If x and y are connected parametrically by x=2at^2, y=at^4, without eliminating the parameter, find dy/dx.
Hint. Compute dx/dt and dy/dt separately, then divide.
dx/dt = 4at. dy/dt = 4at^3. So dy/dx = (dy/dt)/(dx/dt) = 4at^3/(4at) = t^2.
✦ dy/dx = t^2
If x=a.cos(theta), y=b.cos(theta), find dy/dx.
Hint. Compute dx/d(theta) and dy/d(theta) separately, then divide.
dx/d(theta) = -a.sin(theta). dy/d(theta) = -b.sin(theta). So dy/dx = (-b.sin theta)/(-a.sin theta) = b/a, a constant (which makes sense, since y=(b/a)x exactly after eliminating theta).
✦ dy/dx = b/a
If x=sin(t), y=cos(2t), find dy/dx.
Hint. Compute dx/dt and dy/dt separately, then divide.
dx/dt = cos(t). dy/dt = -2.sin(2t) = -4.sin(t)cos(t), using the double angle formula. So dy/dx = -4.sin(t)cos(t)/cos(t) = -4.sin(t).
✦ dy/dx = -4.sin(t)
If x=4t, y=4/t, find dy/dx.
Hint. Compute dx/dt and dy/dt separately, then divide.
Since x=4t is linear, dx/dt = 4. Since y=4/t=4t^{-1}, dy/dt = -4/t^2, by the power rule. So dy/dx = (dy/dt)/(dx/dt) = (-4/t^2)/4 = -1/t^2.
✦ dy/dx = -1/t^2
If x=cos(theta)-cos(2theta), y=sin(theta)-sin(2theta), find dy/dx.
Hint. Compute dx/d(theta) and dy/d(theta) separately, then divide.
dx/d(theta) = -sin(theta)+2.sin(2theta). dy/d(theta) = cos(theta)-2.cos(2theta). So dy/dx = [cos(theta)-2cos(2theta)]/[2sin(2theta)-sin(theta)].
✦ dy/dx = [cos(theta) - 2cos(2theta)]/[2sin(2theta) - sin(theta)]
If x=a(theta-sin(theta)), y=a(1+cos(theta)), find dy/dx.
Hint. Compute dx/d(theta) and dy/d(theta) separately, then divide and simplify using a half-angle identity.
dx/d(theta) = a(1-cos theta). dy/d(theta) = -a.sin(theta). So dy/dx = -sin(theta)/(1-cos theta), which simplifies (using the half-angle identities sin theta=2sin(theta/2)cos(theta/2) and 1-cos theta=2sin^2(theta/2)) to -cot(theta/2).
✦ dy/dx = -cot(theta/2)
If x=sin^3(t)/sqrt(cos 2t), y=cos^3(t)/sqrt(cos 2t), find dy/dx.
Hint. Differentiate both using the product/chain rule, factor out common terms, and recognise the triple-angle sine and cosine formulas hiding in the result.
Differentiating x=sin^3(t).(cos 2t)^{-1/2} by the product rule: dx/dt = 3sin^2(t)cos(t).(cos 2t)^{-1/2} + sin^3(t).sin(2t).(cos 2t)^{-3/2}. Since cos(2t)=1-2sin^2(t) and sin(2t)=2sin(t)cos(t), factoring out cos(t).sin^2(t).(cos 2t)^{-3/2} simplifies this to dx/dt = cos(t).sin(t).sin(3t).(cos 2t)^{-3/2}, because sin(t)(3-4sin^2 t)=sin(3t) is exactly the triple-angle identity. By the symmetric computation, dy/dt = -sin(t).cos(t).cos(3t).(cos 2t)^{-3/2}, since cos(t)(3-4cos^2 t)=-cos(3t) is the matching cosine triple-angle identity. Dividing: dy/dx = -cos(3t)/sin(3t) = -cot(3t).
✦ dy/dx = -cot(3t)
If x=a[cos(t)+log(tan(t/2))], y=a.sin(t), find dy/dx.
Hint. Compute dx/dt and dy/dt separately; the log term's derivative combines with -sin(t) to leave a clean expression.
dx/dt = a.[-sin(t) + (1/tan(t/2)).sec^2(t/2).(1/2)] = a.[-sin(t) + 1/(2.sin(t/2)cos(t/2))] = a.[-sin(t)+1/sin(t)] (using the double-angle identity sin(t)=2sin(t/2)cos(t/2)) = a.[1-sin^2(t)]/sin(t) = a.cos^2(t)/sin(t). dy/dt = a.cos(t). So dy/dx = [a.cos(t)]/[a.cos^2(t)/sin(t)] = sin(t)/cos(t) = tan(t).
✦ dy/dx = tan(t)
If x=a.sec(theta), y=b.tan(theta), find dy/dx.
Hint. Compute dx/d(theta) and dy/d(theta) separately, then divide.
dx/d(theta) = a.sec(theta)tan(theta). dy/d(theta) = b.sec^2(theta). So dy/dx = b.sec^2(theta)/[a.sec(theta)tan(theta)] = b.sec(theta)/[a.tan(theta)] = b/[a.sin(theta)].
✦ dy/dx = b/(a.sin theta)
If x=a(cos(theta)+theta.sin(theta)), y=a(sin(theta)-theta.cos(theta)), find dy/dx.
Hint. Compute dx/d(theta) and dy/d(theta) separately, using the product rule for the theta.sin(theta) and theta.cos(theta) terms.
dx/d(theta) = a.[-sin(theta)+sin(theta)+theta.cos(theta)] = a.theta.cos(theta), since the -sin(theta) and +sin(theta) terms cancel. dy/d(theta) = a.[cos(theta)-cos(theta)+theta.sin(theta)] = a.theta.sin(theta), similarly. So dy/dx = [a.theta.sin(theta)]/[a.theta.cos(theta)] = tan(theta).
✦ dy/dx = tan(theta)
If x=sqrt(a^{sin^{-1} t}), y=sqrt(a^{cos^{-1} t}), show that dy/dx = -y/x.
Hint. Compute dx/dt and dy/dt using the chain rule, noting sin^{-1}(t) and cos^{-1}(t) have derivatives that are negatives of each other in form.
Writing x=a^{(1/2)sin^{-1}t}, dx/dt = x.log(a).(1/2).[1/sqrt(1-t^2)]. Similarly y=a^{(1/2)cos^{-1}t}, dy/dt = y.log(a).(1/2).[-1/sqrt(1-t^2)]. Dividing: dy/dx = (dy/dt)/(dx/dt) = [y.(-1)]/[x.(1)] = -y/x, since the log(a).(1/2)/sqrt(1-t^2) factor is common to both and cancels, leaving only the sign difference.
✦ dy/dx = -y/x, verified.
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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