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.
