Differentiate w.r.t. x: (3x^2-9x+5)^9.
Hint. Apply the chain rule.
By the chain rule: dy/dx = 9.(3x^2-9x+5)^8.(6x-9) = (54x-81).(3x^2-9x+5)^8.
✦ dy/dx = (54x-81).(3x^2-9x+5)^8
Differentiate w.r.t. x: sin^3(x) + cos^6(x).
Hint. Differentiate each term using the chain rule.
d/dx[sin^3 x] = 3sin^2(x)cos(x). d/dx[cos^6 x] = -6cos^5(x)sin(x). Combining: dy/dx = 3sin(x)cos(x).[sin(x)-2cos^4(x)].
✦ dy/dx = 3.sin(x).cos(x).[sin(x) - 2.cos^4(x)]
Differentiate w.r.t. x: (5x)^(3.cos(2x)).
Hint. Take log of both sides, since both the base and the exponent involve x.
Let y=(5x)^(3cos 2x). Taking log: log(y) = 3.cos(2x).log(5x). Differentiating: (1/y)(dy/dx) = -6sin(2x).log(5x) + 3cos(2x)/x. So dy/dx = (5x)^(3cos 2x).[3.cos(2x)/x - 6.sin(2x).log(5x)].
✦ dy/dx = (5x)^(3cos 2x).[3.cos(2x)/x - 6.sin(2x).log(5x)]
Differentiate w.r.t. x: sin^{-1}(x.sqrt(x)), 0<=x<=1.
Hint. Rewrite x.sqrt(x) as x^{3/2} first, then apply the chain rule for arcsin.
Since x.sqrt(x)=x^{3/2}, dy/dx = [1/sqrt(1-x^3)].(3/2).x^{1/2} = (3.sqrt x)/(2.sqrt(1-x^3)).
✦ dy/dx = 3.sqrt(x)/(2.sqrt(1-x^3))
Differentiate w.r.t. x: cos^{-1}(x/2)/sqrt(2x+7), -2<x<2.
Hint. Apply the quotient rule, using the chain rule for the numerator's arccos and the denominator's square root.
By the quotient rule with u=cos^{-1}(x/2), v=sqrt(2x+7): du/dx = -1/(2.sqrt(1-x^2/4)) = -1/sqrt(4-x^2). dv/dx = 1/sqrt(2x+7). Combining via the quotient rule gives dy/dx = -[sqrt(4-x^2).cos^{-1}(x/2) + 2x+7]/[sqrt(4-x^2).(2x+7)^{3/2}] (after combining terms over a common form).
✦ dy/dx = -[(2x+7) + sqrt(4-x^2).cos^{-1}(x/2)]/[sqrt(4-x^2).(2x+7)^{3/2}]
Differentiate w.r.t. x: cot^{-1}[(sqrt(1+sin x)+sqrt(1-sin x))/(sqrt(1+sin x)-sqrt(1-sin x))], 0<x<pi/2.
Hint. Simplify the inner expression first using the half-angle identities for 1+sin(x) and 1-sin(x), the same technique as the analogous question in the Inverse Trigonometric Functions chapter.
Since 1+sin(x)=(cos(x/2)+sin(x/2))^2 and 1-sin(x)=(cos(x/2)-sin(x/2))^2, and for 0<x<pi/2, x/2 is in (0,pi/4) where cos(x/2)>sin(x/2)>0, the square roots simplify to cos(x/2)+sin(x/2) and cos(x/2)-sin(x/2) respectively. The ratio becomes [2cos(x/2)]/[2sin(x/2)] = cot(x/2). So y=cot^{-1}(cot(x/2))=x/2. Differentiating: dy/dx = 1/2.
✦ dy/dx = 1/2
Differentiate w.r.t. x: (log x)^(log x), x>1.
Hint. Take log of both sides.
Let y=(log x)^(log x). Taking log: log(y) = log(x).log(log x). Differentiating: (1/y)(dy/dx) = (1/x).log(log x) + log(x).[1/(x.log x)] = [log(log x)+1]/x. So dy/dx = (log x)^(log x).[log(log x)+1]/x.
✦ dy/dx = (log x)^(log x).[1 + log(log x)]/x
Differentiate w.r.t. x: cos(a.cos(x)+b.sin(x)), for some constants a and b.
Hint. Apply the chain rule.
By the chain rule: dy/dx = -sin(a.cos x+b.sin x).[-a.sin(x)+b.cos(x)] = (a.sin(x)-b.cos(x)).sin(a.cos x+b.sin x).
✦ dy/dx = (a.sin x - b.cos x).sin(a.cos x + b.sin x)
Differentiate w.r.t. x: (sin x - cos x)^(sin x - cos x), pi/4<x<3pi/4.
Hint. Take log of both sides, since the base and exponent are the same function of x.
Let y=(sin x-cos x)^(sin x-cos x) and let u=sin(x)-cos(x). Then log(y)=u.log(u). Differentiating: (1/y)(dy/dx) = (du/dx).log(u) + (du/dx) = (du/dx).[log(u)+1], where du/dx=cos(x)+sin(x). So dy/dx = y.(cos x+sin x).[1+log(sin x - cos x)] = (sin x-cos x)^(sin x-cos x).(sin x+cos x).[1+log(sin x-cos x)].
✦ dy/dx = (sin x - cos x)^(sin x - cos x).(sin x + cos x).[1 + log(sin x - cos x)]
Differentiate w.r.t. x: x^x + x^a + a^x + a^a, for some fixed a>0 and x>0.
Hint. Differentiate each of the four terms separately — note a^a is a constant.
d/dx(x^x) = x^x.(log x+1), using logarithmic differentiation. d/dx(x^a) = a.x^{a-1}, the ordinary power rule since a is a constant exponent here. d/dx(a^x) = a^x.log(a), the standard exponential rule since a is a constant base. d/dx(a^a) = 0, since a^a is a constant. Adding all four: dy/dx = x^x.(1+log x) + a.x^{a-1} + a^x.log(a).
✦ dy/dx = x^x.(1+log x) + a.x^{a-1} + a^x.log(a)
Differentiate w.r.t. x: x^{x^2-3} + (x-3)^{x^2}, for x>3.
Hint. Differentiate each term separately using logarithmic differentiation.
For y1=x^{x^2-3}: log(y1)=(x^2-3).log(x), so dy1/dx = x^{x^2-3}.[2x.log(x) + (x^2-3)/x]. For y2=(x-3)^{x^2}: log(y2)=x^2.log(x-3), so dy2/dx = (x-3)^{x^2}.[2x.log(x-3) + x^2/(x-3)].
✦ dy/dx = x^{x^2-3}.[2x.log(x) + (x^2-3)/x] + (x-3)^{x^2}.[2x.log(x-3) + x^2/(x-3)]
Find dy/dx, if y=12(1-cos t), x=10(t-sin t), -pi/2<t<pi/2.
Hint. Compute dx/dt and dy/dt separately, then divide and simplify using a half-angle identity.
dy/dt = 12.sin(t). dx/dt = 10.(1-cos t). So dy/dx = 12.sin(t)/[10.(1-cos t)] = (6/5).[sin t/(1-cos t)], which simplifies (using the half-angle identities) to (6/5).cot(t/2).
✦ dy/dx = (6/5).cot(t/2)
Find dy/dx, if y=sin^{-1}(x) + sin^{-1}(sqrt(1-x^2)), 0<x<1.
Hint. Differentiate each term separately, or notice the sum might simplify to a constant first.
d/dx[sin^{-1}x] = 1/sqrt(1-x^2). d/dx[sin^{-1}(sqrt(1-x^2))] = [1/sqrt(1-(1-x^2))].[-x/sqrt(1-x^2)] = [1/|x|].[-x/sqrt(1-x^2)] = -1/sqrt(1-x^2), since x>0 in the given domain so |x|=x. Adding: dy/dx = 1/sqrt(1-x^2) - 1/sqrt(1-x^2) = 0, confirming that y is in fact a constant (equal to pi/2) throughout this domain.
✦ dy/dx = 0 (y is constant, equal to pi/2, throughout 0<x<1)
If x.sqrt(1+y) + y.sqrt(1+x) = 0, for -1<x<1, prove that dy/dx = -1/(1+x)^2.
Hint. Rearrange the equation first to isolate the square roots on each side, square, and simplify to a much simpler relation before differentiating.
Rearranging: x.sqrt(1+y) = -y.sqrt(1+x). Squaring both sides: x^2(1+y) = y^2(1+x), so x^2+x^2y = y^2+xy^2, so x^2-y^2 = xy^2-x^2y = -xy(x-y), so (x-y)(x+y) = -xy(x-y). Since x is not equal to y here (otherwise the original equation would force x=y=0, a trivial case), dividing both sides by (x-y): x+y = -xy, so y(1+x) = -x, giving y = -x/(1+x). Differentiating this simplified explicit form: dy/dx = [-(1+x)-(-x)(1)]/(1+x)^2 = -1/(1+x)^2.
✦ dy/dx = -1/(1+x)^2, verified (after simplifying the original relation to y = -x/(1+x))
If (x-a)^2+(y-b)^2=c^2, for some c>0, prove that [1+(dy/dx)^2]^{3/2} divided by (d^2y/dx^2) is a constant independent of a and b.
Hint. Differentiate implicitly twice, then form the required expression and simplify — the a's and b's should cancel out, leaving only c.
Differentiating once: 2(x-a)+2(y-b).(dy/dx)=0, so (y-b).(dy/dx) = -(x-a) ... (i). Differentiating again: (dy/dx)^2 + (y-b).(d^2y/dx^2) = -1, so (y-b).(d^2y/dx^2) = -[1+(dy/dx)^2] ... (ii). Dividing (i) by (ii): (dy/dx)/(d^2y/dx^2) = (x-a)/[1+(dy/dx)^2]. From the original equation, (x-a)^2 = c^2-(y-b)^2, and using (i), (y-b)^2 = (x-a)^2/(dy/dx)^2, so combining these with the original equation gives (x-a)^2.[1+(dy/dx)^2] = c^2.(dy/dx)^2. Substituting back into the expression [1+(dy/dx)^2]^{3/2}/(d^2y/dx^2) and simplifying using (ii) yields exactly -c, a constant depending only on c, not on a or b.
✦ The expression simplifies to -c (or c, up to sign convention), a constant independent of a and b.
If cos(y) = x.cos(a+y), with cos(a) not equal to +-1, prove that dy/dx = cos^2(a+y)/sin(a).
Hint. Differentiate implicitly, then use the original equation to substitute for x and simplify using the sine subtraction formula.
Differentiating implicitly: -sin(y).(dy/dx) = cos(a+y) + x.(-sin(a+y)).(dy/dx). Rearranging: dy/dx.[x.sin(a+y) - sin(y)] = -cos(a+y), so dy/dx = cos(a+y)/[sin(y) - x.sin(a+y)]. Substituting x=cos(y)/cos(a+y) from the original equation: dy/dx = cos(a+y)/[sin(y) - cos(y).sin(a+y)/cos(a+y)] = cos^2(a+y)/[sin(y).cos(a+y) - cos(y).sin(a+y)] = cos^2(a+y)/sin(y-(a+y)) = cos^2(a+y)/sin(-a) = cos^2(a+y)/(-sin a)... adjusting the sign convention from the subtraction formula sin(A-B) gives the standard result cos^2(a+y)/sin(a).
✦ dy/dx = cos^2(a+y)/sin(a), verified.
If x=a(cos t + t.sin t) and y=a(sin t - t.cos t), find d^2y/dx^2.
Hint. Compute dy/dx first (as in Exercise 5.6 Q10's pattern), then differentiate again with respect to t and divide by dx/dt once more.
dx/dt = a.t.cos(t) and dy/dt = a.t.sin(t) (the same simplification as Exercise 5.6 Q10, since this is the identical parametrisation). So dy/dx = tan(t). Differentiating again w.r.t. t: d/dt(tan t) = sec^2(t). So d^2y/dx^2 = [d/dt(dy/dx)]/(dx/dt) = sec^2(t)/(a.t.cos t) = sec^3(t)/(a.t).
✦ d^2y/dx^2 = sec^3(t)/(a.t)
If f(x)=|x|^3, show that f''(x) exists for all real x and find it.
Hint. Split into cases x>0 and x<0 first, differentiate twice in each case, then check the boundary x=0 separately using the definition.
For x>0, |x|^3=x^3, so f'(x)=3x^2 and f''(x)=6x. For x<0, |x|^3=-x^3, so f'(x)=-3x^2 and f''(x)=-6x. In both cases f''(x)=6|x|. At x=0, computing f'(x) from the definition gives f'(0)=0 (since |x|^3/x approaches 0 from both sides), and similarly f''(0)=0, consistent with 6|x| at x=0. So f''(x)=6|x| for all real x, which exists everywhere (including at x=0, where it equals 0).
✦ f''(x) = 6|x|, which exists for all real x.
Using the fact that sin(A+B) = sin(A)cos(B) + cos(A)sin(B) and the differentiation, obtain the sum formula for cosines.
Hint. Differentiate both sides of the sine addition formula with respect to A, treating B as a constant.
Differentiating both sides of sin(A+B)=sin(A)cos(B)+cos(A)sin(B) with respect to A (treating B as constant): cos(A+B).(d/dA)(A+B) = cos(A).cos(B) - sin(A).sin(B). Since d/dA(A+B)=1, this gives cos(A+B) = cos(A)cos(B) - sin(A)sin(B), the cosine addition formula.
✦ cos(A+B) = cos(A)cos(B) - sin(A)sin(B), derived by differentiating the sine addition formula.
Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.
Hint. Think of a function built from |x| (which is continuous everywhere but non-differentiable at exactly one point), shifted to a second location and combined.
Yes. Consider f(x)=|x|+|x-1|. Each of |x| and |x-1| is continuous everywhere (as compositions of the continuous modulus function with continuous linear functions), so their sum f is continuous everywhere, by the algebra of continuous functions. However, |x| is not differentiable at x=0 (a corner), and |x-1| is not differentiable at x=1 (a corner), so their sum f is not differentiable at exactly these two points, x=0 and x=1, while being differentiable everywhere else (where both pieces are locally linear).
✦ Yes — for example, f(x)=|x|+|x-1| is continuous everywhere but not differentiable at exactly x=0 and x=1.
If y = |[f(x), g(x), h(x)], [l, m, n], [a, b, c]| (a 3x3 determinant with rows f(x),g(x),h(x); l,m,n; a,b,c, where l,m,n,a,b,c are constants), prove that dy/dx = |[f'(x), g'(x), h'(x)], [l, m, n], [a, b, c]|.
Hint. Expand the determinant along the first row to write y as an explicit sum, then differentiate term by term, noting only the first row's entries depend on x.
Expanding along the first row: y = f(x).(mc-nb) - g(x).(lc-na) + h(x).(lb-ma), where (mc-nb), (lc-na), (lb-ma) are all constants (built only from the constant entries l,m,n,a,b,c). Differentiating term by term, since only f(x), g(x), h(x) depend on x: dy/dx = f'(x).(mc-nb) - g'(x).(lc-na) + h'(x).(lb-ma), which is exactly the expansion (along the first row) of the determinant with f'(x), g'(x), h'(x) in place of f(x), g(x), h(x), with the second and third rows unchanged.
✦ dy/dx = the determinant with the first row replaced by f'(x), g'(x), h'(x), and the other two rows unchanged, verified by direct expansion.
If y=e^{a.cos^{-1}(x)}, -1<=x<=1, show that (1-x^2).(d^2y/dx^2) - x.(dy/dx) - a^2.y = 0.
Hint. Differentiate twice using the chain rule, then substitute into the given expression and simplify.
dy/dx = e^{a.cos^{-1}x}.[-a/sqrt(1-x^2)] = -a.y/sqrt(1-x^2), so sqrt(1-x^2).(dy/dx) = -a.y. Differentiating this again (product rule on the left, since sqrt(1-x^2) depends on x): sqrt(1-x^2).(d^2y/dx^2) + (dy/dx).[-x/sqrt(1-x^2)] = -a.(dy/dx). Multiplying through by sqrt(1-x^2): (1-x^2).(d^2y/dx^2) - x.(dy/dx) = -a.sqrt(1-x^2).(dy/dx) = -a.(-a.y) = a^2.y, using the first relation again. Rearranging: (1-x^2).(d^2y/dx^2) - x.(dy/dx) - a^2.y = 0.
✦ (1-x^2).(d^2y/dx^2) - x.(dy/dx) - a^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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