Class 12 RD Sharma Solutions Chapter 12 Higher Order Derivatives Exercise 12.1 | Set 2 (original) (raw)

Last Updated : 18 Sep, 2024

Chapter 12 of RD Sharma's Class 12 Mathematics textbook delves into Higher Order Derivatives, an advanced topic in calculus. Exercise 12.1 Set 2 typically focuses on more complex applications of higher order derivatives, including finding nth derivatives of various functions and solving problems related to successive differentiation.

Question 27. If y = [log{x+(√x2+1)}]2, show that (1 + x2)(d2y/dx2) + x(dy/dx) = 2.

**Solution:

We have,

y = [log{x + (√x2 + 1)}]2

_On differentiating both sides w.r.t x,

\frac{dy}{dx}=2log(x+\sqrt{x^2+1}).\frac{d}{dx}(x+\sqrt{x^2+1})

\frac{dy}{dx}=\frac{2log(x+\sqrt{x^2+1})}{(x+\sqrt{x^2+1})}(1+\frac{2x}{2\sqrt{x^2+1}})

\frac{dy}{dx}=\frac{2log(x+\sqrt{x^2+1})}{(x+\sqrt{x^2+1})}(\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}})

dy/dx = 2[log{x + (√x2 + 1)}]/(√x2 + 1)

_Again differentiating both sides w.r.t x,

\frac{d^2y}{dx^2}=\frac{2-\frac{2xlog(x+\sqrt{x^2+1}}{x^2+1}}{x^2+1}

\frac{d^2y}{dx^2}=\frac{2-x\frac{dy}{dx}}{x^2+1}

(x2 + 1)(d2y/dx2) = 2 - x(dy/dx)

(x2 + 1)(d2y/dx2) + x(dy/dx) = 2

_Hence Proved

Question 28. If y = (tan-1x)2, then prove that (1 + x2)2y2 + 2x(1 + x2)y1 = 2

**Solution:

We have,

y = (tan-1x)2

_On differentiating both sides w.r.t x,

y1 = 2(tan-1x)[1/(1 + x2)]

(1 + x2)y1 = 2(tan-1x)

_Again differentiating both sides w.r.t x,

(1 + x2)y2 + 2xy1 = 2/(1 + x2)

(1 + x2)2y2 + 2x(1 + x2)y1 = 2

_Hence Proved

Question 29. If y = cotx, prove that (d2y/dx2) + 2y(dy/dx) = 0.

**Solution:

We have,

y = cotx

_On differentiating both sides w.r.t x,

(dy/dx) = -cosec2x

_Again differentiating both sides w.r.t x,

d2y/dx2 = -(2cosec x) × (-cosec x.cot x)

d2y/dx2 = 2cosec2x.cot x

d2y/dx2 = 2(cot x)(cosec2x)

d2y/dx2 = -2y(dy/dx)

d2y/dx2 + 2y(dy/dx) = 0

Hence Proved

Question 30. Find d2y/dx2 where y = log(x2/e2).

**Solution:

We have,

y = log(x2/e2)

_On differentiating both sides w.r.t x,

\frac{dy}{dx}=\frac{1}{\frac{x^2}{e^2}}×(\frac{2x}{e^2})

(dy/dx) = (2/x)

_Again differentiating both sides w.r.t x,

d2y/dx2 = -(2/x2)

Hence Proved

Question 31. If y = ae2x + be-x, show that (d2y/dx2) - (dy/dx) - 2y = 0.

**Solution:

We have,

y = ae2x + be-x

_On differentiating both sides w.r.t t,

(dy/dx) = 2ae2x - be-x

_Again differentiating both sides w.r.t x,

(d2y/dx2) = 4ae2x + be-x

(d2y/dx2) = 2ae2x - be-x + 2(ae2x + be-x)

(d2y/dx2) = (dy/dx) + 2y

(d2y/dx2) - (dy/dx) - 2y = 0

Hence Proved

Question 32. If y = ex(sinx + cosx), prove that d2y/dx2 - 2(dy/dx) + 2y = 0.

**Solution:

We have,

y = ex(sinx + cosx)

_On differentiating both sides w.r.t x,

(dy/dx) = ex(sinx + cosx) + ex(cosx - sinx)

(dy/dx) = 2excosx

_Again differentiating both sides w.r.t x,

(d2y/dx2) = 2excosx - 2exsinx

Lets take L.H.S,

= d2y/dx2 - 2(dy/dx) + 2y

= 2excosx - 2exsinx - 2(2excosx) + 2ex(sinx + cosx)

= 4excosx - 4excosx - 2exsinx + 2exsinx

= 0

L.H.S = R.H.S

Hence Proved

Question 33. If y = cos-1x, find d2y/dx2 in terms of y alone.

**Solution:

We have,

y = cos-1x

_On differentiating both sides w.r.t x,

(dy/dx) = -1/√(1-x2)

_Again differentiating both sides w.r.t x,

\frac{d^2y}{dx^2}=\frac{-2x}{(2\sqrt{1-x^2})\frac{3}{2}} ...(i)

y = cos-1x

x = cosy

On putting the value of x in equation (i), we get

\frac{d^2y}{dx^2}=\frac{-cosy}{(\sqrt{1-cos^2y})\frac{3}{2}}

\frac{d^2y}{dx^2}=\frac{-cosy}{(sin^2y)\frac{3}{2}}

d2y/dx2 = -cosy/sin3y

d2y/dx2 = -cot y cosec2y

Question 34. If y = e^{acos^{-1}x}, prove that (1 - x2)(d2y/dx2) - x(dy/dx) - a2y = 0.

**Solution:

We have,

y = e^{acos^{-1}x}

Taking log both sides

logy = acos-1x.loge

logy = acos-1x

_On differentiating both sides w.r.t x,

(1/y)(dy/dx) = a×[-1/√(1-x2)]

(dy/dx) = -ay/√(1-x2)

On squaring both sides, we have

(dy/dx)2 = a2y2/(1 - x2)

(1 - x2)(dy/dx)2 = a2y2

_Again differentiating both sides w.r.t x,

2(1 - x2)(dy/dx)(d2y/dx2) - 2x(dy/dx)2 = 2a2y(dy/dx)

(1 - x2)(d2y/dx2) - x(dy/dx) = a2y

(1 - x2)(d2y/dx2) - x(dy/dx) - a2y = 0

Hence Proved

Question 35. If y = 500e7x + 600e-7x, show that d2y/dx2 = 49y.

**Solution:

We have,

y = 500e7x + 600e-7x

_On differentiating both sides w.r.t θ,

(dy/dx) = 7 × (500e7x - 600e-7x)

_Again differentiating both sides w.r.t x,

(d2y/dx2) = 49 × (500e7x + 600e-7x)

(d2y/dx2) = 49y

Hence Proved

Question 36. If x = 2cos t - cos 2t, y = 2sin t - sin 2t, find d2y/dx2 at t = π/2.

**Solution:

We have,

x = 2cos t - cos 2t, and y = 2sin t - sin 2t

_On differentiating both sides w.r.t t,

(dx/dt) = -2sin t + 2sin 2t, (dy/dt) = 2cos t - 2cos 2t

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = (2cos t - 2cos 2t)/(-2sin t + 2sin 2t)

(dy/dx) = (cos t - cos 2t)/(-sin t + sin 2t)

_Again differentiating both sides w.r.t x,

\frac{d^2y}{dx^2}=\frac{(-sin t+2sin 2t)(-sin t+sin 2t)-(cos t-cos 2t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}.\frac{dt}{dx}

\frac{d^2y}{dx^2}=\frac{(-sin t+2sin 2t)(-sin t+sin 2t)-(cos t-cos 2t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2(-2sin t+2sin 2t)}

At t = π/2

\frac{d^2y}{dx^2}=\frac{(-1+0)(-1+0)-(0+1)(0-2)}{(-1+0)^2(-2+0)}

d2y/dx2 = (1 + 2)/-2

d2y/dx2 = -(3/2)

Question 37. If x = 4z2 + 5, y = 6z2 + 7z + 3, find d2y/dx2.

**Solution:

We have,

x = 4z2 + 5, and y = 6z2 + 7z + 3

_On differentiating both sides w.r.t z,

(dx/dz) = 8z, and (dy/dz) = 12z + 7

(dy/dx) = (dy/dz) × (dz/dx)

(dy/dx) = (12z + 7)/8z

_Again differentiating both sides w.r.t x,

\frac{d^2y}{dx^2}=\frac{12×8z-8(12z+7)}{64z^2}.\frac{dz}{dx}

\frac{d^2y}{dx^2}=\frac{96z-96z-56}{64z^2}.\frac{1}{8z}

(d2y/dx2) = -7/64z3

Hence Proved

Question 38. If y = log(1 + cosx), prove that d3y/dx3 + (d2y/dx2).(dy/dx) = 0.

**Solution:

We have,

y = log(1 + cosx)

_On differentiating both sides w.r.t x,

(dy/dx) = -sinx/(1 + cosx)

_Again differentiating both sides w.r.t x,

\frac{d^2y}{dx^2}=\frac{-cosx(1+cosx)-(-sinx)(-cosx)}{(1+cosx)^2}

d2y/dx2 = (-cosx - cos2x - sin2x)/(1 + cosx)2

d2y/dx2 = -(1 + cosx)/(1 + cosx)2

d2y/dx2 = -1/(1 + cosx)

_Again differentiating both sides w.r.t x,

d3y/dx3 = -sinx/(1 + cosx)2

d3y/dx3 + [-1/(1 + cosx)][-sinx/(1 + cosx)] = 0

d3y/dx3 + (d2y/dx2).(dy/dx) = 0

Hence Proved

Question 39. If y = sin(logx), prove that x2(d2y/dx2) + x(dy/dx) + y = 0.

**Solution:

We have,

y = sin(logx)

_On differentiating both sides w.r.t x,

(dy/dx) = cos(logx).(1/x)

x(dy/dx) = cos(logx)

_Again differentiating both sides w.r.t x,

x(d2y/dx2) + (dy/dx) = -sin(logx).(1/x)

x2(d2y/dx2) + x(dy/dx) = -sin(logx)

x2(d2y/dx2) + x(dy/dx) = -y

x2(d2y/dx2) + x(dy/dx) + y = 0

Hence Proved

Question 40. If y = 3e2x + 2e3x, prove that d2y/dx2 - 5(dy/dx) + 6y = 0.

**Solution:

We have,

y = 3e2x + 2e3x

_On differentiating both sides w.r.t x,

(dy/dx) = 6e2x + 6e3x

(dy/dx) = 6(e2x + e3x)

_Again differentiating both sides w.r.t x,

d2y/dx2 = 6(2e2x + 3e3x)

d2y/dx2 = 12e2x + 18e3x

d2y/dx2 = 5(6e2x + 6e3x) - 6(3e2x + 2e3x)

d2y/dx2 = 5(dy/dx) - 6y

d2y/dx2 - 5(dy/dx) + 6y = 0

Hence Proved

Question 41. If y = (cot-1x)2, prove that y2(x2 + 1)2 + 2x(x2 + 1)y1 = 2.

**Solution:

We have,

y = (cot-1x)2

_On differentiating both sides w.r.t x,

y1 = 2(cot-1x) × [-1/(1 + x2)]

(1 + x2)y1 = -2cot-1x

_Again differentiating both sides w.r.t x,

(1 + x2)y2 + 2xy1 = 2/(1 + x2)

(1 + x2)2y2 + 2x(1 + x2)y1 = 2

Hence Proved

Question 42. If y = cosec-1x, then show that x(x2 - 1)d2y/dx2 - (2x2 - 1)(dy/dx) = 0.

**Solution:

We have,

y = cosec-1x

_On differentiating both sides w.r.t x,

(dy/dx) = -1/x√(x2 - 1)

On squaring both sides,

(dy/dx)2 = 1/x2(x2 - 1)

x2(x2 - 1)(dy/x)2 = 1

(x4 - x2)(dy/dx)2 = 1

2(dy/dx)(d2y/dx2)(x4 - x2) + (dy/dx)2(4x3 - 2x) = 0

2x2(x2 - 1)(dy/dx)(d2y/dx2) + 2x(2x2 - 1)(dy/dx)2 = 0

x(x2 - 1)(d2y/dx2) + (2x2 - 1)(dy/dx) = 0

Hence Proved

Question 43. If x = cos t + log(tant/2), y = sin t, then find the value of d2y/dt2 and d2y/dx2 at t = π/4 in terms of y alone.

**Solution:

We have,

y = sin t

_On differentiating both sides w.r.t t,

(dy/dt) = cos t

_Again differentiating both sides w.r.t x,

(d2y/dx2) = -sin t

At t = π/4

(d2y/dx2)t=π/4 = -sin(π/4)

= -1/√2

x = cos t + log(tant/2)

_On differentiating both sides w.r.t t,

\frac{dx}{dt}=-sin t+\frac{1}{tan\frac{t}{2}}.sec^2\frac{t}{2}.\frac{1}{2}

\frac{dx}{dt}=-sin t+\frac{1}{2sin\frac{t}{2}cos\frac{t}{2}}

(dx/dt) = -sin t + (1/sin t)

(dx/dt) = (-sin2t + 1)/sin t

(dx/dt) = cos2t/sint

(dx/dt) = cos t × cot t

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = [cos t] × [1/cos t × cot t]

(dy/dx) = tan t

_Again differentiating both sides w.r.t x,

(d2y/dx2) = sec2t × (dt/dx)

(d2y/dx2) = sec2t × [1/cos t × cot t]

(d2y/dx2) = sin t/cos4t

(d2y/dx2)t=π/4 = sin(π/4)/cos4(π/4)

(d2y/dx2) = 2√2

At t = π/4, (d2y/dx2) = -1/√2 and (d2y/dx2) = 2√2

Question 44. If x = asin t, y = a[cos t + log(tant/2)], find d2y/dx2.

**Solution:

We have,

x = asin t, and y = a[cos t + log(tant/2)]

_On differentiating both sides w.r.t t,

(dx/dt) = acos t and \frac{dy}{dt}=-asin t+a\frac{1}{tan\frac{t}{2}}.sec^2\frac{t}{2}.\frac{1}{2}

\frac{dy}{dt}=-asin t+a\frac{1}{2sin\frac{t}{2}cos\frac{t}{2}}

(dy/dt) = a[-sin t + (1/sin t)]

(dy/dt) = a[(-sin2t + 1)/sin t]

(dy/dt) = a[cos2t/sint]

(dy/dt) = acos t × cot t

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = [acos t × cot t] × [1/acos t]

(dy/dx) = cot t

_Again differentiating both sides w.r.t x,

(d2y/dx2)=-cosec2t × (dt/dx)

(d2y/dx2) = -cosec2t × [1/acos t]

(d2y/dx2) = -(1/asin2t × cos t)

Question 45. If x = a(cos t + tsin t), and y = a(sin t - tcos t), then find the value of d2y/dx2 at t = π/4.

**Solution:

We have,

x = a(cos t + tsin t), and y = a(sin t - tcos t)

_On differentiating both sides w.r.t t,

(dx/dt) = a(-sin t + sin t + tcos t)

(dx/dt) = atcos t

y = a(sin t - tcos t)

_On differentiating both sides w.r.t t,

(dy/dx) = a(cos t - cos t + tsin t)

(dy/dx) = atsin t

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = [atsin t] × [1/atcos t]

(dy/dx) = tan t

_Again differentiating both sides w.r.t x,

(d2y/dx2) = sec2x × (dt/dx)

(d2y/dx2) = sec2x × (1/atcos t)

(d2y/dx2) = 1/atcos3t

\frac{d^2y}{dx^2}=\frac{1}{a.\frac{π}{4}.cos^3\frac{π}{4}}

(d2y/dx2) = (8√2/aπ)

Question 46. If x = a[cos t + log(tant/2)], y = asin t, evaluate (d2y/dx2) at t = π/3.

**Solution:

We have,

y = asin t

_On differentiating both sides w.r.t t,

(dy/dt) = acos t

x = a[cos t + log(tant/2)]

_On differentiating both sides w.r.t t,

\frac{dx}{dt}=a[-sin t+\frac{1}{tan\frac{t}{2}}.sec^2\frac{t}{2}.\frac{1}{2}]

\frac{dx}{dt}=a[-sin t+\frac{1}{2sin\frac{t}{2}cos\frac{t}{2}}]

(dx/dt) = a[-sin t + (1/sin t)]

(dx/dt) = a[(-sin2t + 1)/sin t]

(dx/dt) = a[cos2t/sint]

(dx/dt) = acos t × cot t

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = [cos t] × [1/cos t × cot t]

(dy/dx) = tan t

_Again differentiating both sides w.r.t x,

(d2y/dx2) = sec2t × (dt/dx)

(d2y/dx2) = sec2t × [1/acos t × cot t]

(d2y/dx2) = sin t/acos4t

(d2y/dx2)t=π/3 = sin(π/3)/acos4(π/3)

(d2y/dx2) = (8√3/a)

Question 47. If x = a(cos2t + 2tsin2t), and y = a(sin2t - 2tcos2t), then find d2y/dx2.

**Solution:

We have,

x = a(cos2t + 2tsin2t), and y = a(sin2t - 2tcos2t)

_On differentiating both sides w.r.t t,

(dx/dt) = a(-2sin2t + 2sin2t + 4tcos2t), and (dy/dt) = a(2cos2t - 2cos2t + 4tsin2t)

(dy/dt) = a(4tcos2t), and (dy/dt)=a(4tsin2t)

(dy/dx) = (dy/dz) × (dz/dx)

(dy/dx) = a(4tsin2t)/a(4tcos2t)

(dy/dx) = tan2t

_Again differentiating both sides w.r.t x,

(d2y/dx2) = 2sec22t.(dt/dx)

(d2y/dx2) = 2sec22t/4atcos2t

(d2y/dx2) = 1/2atcos32t

(d2y/dx2) = (1/2at) × (sec3x)

Question 48. If x = asin t - bcos t, y = acos t + bsin t, prove that (d2y/dx2) = -(x2 + y2)/y3

**Solution:

We have,

x = asin t - bcos t

_On differentiating both sides w.r.t t,

(dx/dt) = acos t + bsin t

y = acos t + b sin t

_On differentiating both sides w.r.t t,

(dy/dt) = -asin t + bcos t

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = [-asin t + bcos t] × [1/(acos t + bsin t)]

_Again differentiating both sides w.r.t x,

\frac{d^2y}{dx^2}=\frac{(acost+bsint)(-asint-bcost)-(-asint+bcost)(-asint+bcost)}{(acost+bsint)^2}.\frac{dt}{dx}

\frac{d^2y}{dx^2}=\frac{-(acost+bsint)^2-(-asint+bcost)^2}{(acost+bsint)^3}

\frac{d^2y}{dx^2}=\frac{-(acost+bsint)^2-(asint-bcost)^2}{(acost+bsint)^3}

d2y/dx2 = (-y2 - x2)/y3

d2y/dx2 = -(x2 + y2)/y3

Hence Proved

Question 49. Find A and B so that y = Asin3x + Bcos3x, satisfies the equation d2y/dx2 + 4(dy/dx) + 3y = 10cos3x.

**Solution:

We have,

y = Asin3x + Bcos3x,

_On differentiating both sides w.r.t x,

(dy/dx) = 3Acos3x - 3Bsin3x

_Again differentiating both sides w.r.t x,

d2y/dx2 = -9Asin3x - 9Bcos3x

d2y/dx2 + 4(dy/dx) + 3y = (-9Asin3x - 9Bcos3x) + 4(3Acos3x - 3Bsin3x) + 3(Asin3x + Bcos3x)

= -9Asin3x - 9Bcos3x + 12Acos3x - 12Bsin3x + 3Asin3x + 3Bcos3x

= -6Asin3x - 12Bsin3x - 6Bcos3x + 12Acos3x

= (-6A - 12B)sin3x + (-6B + 12A)cos3x ...(i)

Given that

d2y/dx2 + 4(dy/dx) + 3y = 10cos3x ...(ii)

On comparing the coefficients, we get

(-6A - 12B) = 0 and (-6B + 12A) = 10

Solving equation,

A = (2/3) and B = -(1/3)

Question 50. If y = Ae-ktcos(pt + c), prove that (d2y/dt2) + 2k(dy/dt) + n2y = 0, where n2 = p2 + k2

**Solution:

We have,

y = Ae-ktcos(pt + c)

_On differentiating both sides w.r.t t,

(dy/dt) = -kAe-ktcos(pt + c) - pAe-ktsin(pt + c)

(dy/dt) = -ky - pAe-ktsin(pt + c)

_Again differentiating both sides w.r.t t,

(d2y/dt2) = -k(dy/dt) + pAke-ktsin(pt + c) - p2Ake-ktcos(pt + c)

(d2y/dt2) = -k(dy/dt) + k(-ky - dy/dx) - p2y

(d2y/dt2) = -k(dy/dt) - k2y - k(dy/dt) - p2y

(d2y/dt2) + 2k(dy/dt) + (k2 + p2)y = 0

(d2y/dt2) + 2k(dy/dt) + n2y = 0

Hence Proved

Question 51. If y = xn{acos(logx) + bsin(logx)}, prove that x2(d2y/dt2) + (1 - 2n) x (dy/dt) + (1 + n2)y = 0.

**Solution:

We have,

y = xn{acos(logx) + bsin(logx)} ...(i)

_On differentiating both sides w.r.t x,

(dy/dx) = nxn-1{acos(logx) + bsin(logx)} + xn{-asin(logx).(1/x) + bcos(logx).(1/x)}

x(dy/dx) = nxn{acos(logx) + bsin(logx)} + xn{-asin(logx) + bcos(logx)}

x(dy/dx) = ny + xn{-asin(logx) + bcos(logx)} ...(ii)

xn{-asin(logx) + bcos(logx)} = x(dy/dx) - ny ...(iii)

_Again differentiating both sides w.r.t x,

x(d2y/dx2) + (dy/dx) = n(dy/dx) + nxn-1{-asin(logx) + bcos(logx)} + xn{-acos(logx).(1/x) - bsin(logx).(1/x)}

x2(d2y/dx2) + (dy/dx) = nx(dy/dx) + nxn{-asin(logx) + bcos(logx)} - xn{acos(logx) + bsin(logx)}

x2(d2y/dx2) = nx(dy/dx) + n{x(dy/dx) - ny} - y - (dy/dx) [From equation (ii) and (iii)]

x2(d2y/dx2) = nx(dy/dx) + nx(dy/dx) - (dy/dx) - n2y - y

x2(d2y/dx2) = (dy/dx) x [2n - 1] - (n2 + 1)y

x2(d2y/dt2) + (1 - 2n) x (dy/dt) + (1 + n2)y = 0

Hence Proved

Question 52. y = a[x + \sqrt{x^2+1}]^n+b[x - \sqrt{x^2+1}]^{-n}, prove that (x2+1)d2y/d2x + xdy/dx - ny = 0.

**Solution:

We have y = a[x + \sqrt{x^2+1}]^n+b[x - \sqrt{x^2+1}]^{-n}

_On differentiating both sides w.r.t x,

_dy/dx = na[x + \sqrt{x^2+1}]^{n-1}[1+x(x^2+1)]^{\frac{-1}{2}}+nb[x - \sqrt{x^2+1}]^{-n-1}[1-x(x^2+1)]^{\frac{-1}{2}}

_dy/dx = \frac{na}{\sqrt{x^2+1}}[x + \sqrt{x^2+1}]^{n}+\frac{nb}{\sqrt{x^2+1}}[x - \sqrt{x^2+1}]^{-n}

_dy/dx = \frac{n}{\sqrt{x^2+1}}[a[x + \sqrt{x^2+1}]^{n}+b[x - \sqrt{x^2+1}]^{-n}

_xdy/dx = \frac{nx}{\sqrt{x^2+1}}y

_Again differentiating both sides w.r.t x,

d2y/dx2 = \frac{nx}{\sqrt{x^2+1}}\frac{dy}{dx}+y[\frac{\sqrt{x^2+1}-x^2(x^2+1)^{-\frac{1}{2}}}{x^2+1}]

d2y/dx2 = \frac{n^2x^2}{x^2+1}+y[\frac{1}{(x^2+1)(\sqrt{x^2+1})}]

d2y/dx2 = \frac{n^2x^2(\sqrt{x^2+1})+y}{(x^2+1)(\sqrt{x^2+1})}

(x2+1)d2y/d2x = \frac{n^2x^4(\sqrt{x^2+1})+x^2y}{(x^2+1)(\sqrt{x^2+1})}-\frac{n^2x^2(\sqrt{x^2+1})+y}{(x^2+1)(\sqrt{x^2+1})}

Now put all these values in this equation

(x2+1)d2y/d2x + xdy/dx - ny

\frac{n^2x^4(\sqrt{x^2+1})+x^2y}{(x^2+1)(\sqrt{x^2+1})}-\frac{n^2x^2(\sqrt{x^2+1})+y}{(x^2+1)(\sqrt{x^2+1})}+\frac{nx}{\sqrt{x^2+1}}y-ny=0

Hence Proved

Summary

Exercise 12.1 Set 2 in Chapter 12 of RD Sharma's Class 12 Mathematics provides advanced problems on higher order derivatives. These questions typically involve more complex functions and combinations of functions, requiring students to apply multiple differentiation rules and recognize patterns in successive derivatives. This set helps students deepen their understanding of calculus and prepares them for more sophisticated mathematical analysis.