Continuity and Differentiability Class 12 NCERT Solutions Chapter 5 Ex 5.5

The Continuity and Differentiability Class 12 NCERT Solutions given here cover Exercise 5.5 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 logarithmic differentiation in nearly every shift
• Exercise 5.5 problem count: 18 questions covering f(x)^g(x) forms, products and quotients with mixed bases, and parametric differentiation

These Collegedunia NCERT Solutions for Exercise 5.5 are written by experienced Class 12 Mathematics teachers, follow the latest 2026-27 NCERT print, and lay out each log-on-both-sides step, term-by-term differentiation, and isolation of dydx on a separate line. Every question shows the formula recall, the differentiation, and the boxed final answer.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.5

The the PDF address this in the same order as the NCERT textbook.

Exercise 5.5 introduces logarithmic differentiation, the only systematic method for handling functions where both the base and the exponent vary, such as y = x^x , y = (sin x)^x , y = x^{sin x} .

The core idea is: take log on both sides, use log(a^b) = b log a to bring the exponent down, differentiate implicitly, then solve for dydx .

The 8 variable-base-and-exponent questions form the spine of this exercise. The Collegedunia Expert Solution writes the substitution y = f(x)^g(x) log y = g(x) log f(x) on a fresh line before any differentiation begins, so the method is always reproducible from scratch.

Continuity and Differentiability Ex 5 5 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 5.5 is graded as hard in the 2026-27 NCERT print and is the single most-asked exercise from the this chapter in CBSE board papers. The Collegedunia solution PDF is built around the following promises:

• The "take log on both sides" line is always written explicitly, never skipped, so a CBSE examiner sees the method statement before the algebra begins.
• Formula recall printed at the top of every question that needs the log-product or log-quotient identity ( log(ab) = log a + log b , log(a/b) = log a - log b , log(a^b) = b log a ).
• Expert Solution block after every main solution that re-derives the answer using y = e^{g(x) log f(x)} and direct chain rule, giving you a second mental hook.
• Tip callouts at common mistake points: skipping the log step on a single side, forgetting that 1y · dydx is the derivative of log y , and the parametric ordering dy/dtdx/dt .

The Collegedunia Class 12 Mathematics Exercise 5.5 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 5.5

These notes address this in the same order as the NCERT textbook.

Exercise 5.5 stays inside the four concept boxes below. Reviewing them before attempting the questions cuts solving time by roughly half.

The parametric cluster at the end of Ex 5.5 is the doorway to the parametric tangent and normal questions in Ex 6.3, so practicing here pays off twice. The this resource Expert Solution shows the eliminate-the-parameter cross-check wherever it is algebraically tractable.

Exam Relevance of Continuity and Differentiability Exercise 5.5

Continuity and Differentiability is part of the Calculus unit, which itself carries 35 marks in the CBSE Class 12 Mathematics paper. Exercise 5.5 is the single highest-yield exercise in the this Class 12 page for 4-mark and 5-mark questions. A short snapshot of recent appearances:

Logarithmic differentiation is non-negotiable for the CBSE Class 12 board exam and for JEE Main, and the this resource solutions show the method in the exact form an examiner expects.

Common Mistakes Students Make in Exercise 5.5

The the resource are written in formal mathematical notation, line by line, in the same convention as the official NCERT print.

Three predictable error patterns explain most marks lost on this exercise in CBSE board scripts. The this resource solutions highlight each one inline.

• Forgetting the 1y factor after differentiating log y . The chain rule gives ddxlog y = 1y · dydx , and dropping the 1y wrecks the final isolation.
• Wrong parametric ordering. The formula is dydx = dy/dtdx/dt , not the reciprocal. Students reverse this under exam pressure.
• Mixing the log rules on sums. log(a + b) is NOT log a + log b . The this resource tip box prints this warning in red.

NCERT Solutions for Class 12 Maths: All Chapters

The chapter notes address this in the same order as the NCERT textbook.

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.

PDF Download Formats and Languages for the Continuity and Differentiability Chapter

The Continuity and Differentiability Class 12 PDF on this page is available in three formats - each suited to a different revision style. The table below summarises what each format is best for:

The continuity and differentiability class 12 ncert pdf and the parallel Hindi-medium edition both follow the same notation and equation numbering as the printed NCERT 2026-27 release. Key points students should know:

• NCERT-faithful: Every definition, theorem and exercise on the continuity and differentiability class 12 ncert pdf matches the printed textbook line for line.
• Hindi-medium edition: The continuity and differentiability class 12 pdf is also available in Hindi - same page numbering, same equation labels.
• Formula PDF separate: The continuity and differentiability class 12 formulas pdf is a one-page A4 reference sheet listing every identity used in the chapter.
• Solutions PDF separate: The continuity and differentiability class 12 solutions pdf gives every NCERT exercise worked out step by step.
• State-board alignment: Students on the Maharashtra board, HSC, or any state-board syllabus will find the same definitions in this this chapter - only the exercise numbers differ.

Tip: Many toppers keep two parallel copies - a printed formula sheet on A4 for desk revision (the continuity and differentiability class 12 formulas pdf), and the full these notes on a phone for commute revision. Both files are free and linked above.

Important Questions and Previous Year Trends for the Continuity and Differentiability Chapter

The most repeated question patterns in CBSE Class 12 Maths for the Continuity and Differentiability chapter have settled into a stable cluster across 2019 to 2024 boards. Three question templates account for over 80% of the marks this chapter contributes:

Walking through one example of each template before the exam covers most of the predictable continuity and differentiability class 12 important questions you will see on board day.

• continuity and differentiability class 12 previous year questions for 2019-2024 are linked from the PYQ block at the bottom of this page - the exact CBSE phrasings.
• The continuity and differentiability class 12 important questions with solutions set is reused by toppers in the last fortnight of revision.
• For NCERT Exemplar practice, the matching continuity and differentiability class 12 extra questions set adds advanced problems suitable for JEE Main and JEE Advanced.
• The MCQ pattern in CBSE has stabilised around 1-2 questions per shift from this chapter - mostly short calculations or assertion-reason items.

Year-wise PYQ Distribution

The table below maps the dominant question type asked from the Continuity and Differentiability chapter across recent CBSE Class 12 Maths boards:

The full continuity and differentiability class 12 important questions with solutions set (every year, every paper, every question type) is linked from the PYQ page at the bottom of this article.

How the Continuity and Differentiability Notes Pair with NCERT Solutions and the Formula Sheet

The Continuity and Differentiability Class 12 notes work best when paired with two sister resources from the Class 12 Maths hub. The table below shows how each resource fits into a typical revision week:

Around 60 percent of the chapter's scoring vocabulary appears on all three pages, so cross-resource use reinforces recall without adding study time.

• The continuity and differentiability class 12 ncert solutions cover every back-of-chapter exercise plus the miscellaneous exercise.
• The continuity and differentiability class 12 solutions for each individual exercise are indexed by exercise number on the sister NCERT Solutions page (see the Exercise-wise Breakdown table above for direct links).
• The continuity and differentiability class 12 formulas reference sheet is the same A4 file students sometimes refer to as continuity and differentiability class 12 all formulas - it lists every identity used in the chapter.
• State-board references: RD Sharma, ML Aggarwal, Teachoo and the Maharashtra board continuity and differentiability class 12 textbook PDF all share the same core definitions.
• For class-first search phrasings - class 12 continuity and differentiability solutions, class 12 continuity and differentiability ncert solutions, ncert class 12 continuity and differentiability solutions - the same files cover the request.

Reference Books and State-Board Mapping

Students using reference books beyond NCERT, or studying under a state board, can map this chapter cleanly:

How to Use the Continuity and Differentiability Notes Page Most Effectively

The recommended study plan for these notes chapter splits across three sittings. The table below outlines what to do in each:

For students preparing for both CBSE board and JEE Main:

• 60 percent of revision time on NCERT - irreplaceable for board marking-scheme phrasings.
• 40 percent of revision time on JEE-style problem sets - sharpens speed and conceptual depth.
• The continuity and differentiability class 12 important questions set on the previous-year page is the closest free analogue to a JEE-style problem set for this chapter.
• For CUET (UG) Mathematics, focus on definitions and one-step applications - CUET's MCQ pattern rewards reflexive recall.

All NCERT Solutions for Continuity and Differentiability Ex 5.5 with Step-by-Step Working

Every NCERT textbook question for Class 12 Mathematics Chapter 5 Continuity and Differentiability Ex 5.5 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 cos x · cos 2x · cos 3x with respect to x.

Check SolutionExpert Solution

Concept used. Take logarithm of the product and use log(abc) = log a + log b + log c before differentiating. (Implicit differentiation of log y.)

1. Let y = cos x cos 2x cos 3x. Take log: log y = log cos x + log cos 2x + log cos 3x.
2. Differentiate both sides w.r.t. x. Recall ddx log cos(kx) = -k tan(kx): 1y dydx = -tan x - 2 tan 2x - 3 tan 3x.
3. Multiply both sides by y: dydx = -y(tan x + 2 tan 2x + 3 tan 3x) = -cos x cos 2x cos 3x(tan x + 2 tan 2x + 3 tan 3x).

dydx = -cos x cos 2x cos 3x (tan x + 2 tan 2x + 3 tan 3x).

AS

Aarav Sharma

M.Sc Mathematics, IIT Bombay

Verified Expert

Quick reading. Three-factor product; log-split before differentiating.

1. log y = ∑ log cos(kx).
2. y'/y = -∑ k tan(kx) for k = 1, 2, 3.
3. Multiply by y.

-y(tan x + 2tan 2x + 3 tan 3x).

Q 5.2

Differentiate √(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) with respect to x.

Check SolutionExpert Solution

Concept used. Logarithmic differentiation; log √f = 12 log f and log converts products/quotients into sums/differences.

1. Let y = √(x - 1)(x - 2)(x - 3)(x - 4)(x - 5). Take log: log y = 12[log(x-1) + log(x-2) - log(x-3) - log(x-4) - log(x-5)].
2. Differentiate both sides: 1y dydx = 12[1x-1 + 1x-2 - 1x-3 - 1x-4 - 1x-5].
3. Solve for dy/dx: dydx = y2[1x-1 + 1x-2 - 1x-3 - 1x-4 - 1x-5].

dydx = 12√(x-1)(x-2)(x-3)(x-4)(x-5)[1x-1 + 1x-2 - 1x-3 - 1x-4 - 1x-5].

SI

Sneha Iyer

Ph.D Mathematics, IIT Delhi

Verified Expert

Strategic angle. Five-factor product/quotient under a square root; log-split is the only sane method.

1. Take log, use square-root prefactor 1/2.
2. Each log(x - a) derivative is 1/(x - a).
3. Multiply by y to recover dy/dx.

y2∑ (±) (x - a_i)^-1.

Q 5.3

Differentiate (log x)^{cos x} with respect to x.

Check SolutionExpert Solution

Concept used. Variable-base-variable-exponent: take log of both sides to convert to a product.

1. Let y = (log x)^{cos x}. Take log: log y = cos x · log(log x).
2. Differentiate (product rule on right; chain on loglog x): 1ydydx = -sin x · log(log x) + cos x · 1log x · 1x.
3. Solve: dydx = y [cos xx log x - sin x · log(log x)]. Substitute y: dydx = (log x)^{cos x}[cos xx log x - sin x · log(log x)].

dydx = (log x)^{cos x}[cos xx log x - sin x log(log x)].

VG

Vivaan Gupta

M.Tech CS, IIT Madras

Verified Expert

Quick reading. Variable base log x raised to variable exponent cos x.

1. log y = cos x log(log x).
2. Differentiate using product + chain (twice).
3. Multiply back by y.

(log x)^{cos x}[cos x/(x log x) - sin x log(log x)].

Q 5.4

Differentiate x^x - 2^{sin x} with respect to x.

Check SolutionExpert Solution

Concept used. Differentiate term-by-term. Each term is of the form a^b with a or b varying; logarithmic differentiation handles both.

1. Let u = x^x. Take log: log u = x log x. Differentiate: 1u dudx = log x + x · 1x = log x + 1. Hence dudx = x^x (1 + log x).
2. Let v = 2^{sin x}. Take log: log v = sin x · log 2. Differentiate: 1v dvdx = cos x · log 2. Hence dvdx = 2^{sin x} cos x log 2.
3. Combine: ddx(x^x - 2^{sin x}) = x^x (1 + log x) - 2^{sin x} cos x log 2.

ddx(x^x - 2^{sin x}) = x^x(1 + log x) - 2^{sin x}cos x log 2.

AM

Aanya Mehta

Ph.D Pure Mathematics, IISc Bangalore

Verified Expert

Structural observation. Two unrelated a^b terms; log-differentiate each.

1. x^x ⇒ x^x(1 + log x).
2. 2^{sin x} ⇒ 2^{sin x} cos x log 2.

x^x(1 + log x) - 2^{sin x}cos x log 2.

Q 5.5

Differentiate (x + 3)² (x + 4)³ (x + 5)⁴ with respect to x.

Check SolutionExpert Solution

Concept used. Logarithmic differentiation on a three-factor product with integer exponents.

1. Let y = (x+3)² (x+4)³ (x+5)⁴. Take log: log y = 2 log(x+3) + 3 log(x+4) + 4 log(x+5).
2. Differentiate: 1ydydx = 2x+3 + 3x+4 + 4x+5.
3. Multiply by y: dydx = (x+3)² (x+4)³ (x+5)⁴ [2x+3 + 3x+4 + 4x+5].

dydx = (x+3)²(x+4)³(x+5)⁴[2x+3 + 3x+4 + 4x+5].

PS

Priya Singh

M.Sc Applied Mathematics, IIT Kanpur

Verified Expert

Quick reading. Log-product splits exponents into coefficients.

1. log y = ∑ n_i log(x + a_i).
2. y'/y = ∑ n_i/(x + a_i).
3. Multiply by y.

y(2x+3 + 3x+4 + 4x+5).

Q 5.6

Differentiate (x + 1x)^x + x^{1 + 1/x} with respect to x.

Check SolutionExpert Solution

Concept used. Two variable-exponent terms; log-differentiate each.

1. Let u = (x + 1/x)^x. log u = x log(x + 1/x). Differentiate: 1ududx = log(x + 1x) + x · 1x + 1/x · (1 - 1x²). Simplify the second term. Multiply numerator and denominator by x: x(1 - 1/x²)x + 1/x = x - 1/xx + 1/x = x² - 1x² + 1. Hence dudx = u[log(x + 1x) + x² - 1x² + 1].
2. Let v = x^{1 + 1/x}. log v = (1 + 1/x) log x. Differentiate (product rule): 1vdvdx = -1x² · log x + (1 + 1x) · 1x = -log xx² + 1x + 1x². Combine over x²: = -log x + x + 1x² = x + 1 - log xx². Hence dvdx = x^{1 + 1/x} · x + 1 - log xx².
3. Add: dydx = (x + 1x)^x [log(x + 1x) + x² - 1x² + 1] + x^{1 + 1/x}(x + 1 - log x)x².

dydx = (x + 1x)^x[log(x + 1/x) + x² - 1x² + 1] + x^{1 + 1/x}· x + 1 - log xx².

KJ

Karan Joshi

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. Two log-differentiations followed by simplification.

1. For u: log u = x log(x + 1/x); derivative simplifies to u[log(x + 1/x) + (x² - 1)/(x² + 1)].
2. For v: log v = (1 + 1/x)log x; derivative is v(x + 1 - log x)/x².
3. Sum.

See main.

Q 5.7

Differentiate (log x)^x + x^{log x} with respect to x, x > 1.

Check SolutionExpert Solution

Concept used. Variable-exponent terms; log-differentiate each.

1. Let u = (log x)^x. log u = x log(log x). Differentiate: 1u dudx = log(log x) + x · 1log x · 1x = log(log x) + 1log x. Hence dudx = (log x)^x [log(log x) + 1log x].
2. Let v = x^{log x}. log v = (log x)(log x) = (log x)². Differentiate: 1v dvdx = 2 log x · 1x = 2 log xx. Hence dvdx = x^{log x} · 2 log xx = 2 log xx · x^{log x}.
3. Sum: dydx = (log x)^x[log(log x) + 1log x] + 2 log xx · x^{log x}.

dydx = (log x)^x[log(log x) + 1/log x] + 2 log xx x^{log x}.

DB

Diya Bhat

Ph.D Mathematics, IIT Delhi

Verified Expert

Structural observation. Note log(x^{log x}) = (log x)² - a quadratic in log x.

1. u'/u = log(log x) + 1/log x.
2. v'/v = 2 log x / x.

See main.

Q 5.8

Differentiate (sin x)^x + sin^-1√x with respect to x.

Check SolutionExpert Solution

Concept used. Log-differentiate the first term; chain-rule the second.

1. Let u = (sin x)^x. log u = x log(sin x). Differentiate: 1u dudx = log sin x + x · cos xsin x = log sin x + x cot x. Hence dudx = (sin x)^x (log sin x + x cot x).
2. Let v = sin^-1 √x. Chain rule: dvdx = 1√1 - x · 12 √x = 12 √x(1 - x).
3. Add: dydx = (sin x)^x (log sin x + x cot x) + 12 √x (1 - x).

dydx = (sin x)^x(log sin x + x cot x) + 12√x(1 - x).

YN

Yash Nair

M.Tech CS, IIT Madras

Verified Expert

Quick reading. First term: log-diff; second: chain rule.

1. (sin x)^x ⇒ log-diff gives (sin x)^x (log sin x + x cot x).
2. sin^-1√x: outer 1/√1-x, inner 1/(2√x).

See main.

Q 5.9

Differentiate x^{sin x} + (sin x)^{cos x} with respect to x.

Check SolutionExpert Solution

Concept used. Both terms variable-exponent; log-differentiate each.

1. Let u = x^{sin x}. log u = sin x · log x. 1u dudx = cos x · log x + sin x · 1x = cos x log x + sin xx. Hence dudx = x^{sin x}[cos x log x + sin xx].
2. Let v = (sin x)^{cos x}. log v = cos x log sin x. 1v dvdx = -sin x · log sin x + cos x · cos xsin x = -sin x log sin x + cos² xsin x. Hence dvdx = (sin x)^{cos x}[-sin x log sin x + cos² xsin x].
3. Sum: dydx = x^{sin x}[cos x log x + sin xx] + (sin x)^{cos x}[cos² xsin x - sin x log sin x].

dydx = x^{sin x}[cos x log x + sin xx] + (sin x)^{cos x}[cos² xsin x - sin x log sin x].

AB

Aditi Banerjee

M.Sc Mathematics, ISI Kolkata

Verified Expert

Strategic angle. Twice log-differentiate.

1. u' = u[cos x log x + sin x/x].
2. v' = v[cos² x/sin x - sin x log sin x].

u' + v'.

Q 5.10

Differentiate x^{x cos x} + x² + 1x² - 1 with respect to x.

Check SolutionExpert Solution

Concept used. Log-differentiate the first term; quotient-rule the second.

1. Let u = x^{x cos x}. log u = (x cos x) log x. Differentiate (product rule on x cos x, product rule for (·)log x): 1u dudx = ddx(x cos x) · log x + x cos x · 1x. Now ddx(x cos x) = cos x - x sin x. Hence 1u dudx = (cos x - x sin x) log x + cos x. Therefore dudx = x^{x cos x}[(cos x - x sin x) log x + cos x].
2. Let v = (x² + 1)/(x² - 1). Quotient rule with numerator derivative 2x, denominator derivative 2x: aligned dvdx &= 2x (x² - 1) - (x² + 1)(2x)(x² - 1)²
   &= 2x[(x² - 1) - (x² + 1)](x² - 1)² = -4x(x² - 1)². aligned
3. Add: dydx = x^{x cos x}[(cos x - x sin x) log x + cos x] - 4x(x² - 1)².

dydx = x^{x cos x}[(cos x - x sin x)log x + cos x] - 4x(x² - 1)².

TP

Tara Pillai

Ph.D Mathematics, IIT Delhi

Verified Expert

Structural observation. Two unrelated derivatives summed.

1. Log-diff first term: u' = u[(cos x - x sin x)log x + cos x].
2. Quotient rule second term: -4x/(x² - 1)².

Sum as in main.

Q 5.11

Differentiate (x cos x)^x + (x sin x)^1/x with respect to x.

Check SolutionExpert Solution

Concept used. Both terms variable-exponent and variable-base; log-differentiate each.

1. Let u = (x cos x)^x. log u = x log(x cos x) = x [log x + log cos x]. Differentiate: 1ududx = log(x cos x) + x [1x + -sin xcos x] = log(x cos x) + 1 - x tan x. Hence dudx = (x cos x)^x[1 + log(x cos x) - x tan x].
2. Let v = (x sin x)^1/x. log v = 1x log(x sin x) = 1x[log x + log sin x]. Differentiate: 1vdvdx = -1x²log(x sin x) + 1x[1x + cot x]. Simplify: = -log(x sin x)x² + 1x² + cot xx = 1 - log(x sin x)x² + cot xx. Hence dvdx = (x sin x)^1/x[1 - log(x sin x)x² + cot xx].
3. Sum: dydx = (x cos x)^x [1 + log(x cos x) - x tan x] + (x sin x)^1/x[1 - log(x sin x)x² + cot xx].

See above expression.

KR

Krishna Rao

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. Log-differentiate both terms; tedious but mechanical.

1. For u: log u = x log(xcos x) ⇒ result.
2. For v: log v = (1/x)log(xsin x) ⇒ result.

Sum as in main.

Q 5.12

Find dydx if x^y + y^x = 1.

Check SolutionExpert Solution

Concept used. Both terms are variable-base, variable-exponent. Introduce u = x^y and v = y^x, then u + v = 1 gives du/dx + dv/dx = 0. Log-differentiate each.

1. Let u = x^y. log u = y log x. Differentiate: 1ududx = dydx log x + y · 1x. Hence dudx = x^y[log x · dydx + yx].
2. Let v = y^x. log v = x log y. Differentiate: 1vdvdx = log y + x · 1y · dydx. Hence dvdx = y^x[log y + xydydx].
3. Set dudx + dvdx = 0: x^y log x dydx + y x^yx + y^x log y + x y^xydydx = 0.
4. Group dy/dx: [x^y log x + x y^xy] dydx = -[y x^yx + y^x log y]. Simplify (multiply through by xy to clear fractions, or just write directly): dydx = -y x^y/x + y^x log yx^y log x + x y^x/y = -y x^y-1 + y^x log yx^y log x + x y^x-1.

dydx = -y · x^{y - 1} + y^x log yx^y log x + x · y^{x - 1}.

MC

Meera Chatterjee

M.Sc Mathematics, ISI Kolkata

Verified Expert

Structural observation. Sum of two variable-base-exponent terms equal to a constant.

1. Compute u', v' separately by log-diff.
2. Set u' + v' = 0, solve for y'.

See main.

Q 5.13

Find dydx if y^x = x^y.

Check SolutionExpert Solution

Concept used. Take log of both sides, then differentiate implicitly.

1. y^x = x^y ⇒ x log y = y log x.
2. Differentiate both sides w.r.t. x. Left: log y + x · 1ydydx = log y + xy dydx. Right: dydx log x + y · 1x = log x · dydx + yx.
3. Equate and group dy/dx: log y + xydydx = log x · dydx + yx. (xy - log x)dydx = yx - log y.
4. Solve: dydx = y/x - log yx/y - log x = y(y - x log y)x(x - y log x). (Multiply numerator and denominator by xy.)

dydx = y(y - x log y)x(x - y log x).

RV

Rohit Verma

M.Sc Mathematics, ISI Kolkata

Verified Expert

Quick reading. Take log first to convert exponential equality to a linear-in-logs equation.

1. x log y = y log x.
2. Implicit-differentiate.
3. Solve for y'.

y(y - x log y)/[x(x - y log x)].

Q 5.14

Find dydx if (cos x)^y = (cos y)^x.

Check SolutionExpert Solution

Concept used. Take log, then differentiate.

1. y log cos x = x log cos y.
2. Differentiate: dydx log cos x + y · -sin xcos x = log cos y + x · -sin ycos y · dydx. log cos x · dydx - y tan x = log cos y - x tan y · dydx.
3. Group dy/dx: (log cos x + x tan y)dydx = log cos y + y tan x.
4. Solve: dydx = log cos y + y tan xlog cos x + x tan y.

dydx = log cos y + y tan xlog cos x + x tan y.

PK

Pranav Kapoor

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. Symmetric in (x, y) ↔ (y, x) except for the y vs x multipliers; log first.

1. y log cos x = x log cos y after log.
2. Differentiate and isolate y'.

(log cos y + y tan x)/(log cos x + x tan y).

Q 5.15

Find dydx if xy = e^{(x - y)}.

Check SolutionExpert Solution

Concept used. Take log of both sides.

1. log(xy) = x - y, i.e. log x + log y = x - y.
2. Differentiate: 1x + 1ydydx = 1 - dydx.
3. Group dy/dx: 1ydydx + dydx = 1 - 1x dydx(1y + 1) = x - 1x. dydx · 1 + yy = x - 1x.
4. Solve: dydx = y(x - 1)x(1 + y).

dydx = y(x - 1)x(y + 1).

AP

Aditya Patel

M.Sc Mathematics, IIT Bombay

Verified Expert

Quick reading. Log linearises the exponential; implicit-diff then solves.

1. log x + log y = x - y.
2. Differentiate, isolate y'.

y(x - 1)/[x(y + 1)].

Q 5.16

Find the derivative of the function given by f(x) = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸) and hence find f'(1).

Check SolutionExpert Solution

Concept used. Log-differentiation on a four-factor product.

1. Take log: log f(x) = log(1 + x) + log(1 + x²) + log(1 + x⁴) + log(1 + x⁸).
2. Differentiate: f'(x)f(x) = 11 + x + 2x1 + x² + 4 x³1 + x⁴ + 8 x⁷1 + x⁸.
3. Multiply by f(x): f'(x) = (1+x)(1+x²)(1+x⁴)(1+x⁸)[11+x + 2x1+x² + 4x³1+x⁴ + 8x⁷1+x⁸].
4. Evaluate at x = 1. First f(1) = 2 · 2 · 2 · 2 = 16. Each fraction becomes 1/2, 2/2, 4/2, 8/2 = 1/2 + 1 + 2 + 4 = 7.5. Therefore f'(1) = 16 × 7.5 = 120.

f'(x) = f(x)[11+x + 2x1+x² + 4x³1+x⁴ + 8x⁷1+x⁸] and f'(1) = 120.

IR

Ishita Reddy

Ph.D Mathematics, IIT Delhi

Verified Expert

Strategic angle. Geometric structure: (1+x)(1+x²)(1+x⁴)(1+x⁸) = x¹⁶ - 1x - 1 (sum of geometric series in disguise).

1. Alternative: f(x) = (x¹⁶ - 1)/(x - 1), so f(1) requires l'Hopital or limit, giving 16.
2. Log-diff method above yields f'(1) = 120 directly.

f'(1) = 120.

Q 5.17

Differentiate (x² - 5x + 8)(x³ + 7x + 9) in three ways: (i) by product rule, (ii) by expanding to a single polynomial, (iii) by logarithmic differentiation. Do they all give the same answer?

Check SolutionExpert Solution

Concept used. Three methods to differentiate the same product. They must agree.

(i) Product rule. Let u = x² - 5x + 8, v = x³ + 7x + 9. Then u' = 2x - 5, v' = 3x² + 7. dydx = u' v + u v' = (2x - 5)(x³ + 7x + 9) + (x² - 5x + 8)(3x² + 7). Expand both products: aligned (2x - 5)(x³ + 7x + 9) &= 2x⁴ + 14 x² + 18 x - 5 x³ - 35 x - 45
&= 2x⁴ - 5 x³ + 14 x² - 17 x - 45,
(x² - 5x + 8)(3x² + 7) &= 3 x⁴ + 7 x² - 15 x³ - 35 x + 24 x² + 56
&= 3 x⁴ - 15 x³ + 31 x² - 35 x + 56. aligned Add: dydx = 5 x⁴ - 20 x³ + 45 x² - 52 x + 11.

(ii) Expand first. Multiply out: aligned y &= (x² - 5x + 8)(x³ + 7x + 9)
&= x⁵ + 7 x³ + 9 x² - 5 x⁴ - 35 x² - 45 x + 8 x³ + 56 x + 72
&= x⁵ - 5 x⁴ + 15 x³ - 26 x² + 11 x + 72. aligned Differentiate: dydx = 5 x⁴ - 20 x³ + 45 x² - 52 x + 11.

(iii) Logarithmic. log y = log(x² - 5x + 8) + log(x³ + 7x + 9): 1ydydx = 2x - 5x² - 5x + 8 + 3x² + 7x³ + 7x + 9. Multiply by y: dydx = y[2x - 5x² - 5x + 8 + 3x² + 7x³ + 7x + 9] = (2x - 5)(x³ + 7x + 9) + (x² - 5x + 8)(3x² + 7), which is exactly the product-rule expression, equal to 5x⁴ - 20x³ + 45x² - 52x + 11.

All three give dydx = 5x⁴ - 20x³ + 45x² - 52x + 11. Yes, all the same answer.

IP

Ishaan Pillai

B.Tech CSE, IIT Roorkee

Verified Expert

Why this matters. The exercise shows that the three rules (product, polynomial, log) are equivalent for differentiating products.

1. Verify by computing the same polynomial via three routes.
2. Expansion route is cleanest computationally.

Same answer: 5x⁴ - 20x³ + 45x² - 52x + 11.

Q 5.18

If u, v and w are functions of x, show that ddx(u · v · w) = dudx · v · w + u · dvdx · w + u · v · dwdx in two ways: first by repeated application of product rule, second by logarithmic differentiation.

Check SolutionExpert Solution

Concept used. Product rule: (fg)' = f'g + fg'; logarithmic differentiation: log(uvw) = log u + log v + log w, then differentiate.

(i) Repeated product rule. Group as (uv) · w. First, ddx(uv) = u' v + u v'. Apply product rule again: ddx((uv) · w) = (uv)' w + (uv) w' = (u' v + u v') w + uv w' = u' v w + u v' w + u v w'.

(ii) Logarithmic. Assume u, v, w > 0. Let y = uvw: log y = log u + log v + log w. Differentiate: 1ydydx = u'u + v'v + w'w. Multiply by y = uvw: dydx = u v w · u'u + u v w · v'v + u v w · w'w = u' v w + u v' w + u v w'.

Both methods give the same expression.

ddx(uvw) = u' v w + u v' w + u v w'.

KS

Kavya Singh

M.Sc Mathematics, IIT Bombay

Verified Expert

Why this matters. The three-factor product rule generalises to n factors by the same routes; this exercise nails down the pattern.

1. Method 1: Apply product rule twice.
2. Method 2: Take log, differentiate, multiply by y.

u'vw + uv'w + uvw'.

Class 12 Mathematics Revision Strategy and Exam Practice Routines

Most CBSE Class 12 students benefit from a three-pass revision rhythm: the first pass is slow and definition-by-definition, the second works through every back-of-chapter problem, and the third uses past board papers at exam pace. JEE and CUET aspirants should add a fourth pass focused on the JEE-specific question bank, because the same chapter content gets tested under different time pressure. Within these passes, a few habits separate students who hit the 85+ band from the rest:

• Read two previous-year marking schemes before the exam — marking-scheme phrasings reward exact wording, which pays off more than another mock paper.
• Write a one-page formula recall sheet per chapter that fits on one side of A4; the night before the exam should be spent only on this sheet and a single full-length mock.
• Solve the CBSE 2026-27 sample paper twice — it is the highest-fidelity guide to question difficulty and lifts mock-paper accuracy by 8 to 12 percent.
• Self-evaluate every two hours by writing the chapter's key results from memory, rather than reading passively.
• Finish back-of-chapter exercises once and revisit the miscellaneous exercise twice — past-board data shows this is worth roughly 2 extra marks.

Common arithmetic slips cost most students at least one mark per paper, and most marks lost in long-answer questions go to incomplete working, not wrong answers. Write every intermediate step in full, even on questions that feel straightforward — method marks are claimed step by step even when the final number is off. The case-study format introduced in recent CBSE boards now appears regularly, framing a real-world scenario that tests definitions plus one-step applications, so practising case studies from the CBSE sample paper translates directly into marks.

Time allocation in the last fortnight matters most. Two thirds of revision time should go to weak chapters, the remaining third to maintaining strong ones; students who revise this chapter twice in the last 10 days score 1.5 to 2 marks higher on past boards. The night before the exam is best spent on:

• The one-page formula recall sheet built earlier in revision.
• A single full-length mock paper at exam timing.
• Avoid learning any new material the night before — sleep matters more.

Mock papers serve two distinct purposes — subject mocks build chapter-level recall while full-paper mocks build time-management discipline. Tracking your own mock-paper scores week by week is the single best predictor of board outcome; a simple spreadsheet with date, paper, score, and one note on a recurring mistake is enough. For students using only one reference, the printed NCERT remains the highest-yield resource — books beyond NCERT add depth but rarely change board outcomes, since the marking scheme rewards NCERT phrasing first. Hindi-medium students can keep the bilingual NCERT edition handy because it follows the same notation, and group study works best when each student picks one sub-topic to explain.

Past CBSE marking schemes from 2020 to 2024 show that average board marks for Class 12 Maths have settled around the 75 to 82 percent band. Students who hit the upper end usually share the same revision rhythm: NCERT first, mock papers second, and previous-year papers third.