# Class 12 RD Sharma Solutions – Chapter 12 Higher Order Derivatives – Exercise 12.1 | Set 2

### Question 27. If y = [log{x+(âˆšx^{2}+1)}]^{2}, show that (1 + x^{2})(d^{2}y/dx^{2}) + x(dy/dx) = 2.

**Solution:**

We have,

y = [log{x + (âˆšx

^{2 }+ 1)}]^{2}

On differentiating both sides w.r.t x,dy/dx = 2[log{x + (âˆšx

^{2 }+ 1)}]/(âˆšx^{2 }+ 1)

Again differentiating both sides w.r.t x,(x

^{2 }+ 1)(d^{2}y/dx^{2}) = 2 – x(dy/dx)(x

^{2 }+ 1)(d^{2}y/dx^{2}) + x(dy/dx) = 2

Hence Proved

### Question 28. If y = (tan^{-1}x)^{2}, then prove that (1 + x^{2})^{2}y_{2 }+ 2x(1 + x^{2})y_{1 }= 2

**Solution:**

We have,

y = (tan

^{-1}x)^{2}

On differentiating both sides w.r.t x,y

_{1 }= 2(tan^{-1}x)[1/(1 + x^{2})](1 + x

^{2})y_{1 }= 2(tan^{-1}x)

Again differentiating both sides w.r.t x,(1 + x

^{2})y_{2 }+ 2xy_{1 }= 2/(1 + x^{2})(1 + x

^{2})^{2}y_{2 }+ 2x(1 + x^{2})y_{1 }= 2

Hence Proved

### Question 29. If y = cotx, prove that (d^{2}y/dx^{2}) + 2y(dy/dx) = 0.

**Solution:**

We have,

y = cotx

On differentiating both sides w.r.t x,(dy/dx) = -cosec

^{2}x

Again differentiating both sides w.r.t x,d

^{2}y/dx^{2 }= -(2cosec x) Ã— (-cosec x.cot x)d

^{2}y/dx^{2 }= 2cosec^{2}x.cot xd

^{2}y/dx^{2 }= 2(cot x)(cosec^{2}x)d

^{2}y/dx^{2 }= -2y(dy/dx)d

^{2}y/dx^{2 }+ 2y(dy/dx) = 0Hence Proved

### Question 30. Find d^{2}y/dx^{2} where y = log(x^{2}/e^{2}).

**Solution:**

We have,

y = log(x

^{2}/e^{2})

On differentiating both sides w.r.t x,(dy/dx) = (2/x)

Again differentiating both sides w.r.t x,d

^{2}y/dx^{2 }= -(2/x^{2})Hence Proved

### Question 31. If y = ae^{2x }+ be^{-x}, show that (d^{2}y/dx^{2}) – (dy/dx) – 2y = 0.

**Solution:**

We have,

y = ae

^{2x }+ be^{-x}

On differentiating both sides w.r.tt,(dy/dx) = 2ae

^{2x }– be^{-x}

Again differentiating both sides w.r.t x,(d

^{2}y/dx^{2}) = 4ae^{2x }+ be^{-x}(d

^{2}y/dx^{2}) = 2ae^{2x }– be^{-x }+ 2(ae^{2x }+ be^{-x})(d

^{2}y/dx^{2}) = (dy/dx) + 2y(d

^{2}y/dx^{2}) – (dy/dx) – 2y = 0Hence Proved

### Question 32. If y = e^{x}(sinx + cosx), prove that d^{2}y/dx^{2 }– 2(dy/dx) + 2y = 0.

**Solution:**

We have,

y = e

^{x}(sinx + cosx)

On differentiating both sides w.r.t x,(dy/dx) = e

^{x}(sinx + cosx) + e^{x}(cosx – sinx)(dy/dx) = 2e

^{x}cosx

Again differentiating both sides w.r.t x,(d

^{2}y/dx^{2}) = 2e^{x}cosx – 2e^{x}sinxLets take L.H.S,

= d

^{2}y/dx^{2 }– 2(dy/dx) + 2y= 2e

^{x}cosx – 2e^{x}sinx – 2(2e^{x}cosx) + 2e^{x}(sinx + cosx)= 4e

^{x}cosx – 4e^{x}cosx – 2e^{x}sinx + 2e^{x}sinx= 0

L.H.S = R.H.S

Hence Proved

### Question 33. If y = cos^{-1}x, find d^{2}y/dx^{2} in terms of y alone.

**Solution:**

We have,

y = cos

^{-1}x

On differentiating both sides w.r.t x,(dy/dx) = -1/âˆš(1-x

^{2})

Again differentiating both sides w.r.t x,…(i)

y = cos

^{-1}xx = cosy

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

d

^{2}y/dx^{2 }= -cosy/sin^{3}yd

^{2}y/dx^{2 }= -cot y cosec^{2}y

### Question 34. If y = , prove that (1 – x^{2})(d^{2}y/dx^{2}) – x(dy/dx) – a^{2}y = 0.

**Solution:**

We have,

y =

Taking log both sides

logy = acos

^{-1}x.logelogy = acos

^{-1}x

On differentiating both sides w.r.t x,(1/y)(dy/dx) = aÃ—[-1/âˆš(1-x

^{2})](dy/dx) = -ay/âˆš(1-x

^{2})On squaring both sides, we have

(dy/dx)

^{2 }= a^{2}y^{2}/(1 – x^{2})(1 – x

^{2})(dy/dx)^{2 }= a^{2}y^{2}

Again differentiating both sides w.r.t x,2(1 – x

^{2})(dy/dx)(d^{2}y/dx^{2}) – 2x(dy/dx)^{2 }= 2a^{2}y(dy/dx)(1 – x

^{2})(d^{2}y/dx^{2}) – x(dy/dx) = a^{2}y(1 – x

^{2})(d^{2}y/dx^{2}) – x(dy/dx) – a^{2}y = 0Hence Proved

### Question 35. If y = 500e^{7x }+ 600e^{-7x}, show that d^{2}y/dx^{2 }= 49y.

**Solution:**

We have,

y = 500e

^{7x }+ 600e^{-7x}

On differentiating both sides w.r.tÎ¸,(dy/dx) = 7 Ã— (500e

^{7x }– 600e^{-7x})

Again differentiating both sides w.r.t x,(d

^{2}y/dx^{2}) = 49 Ã— (500e^{7x }+ 600e^{-7x})(d

^{2}y/dx^{2}) = 49yHence Proved

### Question 36. If x = 2cos t – cos 2t, y = 2sin t – sin 2t, find d^{2}y/dx^{2 }at t = Ï€/2.

**Solution:**

We have,

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

On differentiating both sides w.r.tt,(dx/dt) = -2sin t + 2sin 2t, (dy/dt) = 2cos t – 2cos 2t

(dy/dx) = (dy/dt) Ã— (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,At t = Ï€/2

d

^{2}y/dx^{2 }= (1 + 2)/-2d

^{2}y/dx^{2 }= -(3/2)

### Question 37. If x = 4z^{2 }+ 5, y = 6z^{2 }+ 7z + 3, find d^{2}y/dx^{2}.

**Solution:**

We have,

x = 4z

^{2 }+ 5, and y = 6z^{2 }+ 7z + 3

On differentiating both sides w.r.tz,(dx/dz) = 8z, and (dy/dz) = 12z + 7

(dy/dx) = (dy/dz) Ã— (dz/dx)

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

Again differentiating both sides w.r.t x,(d

^{2}y/dx^{2}) = -7/64z^{3}Hence Proved

### Question 38. If y = log(1 + cosx), prove that d^{3}y/dx^{3 }+ (d^{2}y/dx^{2}).(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,d

^{2}y/dx^{2 }= (-cosx – cos^{2}x – sin^{2}x)/(1 + cosx)^{2}d

^{2}y/dx^{2 }= -(1 + cosx)/(1 + cosx)^{2}d

^{2}y/dx^{2 }= -1/(1 + cosx)

Again differentiating both sides w.r.t x,d

^{3}y/dx^{3 }= -sinx/(1 + cosx)^{2}d

^{3}y/dx^{3 }+ [-1/(1 + cosx)][-sinx/(1 + cosx)] = 0d

^{3}y/dx^{3 }+ (d^{2}y/dx^{2}).(dy/dx) = 0Hence Proved

### Question 39. If y = sin(logx), prove that x^{2}(d^{2}y/dx^{2}) + 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(d

^{2}y/dx^{2}) + (dy/dx) = -sin(logx).(1/x)x

^{2}(d^{2}y/dx^{2}) + x(dy/dx) = -sin(logx)x

^{2}(d^{2}y/dx^{2}) + x(dy/dx) = -yx

^{2}(d^{2}y/dx^{2}) + x(dy/dx) + y = 0Hence Proved

### Question 40. If y = 3e^{2x }+ 2e^{3x}, prove that d^{2}y/dx^{2 }– 5(dy/dx) + 6y = 0.

**Solution:**

We have,

y = 3e

^{2x }+ 2e^{3x}

On differentiating both sides w.r.t x,(dy/dx) = 6e

^{2x }+ 6e^{3x}(dy/dx) = 6(e

^{2x }+ e^{3x})

Again differentiating both sides w.r.t x,d

^{2}y/dx^{2 }= 6(2e^{2x }+ 3e^{3x})d

^{2}y/dx^{2 }= 12e^{2x }+ 18e^{3x}d

^{2}y/dx^{2 }= 5(6e^{2x }+ 6e^{3x}) – 6(3e^{2x }+ 2e^{3x})d

^{2}y/dx^{2 }= 5(dy/dx) – 6yd

^{2}y/dx^{2 }– 5(dy/dx) + 6y = 0Hence Proved

### Question 41. If y = (cot^{-1}x)^{2}, prove that y_{2}(x^{2 }+ 1)^{2 }+ 2x(x^{2 }+ 1)y_{1 }= 2.

**Solution:**

We have,

y = (cot

^{-1}x)^{2}

On differentiating both sides w.r.t x,y

_{1 }= 2(cot^{-1}x) Ã— [-1/(1 + x^{2})](1 + x

^{2})y_{1 }= -2cot^{-1}x

Again differentiating both sides w.r.t x,(1 + x

^{2})y_{2 }+ 2xy_{1 }= 2/(1 + x^{2})(1 + x

^{2})^{2}y_{2 }+ 2x(1 + x^{2})y_{1 }= 2Hence Proved

### Question 42. If y = cosec^{-1}x, then show that x(x^{2 }– 1)d^{2}y/dx^{2 }– (2x^{2 }– 1)(dy/dx) = 0.

**Solution:**

We have,

y = cosec

^{-1}x

On differentiating both sides w.r.t x,(dy/dx) = -1/xâˆš(x

^{2 }– 1)On squaring both sides,

(dy/dx)

^{2 }= 1/x^{2}(x^{2 }– 1)x

^{2}(x^{2 }– 1)(dy/x)^{2 }= 1(x

^{4 }– x^{2})(dy/dx)^{2 }= 12(dy/dx)(d

^{2}y/dx^{2})(x^{4 }– x^{2}) + (dy/dx)^{2}(4x^{3 }– 2x) = 02x

^{2}(x^{2 }– 1)(dy/dx)(d^{2}y/dx^{2}) + 2x(2x^{2 }– 1)(dy/dx)^{2 }= 0x(x

^{2 }– 1)(d^{2}y/dx^{2}) + (2x^{2 }– 1)(dy/dx) = 0Hence Proved

### Question 43. If x = cos t + log(tant/2), y = sin t, then find the value of d^{2}y/dt^{2} and d^{2}y/dx^{2} at t = Ï€/4 in terms of y alone.

**Solution:**

We have,

y = sin t

On differentiating both sides w.r.tt,(dy/dt) = cos t

Again differentiating both sides w.r.t x,(d

^{2}y/dx^{2}) = -sin tAt t = Ï€/4

(d

^{2}y/dx^{2})_{t=Ï€/4 }= -sin(Ï€/4)= -1/âˆš2

x = cos t + log(tant/2)

On differentiating both sides w.r.tt,(dx/dt) = -sin t + (1/sin t)

(dx/dt) = (-sin

^{2}t + 1)/sin t(dx/dt) = cos

^{2}t/sint(dx/dt) = cos t Ã— cot t

(dy/dx) = (dy/dt) Ã— (dt/dx)

(dy/dx) = [cos t] Ã— [1/cos t Ã— cot t]

(dy/dx) = tan t

Again differentiating both sides w.r.t x,(d

^{2}y/dx^{2}) = sec^{2}t Ã— (dt/dx)(d

^{2}y/dx^{2}) = sec^{2}t Ã— [1/cos t Ã— cot t](d

^{2}y/dx^{2}) = sin t/cos^{4}t(d

^{2}y/dx^{2})_{t=Ï€/4 }= sin(Ï€/4)/cos^{4}(Ï€/4)(d

^{2}y/dx^{2}) = 2âˆš2At t = Ï€/4, (d

^{2}y/dx^{2}) = -1/âˆš2 and (d^{2}y/dx^{2}) = 2âˆš2

### Question 44. If x = asin t, y = a[cos t + log(tant/2)], find d^{2}y/dx^{2}.

**Solution:**

We have,

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

On differentiating both sides w.r.tt,(dx/dt) = acos t and

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

(dy/dt) = a[(-sin

^{2}t + 1)/sin t](dy/dt) = a[cos

^{2}t/sint](dy/dt) = acos t Ã— cot t

(dy/dx) = (dy/dt) Ã— (dt/dx)

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

(dy/dx) = cot t

Again differentiating both sides w.r.t x,(d

^{2}y/dx^{2})=-cosec^{2}t Ã— (dt/dx)(d

^{2}y/dx^{2}) = -cosec^{2}t Ã— [1/acos t](d

^{2}y/dx^{2}) = -(1/asin^{2}t Ã— cos t)

### Question 45. If x = a(cos t + tsin t), and y = a(sin t – tcos t), then find the value of d^{2}y/dx^{2} at t = Ï€/4.

**Solution:**

We have,

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

On differentiating both sides w.r.tt,(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.tt,(dy/dx) = a(cos t – cos t + tsin t)

(dy/dx) = atsin t

(dy/dx) = (dy/dt) Ã— (dt/dx)

(dy/dx) = [atsin t] Ã— [1/atcos t]

(dy/dx) = tan t

Again differentiating both sides w.r.t x,(d

^{2}y/dx^{2}) = sec^{2}x Ã— (dt/dx)(d

^{2}y/dx^{2}) = sec^{2}x Ã— (1/atcos t)(d

^{2}y/dx^{2}) = 1/atcos^{3}t(d

^{2}y/dx^{2}) = (8âˆš2/aÏ€)

### Question 46. If x = a[cos t + log(tant/2)], y = asin t, evaluate (d^{2}y/dx^{2}) at t = Ï€/3.

**Solution:**

We have,

y = asin t

On differentiating both sides w.r.tt,(dy/dt) = acos t

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

On differentiating both sides w.r.tt,(dx/dt) = a[-sin t + (1/sin t)]

(dx/dt) = a[(-sin

^{2}t + 1)/sin t](dx/dt) = a[cos

^{2}t/sint](dx/dt) = acos t Ã— cot t

(dy/dx) = (dy/dt) Ã— (dt/dx)

(dy/dx) = [cos t] Ã— [1/cos t Ã— cot t]

(dy/dx) = tan t

Again differentiating both sides w.r.t x,(d

^{2}y/dx^{2}) = sec^{2}t Ã— (dt/dx)(d

^{2}y/dx^{2}) = sec^{2}t Ã— [1/acos t Ã— cot t](d

^{2}y/dx^{2}) = sin t/acos^{4}t(d

^{2}y/dx^{2})_{t=Ï€/3 }= sin(Ï€/3)/acos^{4}(Ï€/3)(d

^{2}y/dx^{2}) = (8âˆš3/a)

### Question 47. If x = a(cos2t + 2tsin2t), and y = a(sin2t – 2tcos2t), then find d^{2}y/dx^{2}.

**Solution:**

We have,

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

On differentiating both sides w.r.tt,(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) Ã— (dz/dx)

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

(dy/dx) = tan2t

Again differentiating both sides w.r.t x,(d

^{2}y/dx^{2}) = 2sec^{2}2t.(dt/dx)(d

^{2}y/dx^{2}) = 2sec^{2}2t/4atcos2t(d

^{2}y/dx^{2}) = 1/2atcos^{3}2t(d

^{2}y/dx^{2}) = (1/2at) Ã— (sec^{3}x)

### Question 48. If x = asin t – bcos t, y = acos t + bsin t, prove that (d^{2}y/dx^{2}) = -(x^{2 }+ y^{2})/y^{3}

**Solution:**

We have,

x = asin t – bcos t

On differentiating both sides w.r.tt,(dx/dt) = acos t + bsin t

y = acos t + b sin t

On differentiating both sides w.r.tt,(dy/dt) = -asin t + bcos t

(dy/dx) = (dy/dt) Ã— (dt/dx)

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

Again differentiating both sides w.r.t x,d

^{2}y/dx^{2 }= (-y^{2 }– x^{2})/y^{3}d

^{2}y/dx^{2 }= -(x^{2 }+ y^{2})/y^{3}Hence Proved

### Question 49. Find A and B so that y = Asin3x + Bcos3x, satisfies the equation d^{2}y/dx^{2 }+ 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,d

^{2}y/dx^{2 }= -9Asin3x – 9Bcos3xd

^{2}y/dx^{2 }+ 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

d

^{2}y/dx^{2 }+ 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^{-kt}cos(pt + c), prove that (d^{2}y/dt^{2}) + 2k(dy/dt) + n^{2}y = 0, where n^{2 }= p^{2 }+ k^{2}

**Solution:**

We have,

y = Ae

^{-kt}cos(pt + c)

On differentiating both sides w.r.tt,(dy/dt) = -kAe

^{-kt}cos(pt + c) – pAe^{-kt}sin(pt + c)(dy/dt) = -ky – pAe

^{-kt}sin(pt + c)

Again differentiating both sides w.r.t t,(d

^{2}y/dt^{2}) = -k(dy/dt) + pAke^{-kt}sin(pt + c) – p^{2}Ake^{-kt}cos(pt + c)(d

^{2}y/dt^{2}) = -k(dy/dt) + k(-ky – dy/dx) – p^{2}y(d

^{2}y/dt^{2}) = -k(dy/dt) – k^{2}y – k(dy/dt) – p^{2}y(d

^{2}y/dt^{2}) + 2k(dy/dt) + (k^{2 }+ p^{2})y = 0(d

^{2}y/dt^{2}) + 2k(dy/dt) + n^{2}y = 0Hence Proved

### Question 51. If y = x^{n}{acos(logx) + bsin(logx)}, prove that x^{2}(d^{2}y/dt^{2}) + (1 – 2n) x (dy/dt) + (1 + n^{2})y = 0.

**Solution:**

We have,

y = x

^{n}{acos(logx) + bsin(logx)} …(i)

On differentiating both sides w.r.t x,(dy/dx) = nx

^{n-1}{acos(logx) + bsin(logx)} + x^{n}{-asin(logx).(1/x) + bcos(logx).(1/x)}x(dy/dx) = nx

^{n}{acos(logx) + bsin(logx)} + x^{n}{-asin(logx) + bcos(logx)}x(dy/dx) = ny + x

^{n}{-asin(logx) + bcos(logx)} …(ii)x

^{n}{-asin(logx) + bcos(logx)} = x(dy/dx) – ny …(iii)

Again differentiating both sides w.r.t x,x(d

^{2}y/dx^{2}) + (dy/dx) = n(dy/dx) + nx^{n-1}{-asin(logx) + bcos(logx)} + x^{n}{-acos(logx).(1/x) – bsin(logx).(1/x)}x

^{2}(d^{2}y/dx^{2}) + (dy/dx) = nx(dy/dx) + nx^{n}{-asin(logx) + bcos(logx)} – x^{n}{acos(logx) + bsin(logx)}x

^{2}(d^{2}y/dx^{2}) = n^{x}(dy/dx) + n{x(dy/dx) – ny} – y – (dy/dx) [From equation (ii) and (iii)]x

^{2}(d^{2}y/dx^{2}) = nx(dy/dx) + nx(dy/dx) – (dy/dx) – n^{2}y – yx

^{2}(d^{2}y/dx^{2}) = (dy/dx) x [2n – 1] – (n^{2 }+ 1)yx

^{2}(d^{2}y/dt^{2}) + (1 – 2n) x (dy/dt) + (1 + n^{2})y = 0Hence Proved

### Question 52. y = , prove that (x^{2}+1)d^{2}y/d^{2}x + xdy/dx – ny = 0.

**Solution:**

We have y =

On differentiating both sides w.r.t x,

dy/dx =

dy/dx =

dy/dx =

xdy/dx =

Again differentiating both sides w.r.t x,d

^{2}y/dx^{2 }=d

^{2}y/dx^{2 }=d

^{2}y/dx^{2 }=(x

^{2}+1)d^{2}y/d^{2}x =Now put all these values in this equation

(x

^{2}+1)d^{2}y/d^{2}x + xdy/dx – nyHence Proved

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