Differentiate the function w.r.t. x: sin(x^2+5).
Hint. Apply the chain rule with outer function sin and inner function x^2+5.
By the chain rule, d/dx[sin(x^2+5)] = cos(x^2+5).d/dx(x^2+5) = 2x.cos(x^2+5), since the derivative of the inner function x^2+5 is 2x.
✦ 2x.cos(x^2+5)
Differentiate the function w.r.t. x: cos(sin(x)).
Hint. Apply the chain rule twice, since this is a composition of cos with sin.
By the chain rule, d/dx[cos(sin x)] = -sin(sin x).d/dx(sin x) = -sin(sin x).cos(x), since the derivative of the inner function sin(x) is cos(x).
✦ -cos(x).sin(sin x)
Differentiate the function w.r.t. x: sin(ax+b).
Hint. Apply the chain rule with inner function ax+b.
By the chain rule, d/dx[sin(ax+b)] = cos(ax+b).d/dx(ax+b) = a.cos(ax+b), since the derivative of ax+b is a.
✦ a.cos(ax+b)
Differentiate the function w.r.t. x: sec(tan(sqrt(x))).
Hint. Apply the chain rule three times, working from the outermost function inward.
By the chain rule applied repeatedly: d/dx[sec(tan(sqrt x))] = sec(tan(sqrt x)).tan(tan(sqrt x)).d/dx[tan(sqrt x)] = sec(tan(sqrt x)).tan(tan(sqrt x)).sec^2(sqrt x).(1/(2.sqrt x)), since the derivative of sqrt(x) is 1/(2 sqrt x).
✦ sec(tan(sqrt x)).tan(tan(sqrt x)).sec^2(sqrt x)/(2.sqrt x)
Differentiate the function w.r.t. x: sin(ax+b)/cos(cx+d).
Hint. Apply the quotient rule, using the chain rule for each of the numerator and denominator.
Since this is a quotient of two chain-rule expressions, apply the quotient rule with u=sin(ax+b), v=cos(cx+d): dy/dx = [a.cos(ax+b).cos(cx+d) - sin(ax+b).(-c.sin(cx+d))]/cos^2(cx+d) = [a.cos(ax+b).cos(cx+d) + c.sin(ax+b).sin(cx+d)]/cos^2(cx+d), because the two minus signs from the quotient rule and from differentiating cos(cx+d) cancel.
✦ [a.cos(ax+b).cos(cx+d) + c.sin(ax+b).sin(cx+d)]/cos^2(cx+d)
Differentiate the function w.r.t. x: cos(x^3).sin^2(x^5).
Hint. Apply the product rule, using the chain rule for each factor.
Since this is a product of two chain-rule expressions, apply the product rule with u=cos(x^3), v=sin^2(x^5): dy/dx = -3x^2.sin(x^3).sin^2(x^5) + cos(x^3).2sin(x^5).cos(x^5).5x^4, which simplifies to -3x^2.sin(x^3).sin^2(x^5) + 10x^4.cos(x^3).sin(x^5).cos(x^5), because the factors 2 and 5 from the two chain-rule derivatives multiply together.
✦ -3x^2.sin(x^3).sin^2(x^5) + 10x^4.cos(x^3).sin(x^5).cos(x^5)
Differentiate the function w.r.t. x: 2.sqrt(cot(x^2)).
Hint. Apply the chain rule three times, from the square root inward to x^2.
By the chain rule: d/dx[2.sqrt(cot(x^2))] = 2. (1/(2.sqrt(cot x^2))). d/dx[cot(x^2)] = (1/sqrt(cot x^2)).(-cosec^2(x^2).2x) = -2x.cosec^2(x^2)/sqrt(cot x^2).
✦ -2x.cosec^2(x^2)/sqrt(cot(x^2))
Differentiate the function w.r.t. x: cos(sqrt(x)).
Hint. Apply the chain rule with inner function sqrt(x).
By the chain rule: d/dx[cos(sqrt x)] = -sin(sqrt x).(1/(2.sqrt x)), since the derivative of sqrt(x) is 1/(2 sqrt x).
✦ -sin(sqrt x)/(2.sqrt x)
Prove that the function f given by f(x)=|x-1|, x in R, is not differentiable at x=1.
Hint. Compute the left-hand and right-hand derivatives at x=1 separately using the definition, and show they differ.
For x near 1: if x>1, f(x)=x-1, so the right-hand derivative is lim(h->0+) [f(1+h)-f(1)]/h = lim(h->0+) h/h = 1. If x<1, f(x)=1-x, so the left-hand derivative is lim(h->0-) [f(1+h)-f(1)]/h = lim(h->0-) (-h)/h = -1. Since the left-hand derivative (-1) is not equal to the right-hand derivative (1), f is not differentiable at x=1.
✦ f is not differentiable at x=1, since its left-hand and right-hand derivatives there are -1 and 1 respectively.
Prove that the greatest integer function defined by f(x)=[x], 0<x<3, is not differentiable at x=1 and x=2.
Hint. At each integer, the function jumps, so it's not even continuous there — and a function that isn't continuous at a point can't be differentiable there either.
At x=1: as x approaches 1 from the left, [x]=0; as x approaches 1 from the right (up to 2), [x]=1. So f jumps from 0 to 1 at x=1, meaning f is not even continuous at x=1. The same jump (from 1 to 2) happens at x=2. Since differentiability requires continuity first, and f fails to be continuous at both x=1 and x=2, it cannot be differentiable there.
✦ f is not differentiable at x=1 and x=2, since it is not even continuous at those points.
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.
Pair self-study with a tutor, a live course or a coaching centre.
