Find dy/dx: 2x+3y=sin(x).
Hint. Differentiate both sides w.r.t. x, treating y as a function of x, then isolate dy/dx.
Differentiating both sides: 2+3(dy/dx) = cos(x), since d/dx(sin x)=cos(x). Solving: dy/dx = (cos x - 2)/3.
✦ dy/dx = (cos x - 2)/3
Find dy/dx: 2x+3y=sin(y).
Hint. Differentiate both sides w.r.t. x, treating y as a function of x, then collect dy/dx terms.
Unlike Q1, the sine term here involves y, not x, so its derivative also needs the chain rule: differentiating both sides gives 2+3(dy/dx) = cos(y).(dy/dx), since y is a function of x. Rearranging: dy/dx.[cos(y)-3] = 2, so dy/dx = 2/(cos y - 3).
✦ dy/dx = 2/(cos y - 3)
Find dy/dx: ax+by^2=cos(y).
Hint. Differentiate both sides w.r.t. x, remembering the chain rule for y^2 and cos(y).
Differentiating both sides: a+2by.(dy/dx) = -sin(y).(dy/dx), since d/dx(y^2)=2y.dy/dx and d/dx(cos y)=-sin(y).dy/dx. Rearranging: dy/dx.[2by+sin(y)] = -a, so dy/dx = -a/(2by+sin y).
✦ dy/dx = -a/(2by + sin y)
Find dy/dx: xy+y^2=tan(x)+y.
Hint. Differentiate both sides using the product rule on xy, then collect dy/dx terms.
Differentiating both sides: y + x.(dy/dx) + 2y.(dy/dx) = sec^2(x) + dy/dx, using the product rule on xy. Rearranging: dy/dx.[x+2y-1] = sec^2(x)-y, so dy/dx = (sec^2 x - y)/(x+2y-1).
✦ dy/dx = (sec^2 x - y)/(x + 2y - 1)
Find dy/dx: x^2+xy+y^2=100.
Hint. Differentiate both sides using the product rule on xy.
Differentiating both sides: 2x + (y+x.dy/dx) + 2y.(dy/dx) = 0, using the product rule on xy. Rearranging: dy/dx.[x+2y] = -(2x+y), so dy/dx = -(2x+y)/(x+2y).
✦ dy/dx = -(2x+y)/(x+2y)
Find dy/dx: x^3+x^2y+xy^2+y^3=81.
Hint. Differentiate term by term, using the product rule on the mixed terms.
Differentiating both sides: 3x^2 + (2xy+x^2.dy/dx) + (y^2+2xy.dy/dx) + 3y^2.(dy/dx) = 0. Collecting dy/dx terms: dy/dx.[x^2+2xy+3y^2] = -(3x^2+2xy+y^2), so dy/dx = -(3x^2+2xy+y^2)/(x^2+2xy+3y^2).
✦ dy/dx = -(3x^2+2xy+y^2)/(x^2+2xy+3y^2)
Find dy/dx: sin^2(y)+cos(xy)=K.
Hint. Differentiate both sides, using the product rule inside cos(xy).
Differentiating both sides: 2sin(y)cos(y).(dy/dx) + (-sin(xy)).(y+x.dy/dx) = 0. Collecting dy/dx terms: dy/dx.[2sin(y)cos(y)-x.sin(xy)] = y.sin(xy), so dy/dx = y.sin(xy)/[2sin(y)cos(y)-x.sin(xy)] = -y.sin(xy)/[x.sin(xy)-sin(2y)].
✦ dy/dx = -y.sin(xy)/[x.sin(xy) - sin(2y)]
Find dy/dx: sin^2(x)+cos^2(y)=1.
Hint. Differentiate both sides directly.
Differentiating both sides: 2sin(x)cos(x) + 2cos(y).(-sin y).(dy/dx) = 0, so 2sin(x)cos(x) = 2sin(y)cos(y).(dy/dx). Solving: dy/dx = sin(x)cos(x)/[sin(y)cos(y)] = sin(2x)/sin(2y).
✦ dy/dx = sin(x)cos(x)/[sin(y)cos(y)]
Find dy/dx: y=sin^{-1}(2x/(1+x^2)).
Hint. Substitute x=tan(theta) to collapse the composite expression into a simple multiple of theta.
Let x=tan(theta). Then 2x/(1+x^2) = 2tan(theta)/(1+tan^2 theta) = sin(2theta), a standard identity. So y=sin^{-1}(sin 2theta)=2theta=2.tan^{-1}(x) (valid for the range where this substitution keeps the result in the principal branch). Differentiating: dy/dx = 2/(1+x^2).
✦ dy/dx = 2/(1+x^2)
Find dy/dx: y=tan^{-1}((3x-x^3)/(1-3x^2)), -1/sqrt3 < x < 1/sqrt3.
Hint. Substitute x=tan(theta) and recognise the triple-angle tangent formula.
Let x=tan(theta). Then (3x-x^3)/(1-3x^2) = tan(3theta), the triple-angle tangent formula. So y=tan^{-1}(tan 3theta)=3theta=3.tan^{-1}(x), valid for the given range of x. Differentiating: dy/dx = 3/(1+x^2).
✦ dy/dx = 3/(1+x^2)
Find dy/dx: y=cos^{-1}((1-x^2)/(1+x^2)), 0<x<1.
Hint. Substitute x=tan(theta) and recognise the cosine double-angle formula.
Let x=tan(theta). Then (1-x^2)/(1+x^2) = cos(2theta), the cosine double-angle formula. So y=cos^{-1}(cos 2theta)=2theta=2.tan^{-1}(x), valid for 0<x<1. Differentiating: dy/dx = 2/(1+x^2).
✦ dy/dx = 2/(1+x^2)
Find dy/dx: y=sin^{-1}((1-x^2)/(1+x^2)), 0<x<1.
Hint. Substitute x=tan(theta); this composite expression is the same as Q11's but under arcsin instead of arccos.
This composite is the same as Q11's, but under arcsin instead of arccos, so reuse the same substitution: with x=tan(theta), (1-x^2)/(1+x^2) = cos(2theta) = sin(pi/2-2theta). So y=sin^{-1}(sin(pi/2-2theta))=pi/2-2theta=pi/2-2.tan^{-1}(x), valid for 0<x<1. Differentiating: dy/dx = -2/(1+x^2).
✦ dy/dx = -2/(1+x^2)
Find dy/dx: y=cos^{-1}(2x/(1+x^2)), -1<x<1.
Hint. Substitute x=tan(theta) and recognise the sine double-angle formula, then relate arccos of a sine to a shifted angle.
Substituting x=tan(theta) turns the numerator into 2x/(1+x^2)=sin(2theta)=cos(pi/2-2theta), reusing the same sine double-angle identity as the sin^{-1} version of this composite. So y=cos^{-1}(cos(pi/2-2theta))=pi/2-2theta=pi/2-2.tan^{-1}(x), valid for -1<x<1. Differentiating: dy/dx = -2/(1+x^2).
✦ dy/dx = -2/(1+x^2)
Find dy/dx: y=sin^{-1}(2x.sqrt(1-x^2)), -1/sqrt2 < x < 1/sqrt2.
Hint. Substitute x=sin(theta) and recognise the sine double-angle formula.
Let x=sin(theta). Then 2x.sqrt(1-x^2) = 2sin(theta)cos(theta) = sin(2theta). So y=sin^{-1}(sin 2theta)=2theta=2.sin^{-1}(x), valid for the given range. Differentiating: dy/dx = 2/sqrt(1-x^2).
✦ dy/dx = 2/sqrt(1-x^2)
Find dy/dx: y=sec^{-1}(1/(2x^2-1)), 0<x<1/sqrt2.
Hint. Substitute x=cos(theta) and recognise the cosine double-angle formula.
Let x=cos(theta). Then 2x^2-1=2cos^2(theta)-1=cos(2theta), so 1/(2x^2-1)=sec(2theta). So y=sec^{-1}(sec 2theta)=2theta=2.cos^{-1}(x), valid for 0<x<1/sqrt2. Differentiating: dy/dx = -2/sqrt(1-x^2).
✦ dy/dx = -2/sqrt(1-x^2)
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.
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