# RD Sharma Class 12 Ex 12.1 Solutions Chapter 12 Higher Order Derivatives

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## RD Sharma Class 12 Ex 12.1 Solutions Chapter 12 Higher Order Derivatives

### Question 1(i).  x3 + tanx

Solution:

Let us considered

f(x) = x+ tanx

On differentiating both sides w.r.t x,

f'(x) = 3x+ sec2x

Again differentiating both sides w.r.t x,

f”(x) = 6x + 2(secx)(secx.tanx)

f”(x) = 6x + 2sec2x.tanx

### Question 1(ii). sin(logx)

Solution:

Let us considered

y = sin(logx)

On differentiating both sides w.r.t x,

(dy/dx) = cos(logx) × (1/x)

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

Again differentiating both sides w.r.t x,

d2y/dx= d/dx[cos(logx)/x]

= -[sin(logx) + cos(logx)]/x2

### Question 1(iii). log(sinx)

Solution:

Let us considered

y = log(sinx)

On differentiating both sides w.r.t x,

(dy/dx) = (1/sinx) × (cosx)

(dy/dx) = cotx

Again differentiating both sides w.r.t x,

d2y/dx= -cosec2x

### Question 1(iv). exsin5x

Solution:

Let us considered

y = exsin5x

On differentiating both sides w.r.t x,

(dy/dx) = exsin5x + 5excos5x

Again differentiating both sides w.r.t x,

d2y/dx= exsin5x + 5excos5x + 5(excos5x – 5exsin5x)

d2y/dx= -24exsin5x + 10excos5x

d2y/dx2 = 2ex(5cos5x – 12sinx)

### Question 1(v). e6xcos3x

Solution:

Let us considered

y = e6xcos3x

On differentiating both sides w.r.t x,

(dy/dx) = 6e6xcos3x – 3e6xsin3x

Again differentiating both sides w.r.t x,

d2y/dx= 6(6e6xcos3x – 3e6xsin3x) – 3(6e6xsin3x + 3e6xcos3x)

d2y/dx= 36e6xcos3x – 18e6xsin3x – 18e6xsin3x – 9e6xcos3x

d2y/dx= 27e6xcos3x – 36e6xsin3x

d2y/dx= 9e6x(3cos3x – 4sin3x)

### Question 1(vi). x3logx

Solution:

Let us considered

y = x3logx

On differentiating both sides w.r.t x,

(dy/dx) = logx.3x+ x3(1/x)

(dy/dx) = logx.3x+ x2

(dy/dx) = x2(1 + 3logx)

Again differentiating both sides w.r.t x,

d2y/dx= (1 + 3logx).2x + x2(3/x)

d2y/dx= 2x + 6xlogx + 3x

d2y/dx= x(5 + 6logx)

### Question 1(vii). tan-1x

Solution:

Let us considered

y = tan-1x

On differentiating both sides w.r.t x,

(dy/dx) = 1/(1 + x2)

Again differentiating both sides w.r.t x,

d2y/dx= (-1)(1 + x2)-2.2x

### Question 1(viii). x.cosx

Solution:

Let us considered

y = x.cosx

On differentiating both sides w.r.t x,

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

(dy/dx) = cosx – xsinx

Again differentiating both sides w.r.t x,

d2y/dx= -sinx – (sinx + xcosx)

d2y/dx= -2sinx – xcosx

d2y/dx= -(xcosx + 2sinx)

### Question 1(ix). log(logx)

Solution:

Let us considered

y = log(logx)

On differentiating both sides w.r.t x,

(dy/dx) = (1/logx) × (1/x)

(dy/dx) = 1/xlogx

Again differentiating both sides w.r.t x,

d2y/dx= (-1)(xlogx)-2.[(d/dx)xlogx]

d2y/dx2 = (-1)(xlogx)-2[logx+x.(1/x)]

d2y/dx2= (-1)(xlogx)-2.(logx+1)

### Question 2. If y = e-x.cosx, show that d2y/dx2 = 2e-x.sinx.

Solution:

Let us considered

y = e-x.cosx

On differentiating both sides w.r.t x,

(dy/dx) = -e-x.cosx – e-x.sinx

Again differentiating both sides w.r.t x,

d2y/dx= -(-e-x.cosx – e-x.sinx) – (-e-x.sinx + e-x.cosx)

d2y/dx= e-x.cosx – e-x.cosx + e-x.sinx + e-x.sinx

d2y/dx= 2e-x.sinx

Hence Proved

### Question 3. If y = x + tanx, Show that cos2x(d2y/dx2) – 2y + 2x = 0.

Solution:

Let us considered

y = x + tanx

On differentiating both sides w.r.t x,

(dy/dx) = 1 + sec2x

Again differentiating both sides w.r.t x,

d2y/dx= 0 + (2secx)(secx.tanx)

d2y/dx= 2sec2x.tanx

On multiplying both sides by cos2x

cos2x(d2y/dx2) = 2tanx

cos2x(d2y/dx2) = 2(y – x)   [since, tanx = y – x]

cos2x(d2y/dx2) – 2y + 2x = 0

Hence Proved

### Question 4. If y = x3logx, prove that (d4y/dx4) = (6/x).

Solution:

Let us considered

y = x3logx

On differentiating both sides w.r.t x,

(dy/dx) = logx.3x+ x3(1/x)

(dy/dx) = logx.3x+ x2

(dy/dx) = x2(1 + 3logx)

Again differentiating both sides w.r.t x,

d2y/dx= (1 + 3logx).2x + x2(3/x)

d2y/dx2 = 2x + 6xlogx + 3x

d2y/dx= 5x + 6xlogx

Again differentiating both sides w.r.t x,

d3y/dx= 5 + 6[logx + (x/x)]

d3y/dx= 11 + 6logx

Again differentiating both sides w.r.t x,

d4y/dx= (6/x)

Hence Proved

### Question 5. If y = log(sinx), prove that (d3y/dx3) = 2cosx.cosec3x.

Solution:

Let us considered

y = log(sinx)

On differentiating both sides w.r.t x,

(dy/dx) = (1/sinx) × (cosx)

(dy/dx) = cotx

Again differentiating both sides w.r.t x,

d2y/dx= -cosec2x

Again differentiating both sides w.r.t x,

d3y/dx= -2cosecx.(-cosesx.cotx)

d3y/dx= 2cosec2x.cotx

d3y/dx= 2cosec2x.(cosx/sinx)

d3y/dx= cosx.cosec3x

Hence Proved

### Question 6. If y = 2sinx + 3cosx, show that (d2y/dx2) + y = 0.

Solution:

Let us considered

y = 2sinx + 3cosx

On differentiating both sides w.r.t x,

(dy/dx) = 2cosx – 3sinx

Again differentiating both sides w.r.t x,

d2y/dx= -2sinx – 3cosx

d2y/dx= -(2sinx + 3cosx)

d2y/dx= -y

d2y/dx+ y = 0

Hence Proved

### Question 7. If y = (logx/x), show that (d2y/dx2) = (2logx – 3)/x3

Solution:

Let us considered

y = (logx/x)

On differentiating both sides w.r.t x,

(dy/dx) = (1 – logx)/x2

Again differentiating both sides w.r.t x,

d2y/dx= [-x – 2x(1 – logx)]/x4

d2y/dx= (2xlogx – 3x)/x4

d2y/dx= (2logx – 3)/x3

d2y/dx2 + y = 0

Hence Proved

### Question 8. If x = a secθ, y = b tanθ, show that (d2y/dx2) = -b4/a2y3

Solution:

We have,

x = a secθ and y = b tanθ

On differentiating both sides w.r.t θ,

(dx/dθ) = a secθ.tanθ, (dy/dθ) = b sec2θ

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = (b sec2θ)/(a secθ.tanθ)

(dy/dx) = (b/a).cosecθ

Again differentiating both sides w.r.t x,

(d2y/dx2) = (b/a).(-cosecθ.cotθ).(dθ/dx)

(d2y/dx2) = -(b/a).(cosecθ.cotθ).(1/a secθ.tanθ)

(d2y/dx2) = – (b/a2).(cotθ).(1/tan2θ)

d2y/dx= -(b/a2).(1/tan3θ)

d2y/dx= -(b/a2tan3θ).(b3/b3)

d2y/dx= -(b4/a2y3)

Hence Proved

### Question 9. If x = a(cost + tsint) and y = a(sint – tcost), prove that d2y/dx2 = sec3t/ at 0 < t < π/2.

Solution:

We have,

x = a(cost + tsint)and y=a(sint – tcost)

On differentiating both sides w.r.t t,

(dx/dt) = a(-sint + sint + tcost), (dy/dt) = a(cost – cost + tsint)

(dx/dt) = atcost, (dy/dt) = atsint

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = atsint × [1/atcost]

(dy/dx) = tant

Again differentiating both sides w.r.t x,

(d2y/dx2) = sec2x.(dt/dx)

(d2y/dx2) = sec2x.[1/atcost]

(d2y/dx2) = sec3x/at

Hence Proved

### Question 10. If y = excosx, prove that d2y/dx2 = 2excos(x + π/2).

Solution:

We have,

y = excosx

On differentiating both sides w.r.t x,

(dy/dx) = excosx – exsinx

Again differentiating both sides w.r.t x,

d2y/dx= (excosx – exsinx) – (exsinx + excosx)

d2y/dx= excosx – excosx – exsinx – exsinx

d2y/dx= -2exsinx

d2y/dx= 2excos(x + π/2)

### Question 11. If x = a cosθ, y = b sinθ, show that (d2y/dx2) = -b4/a2y3

Solution:

We have,

x = a cosθ and y = b sinθ

On differentiating both sides w.r.t θ,

(dx/dθ) = -a sinθ, (dy/dθ) = b cosθ

(dy/dx) = (dy/dθ)×(dθ/dx)

(dy/dx) = (b cosθ)/(-a sinθ)

(dy/dx) = -(b/a).cotθ

Again differentiating both sides w.r.t x,

(d2y/dx2) = -(b/a).(-cosec2θ).(dθ/dx)

(d2y/dx2) = (b/a).(cosec2θ).(1/-a sinθ)

(d2y/dx2) = (b/a).(cosec2θ).(1/-a sinθ)

d2y/dx= -(b/a2).(1/sin3θ)

d2y/dx2=-(b/a2sin3θ).(b3/b3)

d2y/dx2 = -(b4/a2y3)   (since y = a sinθ)

Hence Proved

### Question 12. If x = a(1 – cos3θ), y = a sin3θ, show that (d2y/dx2) = 32/27a, at θ = π/6.

Solution:

We have,

x = a(1 – cos3θ) and y = a sin3θ

On differentiating both sides w.r.t θ,

(dx/dθ) = a(3cos2θ.sinθ), (dy/dθ) = a 3sin2θcosθ

(dx/dθ) = 3acos2θ.sinθ, (dy/dθ) = 3asin2θ.cosθ

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = (3asin2θ.cosθ) × (3acos2θ.sinθ)

(dy/dx) = tan2θ/tanθ

(dy/dx) = tanθ

Again differentiating both sides w.r.t x,

(d2y/dx2) = sec2θ(dθ/dx)

(d2y/dx2) = sec2θ.[1/3acos2θ.sinθ]

(d2y/dx2) = sec4θ/3asinθ

At θ = π/6

d2y/dx= sec4(π/6)/3asin(π/6)

d2y/dx= 32/27a

Hence Proved

### Question 13. If x = a(θ + sinθ), y = a(1 + cosθ), prove that (d2y/dx2) = -(a/y2).

Solution:

We have,

x = a(θ + sinθ) and y = a(1 + cosθ)

On differentiating both sides w.r.t θ,

(dx/dθ) = a(1 + cosθ), (dy/dθ) = -asinθ

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = [-asinθ] × [a(1 + cosθ)]

(dy/dx) = -sinθ/(1 + cosθ)

Again differentiating both sides w.r.t x,

(d2y/dx2) = -(1 + cosθ)/a(1 + cosθ)3

(d2y/dx2) = -1/a(1 + cosθ)2

(d2y/dx2) = -[1/a(1 + cosθ)2](a/a)

d2y/dx2 = -a/y2

Hence Proved

### Question 14. If x = a(θ – sinθ), y = a(1 + cosθ), find (d2y/dx2).

Solution:

We have,

x = a(θ – sinθ) and y = a(1 + cosθ)

On differentiating both sides w.r.t θ,

(dx/dθ) = a(1 – cosθ), (dy/dθ) = -asinθ

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = [-asinθ] × [a(1 – cosθ)]

(dy/dx) = -sinθ/(1 – cosθ)

Again differentiating both sides w.r.t x,

(d2y/dx2) = 1/a(1 – cosθ)2

d2y/dx= (1/4a)[cosec4(θ/2)]

### Question 15. If x = a(1 – cosθ), y = a(θ + sinθ), prove that (d2y/dx2) = -1/a at θ = π/2.

Solution:

We have,

x = a(1 – cosθ) and y = a(θ + sinθ)

On differentiating both sides w.r.t θ,

(dx/dθ) = a(sinθ), (dy/dθ) = a(1 + cosθ)

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = [a(1 + cosθ)] × [asinθ)]

(dy/dx) = (1 + cosθ)/sinθ

Again differentiating both sides w.r.t x,

d2y/dx= (-sin2θ – cosθ – cos2θ)/asin3θ

d2y/dx= -(sin2θ + cosθ + cos2θ)/asin3θ

At θ = π/2,

d2y/dx= -(1 + 0)/a

d2y/dx= -(1/a)

Hence Proved

### Question 16. If x = a(1 + cosθ), y = a(θ + sinθ), prove that (d2y/dx2) = -1/a at θ = π/2.

Solution:

We have,

x = a(1 + cosθ) and y = a(θ + sinθ)

On differentiating both sides w.r.t θ,

(dx/dθ) = a(-sinθ), (dy/dθ) = a(1 + cosθ)

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = [a(1 + cosθ)] × [-asinθ)]

(dy/dx) = -(1 + cosθ)/sinθ

Again differentiating both sides w.r.t x,

d2y/dx= (-sin2θ – cosθ – cos2θ)/asin3θ

d2y/dx= -(sin2θ + cosθ + cos2θ)/asin3θ

At θ = π/2,

d2y/dx= -(1 + 0)/a

d2y/dx= -(1/a)

Hence Proved

### Question 17. If x = cosθ, y = sin3θ, prove that y(d2y/dx2) + (dy/dx)2 = 3sin2θ(5cos2θ – 1).

Solution:

We have,

x = cosθ and y = sin3θ

On differentiating both sides w.r.t θ,

(dx/dθ) = -sinθ, (dy/dθ) = 3sin2θ.cosθ

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = [3sin2θ.cosθ] × [-sinθ]

(dy/dx) = -3sinθ.cosθ

Again differentiating both sides w.r.t x,

d2y/dx= -3[sinθ(-sinθ) + cosθ.cosθ](dθ/dx)

d2y/dx= (3sin2θ – 3cos2θ)/-sinθ

d2y/dx= -(3sin2θ – 3cos2θ)/sinθ

L.H.S,

y(d2y/dx2) + (dy/dx)= -sin3θ[(3sin2θ – 3cos2θ)/sinθ] + (-3sinθ.cosθ)2

= 3sin2θ.cos2θ – 3sin4θ + 9sin2θ.cos2θ

= 12sin2θ.cos2θ – 3sin4θ

= 3sin2θ(4cos2θ – sin2θ)

= 3sin2θ(4cos2θ – sin2θ – cos2θ + cos2θ)

= 3sin2θ[5cos2θ – (sin2θ + cos2θ)]

= 3sin2θ(5cos2θ – 1)

= R.H.S

L.H.S = R.H.S

Hence Proved

### Question 18. If y = sin(sinx), prove that (d2y/dx2) + tanx.(dy/dx) + ycos2x = 0

Solution:

We have,

y = sin(sinx)

On differentiating both sides w.r.t x,

(dy/dx) = cos(sinx).cosx

Again differentiating both sides w.r.t x,

d2y/dx= -sin(sinx).cosx.cosx – cos(sinx).sinx

d2y/dx= -sin(sinx).cos2x – cos(sinx).sinx

d2y/dx= -sin(sinx).cos2x – cos(sinx).cosx.tanx

d2y/dx= -ycos2x – (dy/dx)tanx

d2y/dx+ ycos2x + (dy/dx)tanx = 0

Hence Proved

### Question 19. If x = sin t, y = sin pt, prove that (1 – x2)(d2y/dx2) – x.(dy/dx) + p2y = 0

Solution:

We have,

x = sin t, and y = sin pt

On differentiating both sides w.r.t t,

(dx/dt) = cos t, (dy/dt) = pcos pt

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = pcos pt×[1/cos t]

(dy/dx) = pcos pt/cos t

Again differentiating both sides w.r.t x,

d2y/dx= (-p2sin pt.cos t + pcos pt.sin t)/cos3t

d2y/dx= -(p2sin pt)/cos2t + (pcos pt.sin t)/cos3t

cos2t(d2y/dx2) = -p2y + x(dy/dx)

(1 – sin2x)(d2y/dx2) + p2y – x(dy/dx) = 0

(1 – y2)(d2y/dx2) + p2y – x(dy/dx) = 0

### Question 20. If y = (sin-1x)2, prove that (1 – x2)(d2y/dx2) – x.(dy/dx) + p2y = 0.

Solution:

We have,

y = (sin-1x)2,

On differentiating both sides w.r.t t,

Again differentiating both sides w.r.t x,

d2y/dx= [x/(1 – x2)](dy/dx) + 2/(1 – x2)

(1 – x2)d2y/dx= x(dy/dx) + 2

(1 – x2)d2y/dx– x(dy/dx) – 2 = 0

Hence Proved

### Question 21. If y = , prove that (1 + x2)y2 + (2x – 1)y1 = 0.

Solution:

We have,

y =

On differentiating both sides w.r.t t,

y × [1/(1 + x2)]

Again differentiating both sides w.r.t x,

y

(1 + x2)y/(1 + x2) – 2x/(1 + x2)

(1 + x2)y= (dy/dx) – 2x(dy/dx)

(1 + x2)y– (dy/dx) + 2x(dy/dx) = 0

(1 + x2)y+ (2x – 1)(dy/dx) = 0

Hence Proved

### Question 22. If y = 3cos(logx) + 4sin(logx), prove that x2y2 + xy1 + y = 0.

Solution:

We have,

y = 3cos(logx) + 4sin(logx)

On differentiating both sides w.r.t x,

y1 = -3sin(logx) × (1/x) + 4cos(logx) × (1/x)

xy= -3sin(logx) + 4cos(logx)

Again differentiating both sides w.r.t x,

xy+ y= -3cos(logx)×(1/x) – 4sin(logx) × (1/x)

x2y+ xy= -[3cos(logx) + 4sin(logx)]

x2y+ xy= -y

x2y+ xy+ y = 0

Hence Proved

### Question 23. If y = e2x(ax + b), show that y2 – 4y1 + 4y = 0.

Solution:

We have,

y = e2x(ax + b)

On differentiating both sides w.r.t θ,

y= 2e2x(ax + b) + a.e2x

Again differentiating both sides w.r.t x,

y= 4e2x(ax + b) + 2ae2x + 2a.e2x

y= 4e2x(ax + b) + 4a.e2x

Lets take L.H,S,

= y– 4y+ 4y

= 4e2x(ax + b) + 4a.e2x – 4[2e2x(ax + b) + a.e2x] + 4[e2x(ax + b)]

= 8e2x(ax + b) – 8e2x(ax + b) + 4a.e2x – 4a.e2x

= 0

= R.H.S

L.H.S = R.H.S

Hence Proved

### Question 24. If x = sin(logy/a), show that (1 – x2)y2 – xy1 – a2y = 0.

Solution:

We have,

x = sin(logy/a)

(logy/a) = sin-1x

logy = asin-1x

On differentiating both sides w.r.t x,

(1/y)y= a/√(1 – x2)

y= ay/√(1 – x2)

Again differentiating both sides w.r.t x,

y

(1 – x2)y= a√(1 – x2) × y+ axy/√(1 – x2)

(1 – x2)y= a√(1 – x2) × [ay/√(1 – x2)] + x[ay/√(1 – x2)]

(1 – x2)y= a2p + xy

(1 – x2)y– a2p – xy= 0

Hence Proved

### Question 25. If logy = tan-1x, show that (1 + x2)y2 + (2x – 1)y1 = 0.

Solution:

We have,

logy = tan-1x

On differentiating both sides w.r.t θ,

(1/y)y= 1/(1 + x2)

(1 + x2)y= y

Again differentiating both sides w.r.t x,

2xy+ (1 + x2)y= y1

(1 + x2)y+ (2x – 1)y= 0

Hence Proved

### Question 26. If y = tan-1x, show that (1 + x2)(d2y/dx2) + 2x(dy/dx) = 0.

Solution:

We have,

y = tan-1x

On differentiating both sides w.r.t x,

(dy/dx) = 1/(1 + x2)

Again differentiating both sides w.r.t x,

d2y/dx= [-1/(1 + x2)2] × (2x)

d2y/dx= [-2x/(1 + x2)2]

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

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

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

Hence Proved

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 + (√x+ 1)}]2
On differentiating both sides w.r.t x,

dy/dx = 2[log{x + (√x+ 1)}]/(√x+ 1)
Again differentiating both sides w.r.t x,

(x+ 1)(d2y/dx2) = 2 – x(dy/dx)
(x+ 1)(d2y/dx2) + x(dy/dx) = 2
Hence Proved
Question 28. If y = (tan-1x)2, then prove that (1 + x2)2y+ 2x(1 + x2)y= 2
Solution:
We have,
y = (tan-1x)2
On differentiating both sides w.r.t x,
y= 2(tan-1x)[1/(1 + x2)]
(1 + x2)y= 2(tan-1x)
Again differentiating both sides w.r.t x,
(1 + x2)y+ 2xy= 2/(1 + x2)
(1 + x2)2y+ 2x(1 + x2)y= 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/dx= -(2cosec x) × (-cosec x.cot x)
d2y/dx= 2cosec2x.cot x
d2y/dx= 2(cot x)(cosec2x)
d2y/dx= -2y(dy/dx)
d2y/dx+ 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,

(dy/dx) = (2/x)
Again differentiating both sides w.r.t x,
d2y/dx= -(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/dx– 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/dx– 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,
…(i)
y = cos-1x
x = cosy
On putting the value of x in equation (i), we get

d2y/dx= -cosy/sin3y
d2y/dx= -cot y cosec2y
Question 34. If y = , prove that (1 – x2)(d2y/dx2) – x(dy/dx) – a2y = 0.
Solution:
We have,
y =
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)= a2y2/(1 – x2)
(1 – x2)(dy/dx)= a2y2
Again differentiating both sides w.r.t x,
2(1 – x2)(dy/dx)(d2y/dx2) – 2x(dy/dx)= 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/dx= 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/dxat 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,

At t = π/2

d2y/dx= (1 + 2)/-2
d2y/dx= -(3/2)
Question 37. If x = 4z+ 5, y = 6z+ 7z + 3, find d2y/dx2.
Solution:
We have,
x = 4z+ 5, and y = 6z+ 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,

(d2y/dx2) = -7/64z3
Hence Proved
Question 38. If y = log(1 + cosx), prove that d3y/dx+ (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,

d2y/dx= (-cosx – cos2x – sin2x)/(1 + cosx)2
d2y/dx= -(1 + cosx)/(1 + cosx)2
d2y/dx= -1/(1 + cosx)
Again differentiating both sides w.r.t x,
d3y/dx= -sinx/(1 + cosx)2
d3y/dx+ [-1/(1 + cosx)][-sinx/(1 + cosx)] = 0
d3y/dx+ (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/dx– 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/dx= 6(2e2x + 3e3x)
d2y/dx= 12e2x + 18e3x
d2y/dx= 5(6e2x + 6e3x) – 6(3e2x + 2e3x)
d2y/dx= 5(dy/dx) – 6y
d2y/dx– 5(dy/dx) + 6y = 0
Hence Proved
Question 41. If y = (cot-1x)2, prove that y2(x+ 1)+ 2x(x+ 1)y= 2.
Solution:
We have,
y = (cot-1x)2
On differentiating both sides w.r.t x,
y= 2(cot-1x) × [-1/(1 + x2)]
(1 + x2)y= -2cot-1x
Again differentiating both sides w.r.t x,
(1 + x2)y+ 2xy= 2/(1 + x2)
(1 + x2)2y+ 2x(1 + x2)y= 2
Hence Proved
Question 42. If y = cosec-1x, then show that x(x– 1)d2y/dx– (2x– 1)(dy/dx) = 0.
Solution:
We have,
y = cosec-1x
On differentiating both sides w.r.t x,
(dy/dx) = -1/x√(x– 1)
On squaring both sides,
(dy/dx)= 1/x2(x– 1)
x2(x– 1)(dy/x)= 1
(x– x2)(dy/dx)= 1
2(dy/dx)(d2y/dx2)(x– x2) + (dy/dx)2(4x– 2x) = 0
2x2(x– 1)(dy/dx)(d2y/dx2) + 2x(2x– 1)(dy/dx)= 0
x(x– 1)(d2y/dx2) + (2x– 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,

(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

(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

(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,

(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) = -(x+ 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,

d2y/dx= (-y– x2)/y3
d2y/dx= -(x+ y2)/y3
Hence Proved
Question 49. Find A and B so that y = Asin3x + Bcos3x, satisfies the equation d2y/dx+ 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/dx= -9Asin3x – 9Bcos3x
d2y/dx+ 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/dx+ 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 n= p+ 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) + (k+ 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] – (n+ 1)y
x2(d2y/dt2) + (1 – 2n) x (dy/dt) + (1 + n2)y = 0
Hence Proved
Question 52. y = , prove that (x2+1)d2y/d2x + 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,
d2y/dx
d2y/dx
d2y/dx
(x2+1)d2y/d2x =
Now put all these values in this equation
(x2+1)d2y/d2x + xdy/dx – ny

Hence Proved

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