Exercise 5.5 — Continuity and Differentiability

1. 5.5.15 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Differentiate w.r.t. x: cos(x).cos(2x).cos(3x).

   Hint. Take log of both sides to turn the product into a sum, then differentiate term by term.

   Let y=cos(x)cos(2x)cos(3x). Taking log: log(y) = log(cos x)+log(cos 2x)+log(cos 3x). Differentiating: (1/y)(dy/dx) = -tan(x) - 2tan(2x) - 3tan(3x). So dy/dx = -y.[tan(x)+2tan(2x)+3tan(3x)] = -cos(x)cos(2x)cos(3x).[tan(x)+2tan(2x)+3tan(3x)].

   ✦ dy/dx = -cos(x)cos(2x)cos(3x).[tan(x) + 2tan(2x) + 3tan(3x)]

2. 5.5.26 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Differentiate w.r.t. x: sqrt[(x-1)(x-2)/((x-3)(x-4)(x-5))].

   Hint. Take log of both sides first, which turns the square root and product/quotient into a sum, making it much easier to differentiate.

   Let y=sqrt[(x-1)(x-2)/((x-3)(x-4)(x-5))]. Taking log: log(y) = (1/2)[log(x-1)+log(x-2)-log(x-3)-log(x-4)-log(x-5)]. Differentiating: (1/y)(dy/dx) = (1/2)[1/(x-1)+1/(x-2)-1/(x-3)-1/(x-4)-1/(x-5)]. So dy/dx = (y/2).[1/(x-1)+1/(x-2)-1/(x-3)-1/(x-4)-1/(x-5)].

   ✦ dy/dx = (y/2).[1/(x-1)+1/(x-2)-1/(x-3)-1/(x-4)-1/(x-5)], where y is the original function

3. 5.5.35 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Differentiate w.r.t. x: (log x)^cos(x).

   Hint. Take log of both sides, since the base and exponent are both functions of x.

   Let y=(log x)^cos(x). Taking log: log(y) = cos(x).log(log x). Differentiating: (1/y)(dy/dx) = -sin(x).log(log x) + cos(x).[1/(x.log x)]. So dy/dx = y.[cos(x)/(x.log x) - sin(x).log(log x)] = (log x)^cos(x).[cos(x)/(x.log x) - sin(x).log(log x)].

   ✦ dy/dx = (log x)^cos(x).[cos(x)/(x.log x) - sin(x).log(log x)]

4. 5.5.44 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Differentiate w.r.t. x: x^x - 2^sin(x).

   Hint. Differentiate each term separately, using logarithmic differentiation for x^x and the standard a^x rule for 2^sin(x).

   For x^x: taking log, log(y1)=x.log(x), so dy1/dx = x^x.(log x + 1). For 2^sin(x): d/dx[2^sin(x)] = 2^sin(x).log(2).cos(x), using the chain rule with d/dx(a^u)=a^u.log(a).du/dx. Combining: dy/dx = x^x.(log x+1) - 2^sin(x).log(2).cos(x).

   ✦ dy/dx = x^x.(1+log x) - 2^sin(x).log(2).cos(x)

5. 5.5.55 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Differentiate w.r.t. x: (x+3)^2.(x+4)^3.(x+5)^4.

   Hint. Take log of both sides to turn the product into a sum.

   Let y=(x+3)^2(x+4)^3(x+5)^4. Taking log: log(y)=2log(x+3)+3log(x+4)+4log(x+5). Differentiating: (1/y)(dy/dx) = 2/(x+3)+3/(x+4)+4/(x+5). So dy/dx = y.[2/(x+3)+3/(x+4)+4/(x+5)] = (x+3)(x+4)^2(x+5)^3.(9x^2+70x+133), after combining over a common denominator and factoring.

   ✦ dy/dx = (x+3)^2(x+4)^3(x+5)^4.[2/(x+3)+3/(x+4)+4/(x+5)]

6. 5.5.66 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Differentiate w.r.t. x: (x+1/x)^x + x^(1+1/x).

   Hint. Differentiate each term separately using logarithmic differentiation, since each has a variable base and variable exponent.

   For y1=(x+1/x)^x: log(y1)=x.log(x+1/x), so (1/y1)(dy1/dx) = log(x+1/x) + x.(1-1/x^2)/(x+1/x). For y2=x^(1+1/x): log(y2)=(1+1/x).log(x), so (1/y2)(dy2/dx) = -log(x)/x^2 + (1+1/x)/x. Combining both pieces (multiplying each by its own y1 or y2) gives the full derivative.

   ✦ dy/dx = (x+1/x)^x.[log(x+1/x) + x(1-1/x^2)/(x+1/x)] + x^(1+1/x).[(1+1/x)/x - log(x)/x^2]

7. 5.5.75 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Differentiate w.r.t. x: (log x)^x + x^(log x).

   Hint. Differentiate each term separately using logarithmic differentiation.

   For y1=(log x)^x: log(y1)=x.log(log x), so dy1/dx = (log x)^x.[log(log x) + 1/log(x)]. For y2=x^(log x): log(y2)=log(x).log(x)=(log x)^2, so (1/y2)(dy2/dx)=2.log(x)/x, giving dy2/dx = x^(log x).2.log(x)/x = 2.x^(log x - 1).log(x).

   ✦ dy/dx = (log x)^x.[log(log x) + 1/log x] + 2.x^(log x - 1).log(x)

8. 5.5.85 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Differentiate w.r.t. x: (sin x)^x + sin^{-1}(sqrt x).

   Hint. Differentiate each term separately: logarithmic differentiation for the first, the standard arcsin derivative for the second.

   For y1=(sin x)^x: log(y1)=x.log(sin x), so dy1/dx = (sin x)^x.[log(sin x) + x.cot(x)]. For y2=sin^{-1}(sqrt x): dy2/dx = [1/sqrt(1-x)].[1/(2.sqrt x)] = 1/(2.sqrt(x).sqrt(1-x)), using the chain rule.

   ✦ dy/dx = (sin x)^x.[log(sin x) + x.cot(x)] + 1/(2.sqrt(x(1-x)))

9. 5.5.96 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Differentiate w.r.t. x: x^(sin x) + (sin x)^(cos x).

   Hint. Differentiate each term separately using logarithmic differentiation, since both have variable bases and variable exponents.

   For y1=x^(sin x): log(y1)=sin(x).log(x), so dy1/dx = x^(sin x).[cos(x).log(x) + sin(x)/x]. For y2=(sin x)^(cos x): log(y2)=cos(x).log(sin x), so dy2/dx = (sin x)^(cos x).[-sin(x).log(sin x) + cos(x).cot(x)].

   ✦ dy/dx = x^(sin x).[sin(x)/x + cos(x).log x] + (sin x)^(cos x).[cos(x).cot(x) - sin(x).log(sin x)]

10. 5.5.106 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find dy/dx: x^(x.cos x) + (x^2+1)/(x^2-1).

    Hint. Differentiate the first term using logarithmic differentiation, and the second using the quotient rule.

    For y1=x^(x cos x): log(y1)=x.cos(x).log(x), so dy1/dx = x^(x cos x).[cos(x)+cos(x).log(x)-x.sin(x).log(x)], using the product rule inside the log-derivative. For y2=(x^2+1)/(x^2-1): by the quotient rule, dy2/dx = [2x(x^2-1)-(x^2+1)2x]/(x^2-1)^2 = -4x/(x^2-1)^2.

    ✦ dy/dx = x^(x cos x).[cos(x)(1+log x) - x.sin(x).log x] - 4x/(x^2-1)^2

11. 5.5.116 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find dy/dx: (x.cos x)^x + (x.sin x)^(1/x).

    Hint. Differentiate each term separately using logarithmic differentiation.

    For y1=(x cos x)^x: log(y1)=x.log(x cos x)=x[log x+log cos x], so dy1/dx = (x cos x)^x.[log(x cos x) + 1 - x.tan(x)]. For y2=(x sin x)^(1/x): log(y2)=(1/x).log(x sin x), so dy2/dx = (x sin x)^(1/x).[(1/x)(1/x+cot x) - log(x sin x)/x^2].

    ✦ dy/dx = (x cos x)^x.[1 - x.tan(x) + log(x cos x)] + (x sin x)^(1/x).[(1/x^2)(1+x.cot x - log(x sin x))]

12. 5.5.126 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find dy/dx: x^y + y^x = 1.

    Hint. Take log of each term separately (since it's a sum, logarithmic differentiation applies to each power individually), differentiate implicitly, then collect and solve for dy/dx.

    For u=x^y: log(u)=y.log(x), so du/dx = x^y.[y/x + log(x).dy/dx] = y.x^{y-1} + x^y.log(x).dy/dx. For v=y^x: log(v)=x.log(y), so dv/dx = y^x.[log(y) + x.(1/y).dy/dx] = y^x.log(y) + x.y^{x-1}.dy/dx. Since u+v=1, du/dx+dv/dx=0: [y.x^{y-1}+y^x.log y] + dy/dx.[x^y.log x + x.y^{x-1}] = 0. Solving for dy/dx.

    ✦ dy/dx = -[y.x^{y-1} + y^x.log y]/[x^y.log x + x.y^{x-1}]

13. 5.5.135 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find dy/dx: y^x = x^y.

    Hint. Take log of both sides first, turning both sides into simple products, then differentiate implicitly.

    Taking log of both sides: x.log(y) = y.log(x). Differentiating implicitly: log(y) + x.(1/y).dy/dx = (dy/dx).log(x) + y.(1/x). Collecting dy/dx terms: dy/dx.[x/y - log x] = y/x - log(y). Multiplying both sides by xy to clear denominators: dy/dx.[x^2 - xy.log x] = y^2 - xy.log y. So dy/dx = [y^2-xy.log y]/[x^2-xy.log x] = y.(y-x.log y)/[x.(x-y.log x)].

    ✦ dy/dx = y.(y - x.log y)/[x.(x - y.log x)]

14. 5.5.145 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find dy/dx: (cos x)^y = (cos y)^x.

    Hint. Take log of both sides, then differentiate implicitly and collect dy/dx terms.

    Taking log of both sides: y.log(cos x) = x.log(cos y). Differentiating implicitly: (dy/dx).log(cos x) + y.(-tan x) = log(cos y) + x.(-tan y).(dy/dx). Collecting dy/dx terms: dy/dx.[log(cos x)+x.tan(y)] = log(cos y)+y.tan(x). So dy/dx = [log(cos y)+y.tan(x)]/[log(cos x)+x.tan(y)].

    ✦ dy/dx = [log(cos y) + y.tan(x)]/[log(cos x) + x.tan(y)]

15. 5.5.155 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find dy/dx: xy = e^(x-y).

    Hint. Take log of both sides to turn the product and exponential into simpler expressions, then differentiate implicitly.

    Taking log of both sides: log(x)+log(y) = x-y. Differentiating implicitly: 1/x + (1/y).dy/dx = 1 - dy/dx. Collecting dy/dx terms: dy/dx.[1/y+1] = 1-1/x. So dy/dx = y.(1-1/x)/(1+y) = y(x-1)/[x(1+y)].

    ✦ dy/dx = y(x-1)/[x(1+y)]

16. 5.5.165 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find the derivative of the function given by f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8) and hence find f'(1).

    Hint. Take log of both sides to turn the product into a sum, differentiate, then substitute x=1 at the end.

    Let y=(1+x)(1+x^2)(1+x^4)(1+x^8). Taking log: log(y) = log(1+x)+log(1+x^2)+log(1+x^4)+log(1+x^8). Differentiating: (1/y)(dy/dx) = 1/(1+x) + 2x/(1+x^2) + 4x^3/(1+x^4) + 8x^7/(1+x^8). So f'(x) = y.[1/(1+x)+2x/(1+x^2)+4x^3/(1+x^4)+8x^7/(1+x^8)]. At x=1: y=(2)(2)(2)(2)=16, and the bracket = 1/2+2/2+4/2+8/2 = 0.5+1+2+4 = 7.5. So f'(1) = 16 x 7.5 = 120.

    ✦ f'(x) = (1+x)(1+x^2)(1+x^4)(1+x^8).[1/(1+x)+2x/(1+x^2)+4x^3/(1+x^4)+8x^7/(1+x^8)]; f'(1) = 120

17. 5.5.175 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Differentiate (x^2-5x+8)(x^3+7x+9) in three ways: (i) using product rule (ii) by expanding the product to obtain a single polynomial (iii) by logarithmic differentiation. Do they all give the same answer?

    Hint. Compute the derivative all three ways independently and compare the final expanded polynomials.

    (i) Product rule: d/dx[(x^2-5x+8)(x^3+7x+9)] = (2x-5)(x^3+7x+9) + (x^2-5x+8)(3x^2+7). Expanding this gives 5x^4-20x^3+45x^2-52x+11. (ii) Expanding the original product first gives x^5-5x^4+15x^3-26x^2+11x+72, whose derivative is 5x^4-20x^3+45x^2-52x+11. (iii) Logarithmic differentiation (valid where the product is positive) gives the same result after multiplying back through by y. All three methods agree.

    ✦ All three methods give the same derivative: 5x^4 - 20x^3 + 45x^2 - 52x + 11

18. 5.5.186 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    If u, v and w are functions of x, then show that d/dx(u.v.w) = (du/dx).v.w + u.(dv/dx).w + u.v.(dw/dx), in two ways: first by repeated application of the product rule, second by logarithmic differentiation.

    Hint. For the first method, group the triple product as u times (v.w) and apply the two-factor product rule twice. For the second, take log of the product to turn it into a sum before differentiating.

    Method 1 (repeated product rule): treat u.v.w as u.(vw). By the product rule, d/dx[u.(vw)] = (du/dx).(vw) + u.d/dx(vw). Applying the product rule again to vw: d/dx(vw) = (dv/dx).w + v.(dw/dx). Substituting back: d/dx(uvw) = (du/dx).vw + u.[(dv/dx).w + v.(dw/dx)] = (du/dx).vw + u.(dv/dx).w + u.v.(dw/dx). Method 2 (logarithmic differentiation): let y=uvw, so log(y)=log(u)+log(v)+log(w). Differentiating: (1/y)(dy/dx) = (1/u)(du/dx) + (1/v)(dv/dx) + (1/w)(dw/dx). Multiplying through by y=uvw gives dy/dx = (du/dx).vw + u.(dv/dx).w + uv.(dw/dx), the identical result.

    ✦ Both methods give the same identity: d/dx(uvw) = (du/dx).vw + u.(dv/dx).w + uv.(dw/dx)

Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh105.pdf) — Exercise 5.1 (34 questions), Exercise 5.2 (10 questions), Exercise 5.3 (15 questions), Exercise 5.4 (10 questions), Exercise 5.5 (18 questions), Exercise 5.6 (11 questions), Exercise 5.7 (11 questions), plus the chapter's Miscellaneous Exercise (22 questions), 131 questions total — the largest chapter in the project so far. The old stub taught Rolle's Theorem and the Mean Value Theorem as a full chapter section, but neither appears anywhere in the current book — confirmed by a full-text search finding zero mentions of 'mean value' or 'Rolle' anywhere in the chapter, consistent with the syllabus line itself, which never mentions MVT. The old stub also collapsed all seven real exercises into one invented 45-question group with no solutions file behind it. Every derivative in this file (all 131 questions) was independently computed and verified with sympy, including the second-order-derivative proofs, the parametric-form questions, and the implicit-differentiation identities in the Miscellaneous Exercise; several answers were re-expressed in the clean logarithmic-differentiation form the book itself uses (e.g. dy/dx = y.[...]) rather than sympy's fully-expanded single-fraction form, since that is the technique the exercise is testing.. Questions are referenced from the NCERT textbook for identification.