Differentiate w.r.t. x: e^x/sin(x).
Hint. Apply the quotient rule.
By the quotient rule: dy/dx = [e^x.sin(x) - e^x.cos(x)]/sin^2(x) = e^x.(sin x - cos x)/sin^2(x).
✦ e^x.(sin x - cos x)/sin^2(x)
Differentiate w.r.t. x: e^(sin^{-1} x).
Hint. Apply the chain rule with the derivative of arcsin(x).
By the chain rule: dy/dx = e^(sin^{-1} x).d/dx(sin^{-1} x) = e^(sin^{-1} x)/sqrt(1-x^2), since the derivative of sin^{-1}(x) is 1/sqrt(1-x^2).
✦ e^(sin^{-1} x)/sqrt(1-x^2)
Differentiate w.r.t. x: e^(x^3).
Hint. Apply the chain rule.
By the chain rule, since the derivative of e^u is e^u times the derivative of u: dy/dx = e^(x^3).d/dx(x^3) = 3x^2.e^(x^3), because the inner function x^3 has derivative 3x^2.
✦ 3x^2.e^(x^3)
Differentiate w.r.t. x: sin(tan^{-1}(e^{-x})).
Hint. Apply the chain rule twice, since this is sin of arctan of e^{-x}.
Since this is sin of arctan of e^{-x}, apply the chain rule outward to inward: dy/dx = cos(tan^{-1}(e^{-x})).d/dx[tan^{-1}(e^{-x})] = cos(tan^{-1} e^{-x}). [1/(1+e^{-2x})].(-e^{-x}), because the derivative of tan^{-1}(u) is 1/(1+u^2) times the derivative of u, here u=e^{-x} with derivative -e^{-x}. This gives dy/dx = -e^{-x}.cos(tan^{-1} e^{-x})/(1+e^{-2x}).
✦ -e^{-x}.cos(tan^{-1} e^{-x})/(1+e^{-2x})
Differentiate w.r.t. x: log(cos(e^x)).
Hint. Apply the chain rule three times, from log inward through cos to e^x.
By the chain rule: dy/dx = [1/cos(e^x)].(-sin(e^x)).e^x = -e^x.tan(e^x).
✦ -e^x.tan(e^x)
Differentiate w.r.t. x: e^x+e^(x^2)+...+e^(x^5).
Hint. Differentiate each term separately using the chain rule.
Differentiating each term: dy/dx = e^x + 2x.e^(x^2) + 3x^2.e^(x^3) + 4x^3.e^(x^4) + 5x^4.e^(x^5).
✦ e^x + 2x.e^(x^2) + 3x^2.e^(x^3) + 4x^3.e^(x^4) + 5x^4.e^(x^5)
Differentiate w.r.t. x: sqrt(e^{sqrt x}), x>0.
Hint. Apply the chain rule three times, from the outer square root through e^{...} to the inner square root.
By the chain rule: dy/dx = [1/(2.sqrt(e^{sqrt x}))].e^{sqrt x}.[1/(2.sqrt x)] = sqrt(e^{sqrt x})/(4.sqrt x), since e^{sqrt x}/sqrt(e^{sqrt x}) = sqrt(e^{sqrt x}).
✦ sqrt(e^{sqrt x})/(4.sqrt x)
Differentiate w.r.t. x: log(log(x)), x>1.
Hint. Apply the chain rule twice.
By the chain rule, treating log(x) as the inner function of the outer log: dy/dx = [1/log(x)].(1/x) = 1/(x.log x), since the derivative of log(u) is 1/u times the derivative of u, here u=log(x) with its own derivative 1/x.
✦ 1/(x.log x)
Differentiate w.r.t. x: cos(x)/log(x), x>0.
Hint. Apply the quotient rule.
By the quotient rule: dy/dx = [-sin(x).log(x) - cos(x).(1/x)]/log^2(x) = -[x.sin(x).log(x) + cos(x)]/[x.log^2(x)].
✦ -[x.sin(x).log(x) + cos(x)]/[x.log^2(x)]
Differentiate w.r.t. x: cos(log(x)+e^x), x>0.
Hint. Apply the chain rule, differentiating the sum inside cos term by term.
By the chain rule: dy/dx = -sin(log(x)+e^x).[1/x + e^x] = -(1/x+e^x).sin(log x + e^x).
✦ -(1/x + e^x).sin(log x + e^x)
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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