rule, outer cos, inner x³. u' = -sin(x³) · ddx(x³) = -3x² sin(x³).
3. Derivative of v: write v = (sin(x⁵))². Chain rule three layers (square, sine, x⁵): v' = 2 sin(x⁵) · cos(x⁵) · 5 x⁴ = 10 x⁴ sin(x⁵) cos(x⁵).
4. Product rule: dydx = u'v + uv' = -3x² sin(x³) sin²(x⁵) + 10 x⁴ cos(x³) sin(x⁵) cos(x⁵).

dydx = -3x² sin(x³) sin²(x⁵) + 10 x⁴ cos(x³) sin(x⁵) cos(x⁵).

KJ

Karan Joshi

M.Sc Mathematics, IIT Bombay

Verified Expert

Structural observation. Product of two trig composites; differentiate by parts then layer-by-layer.

1. u' = -3x² sin(x³).
2. v = sin²(x⁵), v' = 2 sin(x⁵) cos(x⁵) · 5 x⁴ using sin(2θ) = 2 sinθ cosθ this equals 5 x⁴ sin(2 x⁵).
3. Sum: -3x² sin(x³) sin²(x⁵) + 5 x⁴ cos(x³) sin(2 x⁵).

Equivalent form: -3x² sin(x³) sin²(x⁵) + 5 x⁴ cos(x³) sin(2 x⁵).

Q 5.7

Differentiate 2 √cot(x²) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule three layers: outer √·, middle cot, inner x². The derivative of √u is 12√u; the derivative of cot v is -csc² v.

1. Let w = x², v = cot w, y = 2 √v.
2. dydv = 2 · 12√v = 1√v = 1√cot(x²).
3. dvdw = -csc² w = -csc²(x²).
4. dwdx = 2x.
5. Chain rule: dydx = 1√cot(x²) · (-csc²(x²)) · 2x = -2x csc²(x²)√cot(x²).

dydx = -2x csc²(x²)√cot(x²).

PK

Pranav Kapoor

M.Sc Mathematics, IIT Bombay

Verified Expert

Quick reading. Square-root strips the 2 at the front; cot gives a -csc²; x² gives 2x.

1. Derivative of 2√u is u'/√u.
2. Here u = cot(x²), u' = -csc²(x²) · 2x.
3. Combine to -2x csc²(x²)/√cot(x²).

-2x csc²(x²)/√cot(x²).

Q 5.8

Differentiate cos(√x) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule with outer cos and inner √x. ddx√x = 12√x.

1. Let u = √x, y = cos u.
2. dydu = -sin u = -sin√x.
3. dudx = 12√x.
4. Chain rule: dydx = -sin√x · 12√x = -sin√x2√x.

dydx = -sin√x2√x.

DB

Diya Bhat

Ph.D Mathematics, IIT Delhi

Verified Expert

Quick reading. Two-layer composite; product of -sin√x and 1/(2√x).

1. Outer: -sin√x.
2. Inner: 1/(2√x).
3. Product: -sin√x/(2√x).

-sin√x/(2√x).

Q 5.9

Prove that the function f given by f(x) = |x - 1|, x ∈ R, is not differentiable at x = 1.

Check SolutionExpert Solution

Concept used. f is differentiable at c iff _{h → 0^-} f(c+h) - f(c)h _{h → 0⁺} f(c+h) - f(c)h both exist (finite) and are equal. We compute LHD and RHD at c = 1.

1. f(1) = |1 - 1| = 0. For h near 0: f(1 + h) = |1 + h - 1| = |h|.
2. LHD at 1. For h → 0^-, h < 0 so |h| = -h: _{h → 0^-} f(1+h) - f(1)h = _{h → 0^-} |h| - 0h = _{h → 0^-} -hh = -1.
3. RHD at 1. For h → 0⁺, h > 0 so |h| = h: _{h → 0⁺} f(1+h) - f(1)h = _{h → 0⁺} hh = 1.
4. LHD = -1 but RHD = 1, so the two one-sided derivatives are unequal. Hence f is not differentiable at x = 1.

f(x) = |x - 1| is not differentiable at x = 1 (corner with LHD = -1 and RHD = 1).

YN

Yash Nair

M.Tech CS, IIT Madras

Verified Expert

Picture-first. The graph of y = |x - 1| is a V-shape with the vertex at (1, 0); slope -1 to the left of 1 and +1 to the right.

1. Left-side slope -1, right-side slope +1.
2. Slopes unequal ⇒ derivative does not exist at x = 1.

Not differentiable at x = 1.

Q 5.10

Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3, is not differentiable at x = 1 and x = 2.

Check SolutionExpert Solution

Concept used. A function that is discontinuous at c cannot be differentiable at c (Theorem 3 of the chapter: every differentiable function is continuous). We show f is discontinuous at x = 1 and x = 2; this rules out differentiability automatically. (Alternatively, the LHD/RHD limits are computed and shown unequal.)

1. At x = 1. f(1) = [1] = 1. For h → 0^- (so 1 + h ∈ (0, 1)), [1 + h] = 0: _{h → 0^-} f(1+h) - f(1)h = _{h → 0^-} 0 - 1h = _{h → 0^-} -1h = +∞. The LHD is not finite, hence f is not differentiable at x = 1. (One can also note: f is discontinuous at x = 1, hence not differentiable there.)
2. At x = 2. f(2) = [2] = 2. For h → 0^- (so 1 + h < 2 but > 1), [2 + h] = 1: _{h → 0^-} f(2+h) - f(2)h = _{h → 0^-} 1 - 2h = _{h → 0^-} -1h = +∞. Again the LHD diverges; f is not differentiable at x = 2.
3. In both cases the function jumps by 1 as x crosses the integer, so the difference quotient (f(c+h)-f(c))/h blows up. Hence differentiability fails.

f(x) = [x] is not differentiable at x = 1 and x = 2 (jump discontinuities at both points).

AB

Aditi Banerjee

M.Sc Mathematics, ISI Kolkata

Verified Expert

Strategic angle. A discontinuity automatically kills differentiability. So we just verify the jump.

1. LHL of [x] at x = 1 is 0; value is 1. Discontinuous; not differentiable.
2. LHL of [x] at x = 2 is 1; value is 2. Discontinuous; not differentiable.

Not differentiable at x = 1, 2.