# Class 12 RD Sharma Solutions – Chapter 12 Higher Order Derivatives – Exercise 12.1 | Set 1

### Question 1(i).  x3 + tanx

Solution:

Let us considered

f(x) = x3 + tanx

On differentiating both sides w.r.t x,

f'(x) = 3x2 + 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) Ã— (1/x)

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

Again differentiating both sides w.r.t x,

d2y/dx2 = 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) Ã— (cosx)

(dy/dx) = cotx

Again differentiating both sides w.r.t x,

d2y/dx2 = -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/dx2 = exsin5x + 5excos5x + 5(excos5x – 5exsin5x)

d2y/dx2 = -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/dx2 = 6(6e6xcos3x – 3e6xsin3x) – 3(6e6xsin3x + 3e6xcos3x)

d2y/dx2 = 36e6xcos3x – 18e6xsin3x – 18e6xsin3x – 9e6xcos3x

d2y/dx2 = 27e6xcos3x – 36e6xsin3x

d2y/dx2 = 9e6x(3cos3x – 4sin3x)

### Question 1(vi). x3logx

Solution:

Let us considered

y = x3logx

On differentiating both sides w.r.t x,

(dy/dx) = logx.3x2 + x3(1/x)

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

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

Again differentiating both sides w.r.t x,

d2y/dx2 = (1 + 3logx).2x + x2(3/x)

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

d2y/dx2 = 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/dx2 = (-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/dx2 = -sinx – (sinx + xcosx)

d2y/dx2 = -2sinx – xcosx

d2y/dx2 = -(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) Ã— (1/x)

(dy/dx) = 1/xlogx

Again differentiating both sides w.r.t x,

d2y/dx2 = (-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/dx2 = -(-e-x.cosx – e-x.sinx) – (-e-x.sinx + e-x.cosx)

d2y/dx2 = e-x.cosx – e-x.cosx + e-x.sinx + e-x.sinx

d2y/dx2 = 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/dx2 = 0 + (2secx)(secx.tanx)

d2y/dx2 = 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.3x2 + x3(1/x)

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

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

Again differentiating both sides w.r.t x,

d2y/dx2 = (1 + 3logx).2x + x2(3/x)

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

d2y/dx2 = 5x + 6xlogx

Again differentiating both sides w.r.t x,

d3y/dx3 = 5 + 6[logx + (x/x)]

d3y/dx3 = 11 + 6logx

Again differentiating both sides w.r.t x,

d4y/dx4 = (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) Ã— (cosx)

(dy/dx) = cotx

Again differentiating both sides w.r.t x,

d2y/dx2 = -cosec2x

Again differentiating both sides w.r.t x,

d3y/dx3 = -2cosecx.(-cosesx.cotx)

d3y/dx3 = 2cosec2x.cotx

d3y/dx3 = 2cosec2x.(cosx/sinx)

d3y/dx3 = 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/dx2 = -2sinx – 3cosx

d2y/dx2 = -(2sinx + 3cosx)

d2y/dx2 = -y

d2y/dx2 + 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/dx2 = [-x – 2x(1 – logx)]/x4

d2y/dx2 = (2xlogx – 3x)/x4

d2y/dx2 = (2logx – 3)/x3

d2y/dx2 + y = 0

Hence Proved

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

Solution:

We have,

x = a secÎ¸ and y = b tanÎ¸

On differentiating both sides w.r.t Î¸,

(dx/dÎ¸) = a secÎ¸.tanÎ¸, (dy/dÎ¸) = b sec2Î¸

(dy/dx) = (dy/dÎ¸) Ã— (dÎ¸/dx)

(dy/dx) = (b sec2Î¸)/(a secÎ¸.tanÎ¸)

(dy/dx) = (b/a).cosecÎ¸

Again differentiating both sides w.r.t x,

(d2y/dx2) = (b/a).(-cosecÎ¸.cotÎ¸).(dÎ¸/dx)

(d2y/dx2) = -(b/a).(cosecÎ¸.cotÎ¸).(1/a secÎ¸.tanÎ¸)

(d2y/dx2) = – (b/a2).(cotÎ¸).(1/tan2Î¸)

d2y/dx2 = -(b/a2).(1/tan3Î¸)

d2y/dx2 = -(b/a2tan3Î¸).(b3/b3)

d2y/dx2 = -(b4/a2y3)

Hence Proved

### Question 9. If x = a(cost + tsint) and y = a(sint – tcost), prove that d2y/dx2 = sec3t/ at 0 < t < Ï€/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) Ã— (dt/dx)

(dy/dx) = atsint Ã— [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 + Ï€/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/dx2 = (excosx – exsinx) – (exsinx + excosx)

d2y/dx2 = excosx – excosx – exsinx – exsinx

d2y/dx2 = -2exsinx

d2y/dx2 = 2excos(x + Ï€/2)

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

Solution:

We have,

x = a cosÎ¸ and y = b sinÎ¸

On differentiating both sides w.r.t Î¸,

(dx/dÎ¸) = -a sinÎ¸, (dy/dÎ¸) = b cosÎ¸

(dy/dx) = (dy/dÎ¸)Ã—(dÎ¸/dx)

(dy/dx) = (b cosÎ¸)/(-a sinÎ¸)

(dy/dx) = -(b/a).cotÎ¸

Again differentiating both sides w.r.t x,

(d2y/dx2) = -(b/a).(-cosec2Î¸).(dÎ¸/dx)

(d2y/dx2) = (b/a).(cosec2Î¸).(1/-a sinÎ¸)

(d2y/dx2) = (b/a).(cosec2Î¸).(1/-a sinÎ¸)

d2y/dx2 = -(b/a2).(1/sin3Î¸)

d2y/dx2=-(b/a2sin3Î¸).(b3/b3)

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

Hence Proved

### Question 12. If x = a(1 – cos3Î¸), y = a sin3Î¸, show that (d2y/dx2) = 32/27a, at Î¸ = Ï€/6.

Solution:

We have,

x = a(1 – cos3Î¸) and y = a sin3Î¸

On differentiating both sides w.r.t Î¸,

(dx/dÎ¸) = a(3cos2Î¸.sinÎ¸), (dy/dÎ¸) = a 3sin2Î¸cosÎ¸

(dx/dÎ¸) = 3acos2Î¸.sinÎ¸, (dy/dÎ¸) = 3asin2Î¸.cosÎ¸

(dy/dx) = (dy/dÎ¸) Ã— (dÎ¸/dx)

(dy/dx) = (3asin2Î¸.cosÎ¸) Ã— (3acos2Î¸.sinÎ¸)

(dy/dx) = tan2Î¸/tanÎ¸

(dy/dx) = tanÎ¸

Again differentiating both sides w.r.t x,

(d2y/dx2) = sec2Î¸(dÎ¸/dx)

(d2y/dx2) = sec2Î¸.[1/3acos2Î¸.sinÎ¸]

(d2y/dx2) = sec4Î¸/3asinÎ¸

At Î¸ = Ï€/6

d2y/dx2 = sec4(Ï€/6)/3asin(Ï€/6)

d2y/dx2 = 32/27a

Hence Proved

### Question 13. If x = a(Î¸ + sinÎ¸), y = a(1 + cosÎ¸), prove that (d2y/dx2) = -(a/y2).

Solution:

We have,

x = a(Î¸ + sinÎ¸) and y = a(1 + cosÎ¸)

On differentiating both sides w.r.t Î¸,

(dx/dÎ¸) = a(1 + cosÎ¸), (dy/dÎ¸) = -asinÎ¸

(dy/dx) = (dy/dÎ¸) Ã— (dÎ¸/dx)

(dy/dx) = [-asinÎ¸] Ã— [a(1 + cosÎ¸)]

(dy/dx) = -sinÎ¸/(1 + cosÎ¸)

Again differentiating both sides w.r.t x,

(d2y/dx2) = -(1 + cosÎ¸)/a(1 + cosÎ¸)3

(d2y/dx2) = -1/a(1 + cosÎ¸)2

(d2y/dx2) = -[1/a(1 + cosÎ¸)2](a/a)

d2y/dx2 = -a/y2

Hence Proved

### Question 14. If x = a(Î¸ – sinÎ¸), y = a(1 + cosÎ¸), find (d2y/dx2).

Solution:

We have,

x = a(Î¸ – sinÎ¸) and y = a(1 + cosÎ¸)

On differentiating both sides w.r.t Î¸,

(dx/dÎ¸) = a(1 – cosÎ¸), (dy/dÎ¸) = -asinÎ¸

(dy/dx) = (dy/dÎ¸) Ã— (dÎ¸/dx)

(dy/dx) = [-asinÎ¸] Ã— [a(1 – cosÎ¸)]

(dy/dx) = -sinÎ¸/(1 – cosÎ¸)

Again differentiating both sides w.r.t x,

(d2y/dx2) = 1/a(1 – cosÎ¸)2

d2y/dx2 = (1/4a)[cosec4(Î¸/2)]

### Question 15. If x = a(1 – cosÎ¸), y = a(Î¸ + sinÎ¸), prove that (d2y/dx2) = -1/a at Î¸ = Ï€/2.

Solution:

We have,

x = a(1 – cosÎ¸) and y = a(Î¸ + sinÎ¸)

On differentiating both sides w.r.t Î¸,

(dx/dÎ¸) = a(sinÎ¸), (dy/dÎ¸) = a(1 + cosÎ¸)

(dy/dx) = (dy/dÎ¸) Ã— (dÎ¸/dx)

(dy/dx) = [a(1 + cosÎ¸)] Ã— [asinÎ¸)]

(dy/dx) = (1 + cosÎ¸)/sinÎ¸

Again differentiating both sides w.r.t x,

d2y/dx2 = (-sin2Î¸ – cosÎ¸ – cos2Î¸)/asin3Î¸

d2y/dx2 = -(sin2Î¸ + cosÎ¸ + cos2Î¸)/asin3Î¸

At Î¸ = Ï€/2,

d2y/dx2 = -(1 + 0)/a

d2y/dx2 = -(1/a)

Hence Proved

### Question 16. If x = a(1 + cosÎ¸), y = a(Î¸ + sinÎ¸), prove that (d2y/dx2) = -1/a at Î¸ = Ï€/2.

Solution:

We have,

x = a(1 + cosÎ¸) and y = a(Î¸ + sinÎ¸)

On differentiating both sides w.r.t Î¸,

(dx/dÎ¸) = a(-sinÎ¸), (dy/dÎ¸) = a(1 + cosÎ¸)

(dy/dx) = (dy/dÎ¸) Ã— (dÎ¸/dx)

(dy/dx) = [a(1 + cosÎ¸)] Ã— [-asinÎ¸)]

(dy/dx) = -(1 + cosÎ¸)/sinÎ¸

Again differentiating both sides w.r.t x,

d2y/dx2 = (-sin2Î¸ – cosÎ¸ – cos2Î¸)/asin3Î¸

d2y/dx2 = -(sin2Î¸ + cosÎ¸ + cos2Î¸)/asin3Î¸

At Î¸ = Ï€/2,

d2y/dx2 = -(1 + 0)/a

d2y/dx2 = -(1/a)

Hence Proved

### Question 17. If x = cosÎ¸, y = sin3Î¸, prove that y(d2y/dx2) + (dy/dx)2 = 3sin2Î¸(5cos2Î¸ – 1).

Solution:

We have,

x = cosÎ¸ and y = sin3Î¸

On differentiating both sides w.r.t Î¸,

(dx/dÎ¸) = -sinÎ¸, (dy/dÎ¸) = 3sin2Î¸.cosÎ¸

(dy/dx) = (dy/dÎ¸) Ã— (dÎ¸/dx)

(dy/dx) = [3sin2Î¸.cosÎ¸] Ã— [-sinÎ¸]

(dy/dx) = -3sinÎ¸.cosÎ¸

Again differentiating both sides w.r.t x,

d2y/dx2 = -3[sinÎ¸(-sinÎ¸) + cosÎ¸.cosÎ¸](dÎ¸/dx)

d2y/dx2 = (3sin2Î¸ – 3cos2Î¸)/-sinÎ¸

d2y/dx2 = -(3sin2Î¸ – 3cos2Î¸)/sinÎ¸

L.H.S,

y(d2y/dx2) + (dy/dx)2 = -sin3Î¸[(3sin2Î¸ – 3cos2Î¸)/sinÎ¸] + (-3sinÎ¸.cosÎ¸)2

= 3sin2Î¸.cos2Î¸ – 3sin4Î¸ + 9sin2Î¸.cos2Î¸

= 12sin2Î¸.cos2Î¸ – 3sin4Î¸

= 3sin2Î¸(4cos2Î¸ – sin2Î¸)

= 3sin2Î¸(4cos2Î¸ – sin2Î¸ – cos2Î¸ + cos2Î¸)

= 3sin2Î¸[5cos2Î¸ – (sin2Î¸ + cos2Î¸)]

= 3sin2Î¸(5cos2Î¸ – 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/dx2 = -sin(sinx).cosx.cosx – cos(sinx).sinx

d2y/dx2 = -sin(sinx).cos2x – cos(sinx).sinx

d2y/dx2 = -sin(sinx).cos2x – cos(sinx).cosx.tanx

d2y/dx2 = -ycos2x – (dy/dx)tanx

d2y/dx2 + 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) Ã— (dt/dx)

(dy/dx) = pcos ptÃ—[1/cos t]

(dy/dx) = pcos pt/cos t

Again differentiating both sides w.r.t x,

d2y/dx2 = (-p2sin pt.cos t + pcos pt.sin t)/cos3t

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

(1 – x2)d2y/dx2 = x(dy/dx) + 2

(1 – x2)d2y/dx2 – 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,

y1  Ã— [1/(1 + x2)]

Again differentiating both sides w.r.t x,

y2

(1 + x2)y2 /(1 + x2) – 2x/(1 + x2)

(1 + x2)y2 = (dy/dx) – 2x(dy/dx)

(1 + x2)y2 – (dy/dx) + 2x(dy/dx) = 0

(1 + x2)y2 + (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) Ã— (1/x) + 4cos(logx) Ã— (1/x)

xy1 = -3sin(logx) + 4cos(logx)

Again differentiating both sides w.r.t x,

xy2 + y1 = -3cos(logx)Ã—(1/x) – 4sin(logx) Ã— (1/x)

x2y2 + xy1 = -[3cos(logx) + 4sin(logx)]

x2y2 + xy1 = -y

x2y2 + xy1 + 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 Î¸,

y1 = 2e2x(ax + b) + a.e2x

Again differentiating both sides w.r.t x,

y2 = 4e2x(ax + b) + 2ae2x + 2a.e2x

y2 = 4e2x(ax + b) + 4a.e2x

Lets take L.H,S,

= y2 – 4y1 + 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)y1 = a/âˆš(1 – x2)

y1 = ay/âˆš(1 – x2)

Again differentiating both sides w.r.t x,

y

(1 – x2)y2 = aâˆš(1 – x2) Ã— y1 + axy/âˆš(1 – x2)

(1 – x2)y2 = aâˆš(1 – x2) Ã— [ay/âˆš(1 – x2)] + x[ay/âˆš(1 – x2)]

(1 – x2)y2 = a2p + xy

(1 – x2)y2 – a2p – xy1 = 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 Î¸,

(1/y)y1 = 1/(1 + x2)

(1 + x2)y1 = y

Again differentiating both sides w.r.t x,

2xy1 + (1 + x2)y2 = y1

(1 + x2)y2 + (2x – 1)y1 = 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/dx2 = [-1/(1 + x2)2] Ã— (2x)

d2y/dx2 = [-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

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