Find the second order derivative of x^2+3x+2.
Hint. Differentiate twice.
Since this is a polynomial, differentiate term by term twice: first derivative is 2x+3, and differentiating that again (since the derivative of 2x+3 is just the coefficient of x) gives the second derivative, 2.
✦ d^2y/dx^2 = 2
Find the second order derivative of x^20.
Hint. Differentiate twice using the power rule.
First derivative: 20x^19. Second derivative: 20.19.x^18 = 380x^18.
✦ d^2y/dx^2 = 380x^18
Find the second order derivative of x.cos(x).
Hint. Differentiate twice using the product rule.
First derivative (product rule): cos(x) - x.sin(x). Second derivative (product rule again on x.sin(x)): -sin(x) - [sin(x)+x.cos(x)] = -2sin(x) - x.cos(x).
✦ d^2y/dx^2 = -2sin(x) - x.cos(x)
Find the second order derivative of log(x).
Hint. Differentiate twice using the power rule on 1/x.
First derivative: 1/x = x^{-1}. Second derivative: -x^{-2} = -1/x^2.
✦ d^2y/dx^2 = -1/x^2
Find the second order derivative of x^3.log(x).
Hint. Differentiate twice using the product rule.
First derivative (product rule): 3x^2.log(x) + x^2. Second derivative (product rule again): 6x.log(x) + 3x + 2x = 6x.log(x) + 5x = x.(6.log x + 5).
✦ d^2y/dx^2 = x.(6.log x + 5)
Find the second order derivative of e^x.sin(5x).
Hint. Differentiate twice using the product rule.
Since this is a product of e^x and sin(5x), apply the product rule: first derivative = e^x.sin(5x) + 5.e^x.cos(5x) = e^x.[sin(5x)+5cos(5x)]. Applying the product rule again to this new product gives the second derivative: e^x.[sin(5x)+5cos(5x)] + e^x.[5cos(5x)-25sin(5x)] = e^x.[10cos(5x)-24sin(5x)], because the two cos(5x) terms combine and the sin(5x) terms combine separately.
✦ d^2y/dx^2 = e^x.[10.cos(5x) - 24.sin(5x)]
Find the second order derivative of e^{6x}.cos(3x).
Hint. Differentiate twice using the product rule.
Since this is a product of e^{6x} and cos(3x), apply the product rule: first derivative = 6.e^{6x}.cos(3x) - 3.e^{6x}.sin(3x) = e^{6x}.[6cos(3x)-3sin(3x)]. Applying the product rule again to this new product gives the second derivative: 6.e^{6x}.[6cos(3x)-3sin(3x)] + e^{6x}.[-18sin(3x)-9cos(3x)] = e^{6x}.[27cos(3x)-36sin(3x)], because the cos(3x) terms combine and the sin(3x) terms combine separately.
✦ d^2y/dx^2 = e^{6x}.[27.cos(3x) - 36.sin(3x)]
Find the second order derivative of tan^{-1}(x).
Hint. Differentiate twice, using the quotient rule on the first derivative.
First derivative: 1/(1+x^2). Second derivative (quotient rule, or power rule on (1+x^2)^{-1}): -2x/(1+x^2)^2.
✦ d^2y/dx^2 = -2x/(1+x^2)^2
Find the second order derivative of log(log(x)).
Hint. Differentiate twice using the quotient rule on the first derivative.
First derivative: 1/(x.log x). Second derivative (quotient rule on 1/(x log x)): -(log x + 1)/(x^2.log^2 x).
✦ d^2y/dx^2 = -(1+log x)/(x^2.log^2 x)
Find the second order derivative of sin(log(x)).
Hint. Differentiate twice using the product/quotient rule.
First derivative: cos(log x)/x. Second derivative (quotient rule): [-sin(log x)/x . x - cos(log x)]/x^2 = -[sin(log x)+cos(log x)]/x^2.
✦ d^2y/dx^2 = -[sin(log x) + cos(log x)]/x^2
If y=5.cos(x)-3.sin(x), prove that d^2y/dx^2 + y = 0.
Hint. Differentiate twice and add back the original y.
First derivative: -5.sin(x)-3.cos(x). Second derivative: -5.cos(x)+3.sin(x) = -(5cos x - 3sin x) = -y. So d^2y/dx^2 + y = -y+y = 0.
✦ d^2y/dx^2 + y = 0, 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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