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

### Find the second order derivative of following function

### Question 1(i). x^{3 }+ tanx

**Solution:**

Let us considered

f(x) = x

^{3 }+ tanxOn differentiating both sides w.r.t x,

f'(x) = 3x

^{2 }+ sec^{2}xAgain differentiating both sides w.r.t x,

f”(x) = 6x + 2(secx)(secx.tanx)

f”(x) = 6x + 2sec

^{2}x.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,

d

^{2}y/dx^{2 }= d/dx[cos(logx)/x]=

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

^{2}

### 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,

d

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

### Question 1(iv). e^{x}sin5x

**Solution:**

Let us considered

y = e

^{x}sin5xOn differentiating both sides w.r.t x,

(dy/dx) = e

^{x}sin5x + 5e^{x}cos5xAgain differentiating both sides w.r.t x,

d

^{2}y/dx^{2 }= e^{x}sin5x + 5e^{x}cos5x + 5(e^{x}cos5x – 5e^{x}sin5x)d

^{2}y/dx^{2 }= -24e^{x}sin5x + 10e^{x}cos5xd

^{2}y/dx^{2}= 2e^{x}(5cos5x – 12sinx)

### Question 1(v). e^{6x}cos3x

**Solution:**

Let us considered

y = e

^{6x}cos3xOn differentiating both sides w.r.t x,

(dy/dx) = 6e

^{6x}cos3x – 3e^{6x}sin3xAgain differentiating both sides w.r.t x,

d

^{2}y/dx^{2 }= 6(6e^{6x}cos3x – 3e^{6x}sin3x) – 3(6e^{6x}sin3x + 3e^{6x}cos3x)d

^{2}y/dx^{2 }= 36e^{6x}cos3x – 18e^{6x}sin3x – 18e^{6x}sin3x – 9e^{6x}cos3xd

^{2}y/dx^{2 }= 27e^{6x}cos3x – 36e^{6x}sin3xd

^{2}y/dx^{2 }= 9e^{6x}(3cos3x – 4sin3x)

### Question 1(vi). x^{3}logx

**Solution:**

Let us considered

y = x

^{3}logxOn differentiating both sides w.r.t x,

(dy/dx) = logx.3x

^{2 }+ x^{3}(1/x)(dy/dx) = logx.3x

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

^{2}(1 + 3logx)Again differentiating both sides w.r.t x,

d

^{2}y/dx^{2 }= (1 + 3logx).2x + x^{2}(3/x)d

^{2}y/dx^{2 }= 2x + 6xlogx + 3xd

^{2}y/dx^{2 }= x(5 + 6logx)

### Question 1(vii). tan^{-1}x

**Solution:**

Let us considered

y = tan

^{-1}xOn differentiating both sides w.r.t x,

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

^{2})Again differentiating both sides w.r.t x,

d

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

d

^{2}y/dx^{2 }= -sinx – (sinx + xcosx)d

^{2}y/dx^{2 }= -2sinx – xcosxd

^{2}y/dx^{2 }= -(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,

d

^{2}y/dx^{2 }= (-1)(xlogx)^{-2}.[(d/dx)xlogx]d

^{2}y/dx^{2}= (-1)(xlogx)^{-2}[logx+x.(1/x)]d

^{2}y/dx^{2}= (-1)(xlogx)^{-2}.(logx+1)

### Question 2. If y = e^{-x}.cosx, show that d^{2}y/dx^{2 }= 2e^{-x}.sinx.

**Solution:**

Let us considered

y = e

^{-x}.cosxOn differentiating both sides w.r.t x,

(dy/dx) = -e

^{-x}.cosx – e^{-x}.sinxAgain differentiating both sides w.r.t x,

d

^{2}y/dx^{2 }= -(-e^{-x}.cosx – e^{-x}.sinx) – (-e^{-x}.sinx + e^{-x}.cosx)d

^{2}y/dx^{2 }= e^{-x}.cosx – e^{-x}.cosx + e^{-x}.sinx + e^{-x}.sinxd

^{2}y/dx^{2 }= 2e^{-x.}sinxHence Proved

### Question 3. If y = x + tanx, Show that cos^{2}x(d^{2}y/dx^{2}) – 2y + 2x = 0.

**Solution:**

Let us considered

y = x + tanx

On differentiating both sides w.r.t x,

(dy/dx) = 1 + sec

^{2}xAgain differentiating both sides w.r.t x,

d

^{2}y/dx^{2 }= 0 + (2secx)(secx.tanx)d

^{2}y/dx^{2 }= 2sec^{2}x.tanxOn multiplying both sides by cos

^{2}xcos

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

^{2}x(d^{2}y/dx^{2}) = 2(y – x) [since, tanx = y – x]cos

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

### Question 4. If y = x^{3}logx, prove that (d^{4}y/dx^{4}) = (6/x).

**Solution:**

Let us considered

y = x

^{3}logxOn differentiating both sides w.r.t x,

(dy/dx) = logx.3x

^{2 }+ x^{3}(1/x)(dy/dx) = logx.3x

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

^{2}(1 + 3logx)Again differentiating both sides w.r.t x,

d

^{2}y/dx^{2 }= (1 + 3logx).2x + x^{2}(3/x)d

^{2}y/dx2 = 2x + 6xlogx + 3xd

^{2}y/dx^{2 }= 5x + 6xlogxAgain differentiating both sides w.r.t x,

d

^{3}y/dx^{3 }= 5 + 6[logx + (x/x)]d

^{3}y/dx^{3 }= 11 + 6logxAgain differentiating both sides w.r.t x,

d

^{4}y/dx^{4 }= (6/x)Hence Proved

### Question 5. If y = log(sinx), prove that (d^{3}y/dx^{3}) = 2cosx.cosec^{3}x.

**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,

d

^{2}y/dx^{2 }= -cosec^{2}xAgain differentiating both sides w.r.t x,

d

^{3}y/dx^{3 }= -2cosecx.(-cosesx.cotx)d

^{3}y/dx^{3 }= 2cosec^{2}x.cotxd

^{3}y/dx^{3 }= 2cosec^{2}x.(cosx/sinx)d

^{3}y/dx^{3 }= cosx.cosec^{3}xHence Proved

### Question 6. If y = 2sinx + 3cosx, show that (d^{2}y/dx^{2}) + 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,

d

^{2}y/dx^{2 }= -2sinx – 3cosxd

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

^{2}y/dx^{2 }= -yd

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

### Question 7. If y = (logx/x), show that (d^{2}y/dx^{2}) = (2logx – 3)/x^{3}

**Solution:**

Let us considered

y = (logx/x)

On differentiating both sides w.r.t x,

(dy/dx) = (1 – logx)/x

^{2}Again differentiating both sides w.r.t x,

d

^{2}y/dx^{2 }= [-x – 2x(1 – logx)]/x^{4}d

^{2}y/dx^{2 }= (2xlogx – 3x)/x^{4}d

^{2}y/dx^{2 }= (2logx – 3)/x^{3}d2y/dx2 + y = 0

Hence Proved

### Question 8. If x = a secθ, y = b tanθ, show that (d^{2}y/dx^{2}) = -b^{4}/a^{2}y^{3}

**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 sec

^{2}θ(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = (b sec

^{2}θ)/(a secθ.tanθ)(dy/dx) = (b/a).cosecθ

Again differentiating both sides w.r.t x,

(d

^{2}y/dx^{2}) = (b/a).(-cosecθ.cotθ).(dθ/dx)(d

^{2}y/dx^{2}) = -(b/a).(cosecθ.cotθ).(1/a secθ.tanθ)(d

^{2}y/dx^{2}) = – (b/a^{2}).(cotθ).(1/tan^{2}θ)d

^{2}y/dx^{2 }= -(b/a^{2}).(1/tan^{3}θ)d

^{2}y/dx^{2 }= -(b/a^{2}tan^{3}θ).(b^{3}/b^{3})d

^{2}y/dx^{2 }= -(b^{4}/a^{2}y^{3})Hence Proved

### Question 9. If x = a(cost + tsint) and y = a(sint – tcost), prove that d^{2}y/dx^{2 }= sec^{3}t/ 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,

(d

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

^{2}y/dx^{2}) = sec2x.[1/atcost](d

^{2}y/dx^{2}) = sec^{3}x/atHence Proved

### Question 10. If y = e^{x}cosx, prove that d^{2}y/dx^{2 }= 2e^{x}cos(x + π/2).

**Solution:**

We have,

y = e

^{x}cosxOn differentiating both sides w.r.t x,

(dy/dx) = e

^{x}cosx – e^{x}sinxAgain differentiating both sides w.r.t x,

d

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

^{2}y/dx^{2 }= e^{x}cosx – e^{x}cosx – e^{x}sinx – e^{x}sinxd

^{2}y/dx^{2 }= -2e^{x}sinxd

^{2}y/dx^{2 }= 2e^{x}cos(x + π/2)

### Question 11. If x = a cosθ, y = b sinθ, show that (d^{2}y/dx^{2}) = -b^{4}/a^{2}y^{3}

**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,

(d

^{2}y/dx^{2}) = -(b/a).(-cosec^{2}θ).(dθ/dx)(d

^{2}y/dx^{2}) = (b/a).(cosec^{2}θ).(1/-a sinθ)(d

^{2}y/dx^{2}) = (b/a).(cosec^{2}θ).(1/-a sinθ)d

^{2}y/dx^{2 }= -(b/a^{2}).(1/sin^{3}θ)d

^{2}y/dx^{2}=-(b/a^{2}sin^{3}θ).(b^{3}/b^{3})d

^{2}y/dx2 = -(b^{4}/a^{2}y^{3}) (since y = a sinθ)Hence Proved

### Question 12. If x = a(1 – cos^{3}θ), y = a sin^{3}θ, show that (d^{2}y/dx^{2}) = 32/27a, at θ = π/6.

**Solution:**

We have,

x = a(1 – cos

^{3}θ) and y = a sin^{3}θOn differentiating both sides w.r.t θ,

(dx/dθ) = a(3cos

^{2}θ.sinθ), (dy/dθ) = a 3sin^{2}θcosθ(dx/dθ) = 3acos

^{2}θ.sinθ, (dy/dθ) = 3asin^{2}θ.cosθ(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = (3asin

^{2}θ.cosθ) × (3acos^{2}θ.sinθ)(dy/dx) = tan

^{2}θ/tanθ(dy/dx) = tanθ

Again differentiating both sides w.r.t x,

(d

^{2}y/dx^{2}) = sec^{2}θ(dθ/dx)(d

^{2}y/dx^{2}) = sec^{2}θ.[1/3acos^{2}θ.sinθ](d

^{2}y/dx^{2}) = sec^{4}θ/3asinθAt θ = π/6

d

^{2}y/dx^{2 }= sec^{4}(π/6)/3asin(π/6)d

^{2}y/dx^{2 }= 32/27aHence Proved

### Question 13. If x = a(θ + sinθ), y = a(1 + cosθ), prove that (d^{2}y/dx^{2}) = -(a/y^{2}).

**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,

(d

^{2}y/dx^{2}) = -(1 + cosθ)/a(1 + cosθ)^{3}(d

^{2}y/dx^{2}) = -1/a(1 + cosθ)^{2}(d

^{2}y/dx^{2}) = -[1/a(1 + cosθ)^{2}](a/a)d2y/dx2 = -a/y

^{2}Hence Proved

### Question 14. If x = a(θ – sinθ), y = a(1 + cosθ), find (d^{2}y/dx^{2}).

**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,

(d

^{2}y/dx^{2}) = 1/a(1 – cosθ)^{2}d

^{2}y/dx^{2 }= (1/4a)[cosec^{4}(θ/2)]

### Question 15. If x = a(1 – cosθ), y = a(θ + sinθ), prove that (d^{2}y/dx^{2}) = -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,

d

^{2}y/dx^{2 }= (-sin^{2}θ – cosθ – cos^{2}θ)/asin^{3}θd

^{2}y/dx^{2 }= -(sin^{2}θ + cosθ + cos^{2}θ)/asin^{3}θAt θ = π/2,

d

^{2}y/dx^{2 }= -(1 + 0)/ad

^{2}y/dx^{2 }= -(1/a)Hence Proved

### Question 16. If x = a(1 + cosθ), y = a(θ + sinθ), prove that (d^{2}y/dx^{2}) = -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,

d

^{2}y/dx^{2 }= (-sin^{2}θ – cosθ – cos^{2}θ)/asin^{3}θd

^{2}y/dx^{2 }= -(sin^{2}θ + cosθ + cos^{2}θ)/asin^{3}θAt θ = π/2,

d

^{2}y/dx^{2 }= -(1 + 0)/ad

^{2}y/dx^{2 }= -(1/a)Hence Proved

### Question 17. If x = cosθ, y = sin^{3}θ, prove that y(d^{2}y/dx^{2}) + (dy/dx)^{2 }= 3sin^{2}θ(5cos^{2}θ – 1).

**Solution:**

We have,

x = cosθ and y = sin

^{3}θOn differentiating both sides w.r.t θ,

(dx/dθ) = -sinθ, (dy/dθ) = 3sin

^{2}θ.cosθ(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = [3sin

^{2}θ.cosθ] × [-sinθ](dy/dx) = -3sinθ.cosθ

Again differentiating both sides w.r.t x,

d

^{2}y/dx^{2 }= -3[sinθ(-sinθ) + cosθ.cosθ](dθ/dx)d

^{2}y/dx^{2 }= (3sin^{2}θ – 3cos^{2}θ)/-sinθd

^{2}y/dx^{2 }= -(3sin^{2}θ – 3cos^{2}θ)/sinθL.H.S,

y(d

^{2}y/dx^{2}) + (dy/dx)^{2 }= -sin^{3}θ[(3sin^{2}θ – 3cos^{2}θ)/sinθ] + (-3sinθ.cosθ)^{2}= 3sin

^{2}θ.cos^{2}θ – 3sin^{4}θ + 9sin^{2}θ.cos^{2}θ= 12sin

^{2}θ.cos^{2}θ – 3sin^{4}θ= 3sin

^{2}θ(4cos^{2}θ – sin^{2}θ)= 3sin

^{2}θ(4cos^{2}θ – sin^{2}θ – cos^{2}θ + cos^{2}θ)= 3sin

^{2}θ[5cos^{2}θ – (sin^{2}θ + cos^{2}θ)]= 3sin

^{2}θ(5cos^{2}θ – 1)= R.H.S

L.H.S = R.H.S

Hence Proved

### Question 18. If y = sin(sinx), prove that (d^{2}y/dx^{2}) + tanx.(dy/dx) + ycos^{2}x = 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,

d

^{2}y/dx^{2 }= -sin(sinx).cosx.cosx – cos(sinx).sinxd

^{2}y/dx^{2 }= -sin(sinx).cos^{2}x – cos(sinx).sinxd

^{2}y/dx^{2 }= -sin(sinx).cos^{2}x – cos(sinx).cosx.tanxd

^{2}y/dx^{2 }= -ycos^{2}x – (dy/dx)tanxd

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

### Question 19. If x = sin t, y = sin pt, prove that (1 – x^{2})(d^{2}y/dx^{2}) – x.(dy/dx) + p^{2}y = 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,

d

^{2}y/dx^{2 }= (-p^{2}sin pt.cos t + pcos pt.sin t)/cos^{3}td

^{2}y/dx^{2 }= -(p^{2}sin pt)/cos^{2}t + (pcos pt.sin t)/cos^{3}tcos

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

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

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

### Question 20. If y = (sin^{-1}x)^{2}, prove that (1 – x^{2})(d^{2}y/dx^{2}) – x.(dy/dx) + p^{2}y = 0.

**Solution:**

We have,

y = (sin

^{-1}x)^{2},On differentiating both sides w.r.t t,

Again differentiating both sides w.r.t x,

d

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

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

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

**Question 21. **If y = , prove that (1 + x^{2})y_{2 }+ (2x – 1)y_{1 }= 0.

**Solution:**

We have,

y =

On differentiating both sides w.r.t t,

y

_{1 }= × [1/(1 + x^{2})]Again differentiating both sides w.r.t x,

y

_{2 }=(1 + x

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

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

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

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

### Question 22. If y = 3cos(logx) + 4sin(logx), prove that x^{2}y_{2 }+ xy_{1 }+ y = 0.

**Solution:**

We have,

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

On differentiating both sides w.r.t x,

y

_{1}= -3sin(logx) × (1/x) + 4cos(logx) × (1/x)xy

_{1 }= -3sin(logx) + 4cos(logx)Again differentiating both sides w.r.t x,

xy

_{2 }+ y_{1 }= -3cos(logx)×(1/x) – 4sin(logx) × (1/x)x

^{2}y_{2 }+ xy_{1 }= -[3cos(logx) + 4sin(logx)]x

^{2}y_{2 }+ xy_{1 }= -yx

^{2}y_{2 }+ xy_{1 }+ y = 0Hence Proved

### Question 23. If y = e^{2x}(ax + b), show that y_{2 }– 4y_{1 }+ 4y = 0.

**Solution:**

We have,

y = e

^{2x}(ax + b)On differentiating both sides w.r.t θ,

y

_{1 }= 2e^{2x}(ax + b) + a.e^{2x}Again differentiating both sides w.r.t x,

y

_{2 }= 4e^{2x}(ax + b) + 2ae^{2x }+ 2a.e^{2x}y

_{2 }= 4e^{2x}(ax + b) + 4a.e^{2x}Lets take L.H,S,

= y

_{2 }– 4y_{1 }+ 4y= 4e

^{2x}(ax + b) + 4a.e^{2x }– 4[2e^{2x}(ax + b) + a.e^{2x}] + 4[e^{2x}(ax + b)]= 8e

^{2x}(ax + b) – 8e^{2x}(ax + b) + 4a.e^{2x }– 4a.e^{2x}= 0

= R.H.S

L.H.S = R.H.S

Hence Proved

### Question 24. If x = sin(logy/a), show that (1 – x^{2})y_{2 }– xy_{1 }– a^{2}y = 0.

**Solution:**

We have,

x = sin(logy/a)

(logy/a) = sin

^{-1}xlogy = asin

^{-1}xOn differentiating both sides w.r.t x,

(1/y)y

_{1 }= a/√(1 – x^{2})y

_{1 }= ay/√(1 – x^{2})Again differentiating both sides w.r.t x,

y

_{2 }(1 – x

^{2})y_{2 }= a√(1 – x^{2}) × y_{1 }+ axy/√(1 – x^{2})(1 – x

^{2})y_{2 }= a√(1 – x^{2}) × [ay/√(1 – x^{2})] + x[ay/√(1 – x^{2})](1 – x

^{2})y_{2 }= a^{2}p + xy(1 – x

^{2})y_{2 }– a^{2}p – xy_{1 }= 0Hence Proved

### Question 25. If logy = tan^{-1}x, show that (1 + x^{2})y_{2 }+ (2x – 1)y_{1 }= 0.

**Solution:**

We have,

logy = tan

^{-1}xOn differentiating both sides w.r.t θ,

(1/y)y

_{1 }= 1/(1 + x^{2})(1 + x

^{2})y_{1 }= yAgain differentiating both sides w.r.t x,

2xy

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

^{2})y_{2 }+ (2x – 1)y_{1 }= 0Hence Proved

### Question 26. If y = tan^{-1}x, show that (1 + x^{2})(d^{2}y/dx^{2}) + 2x(dy/dx) = 0.

**Solution:**

We have,

y = tan

^{-1}xOn differentiating both sides w.r.t x,

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

^{2})Again differentiating both sides w.r.t x,

d

^{2}y/dx^{2 }= [-1/(1 + x^{2})^{2}] × (2x)d

^{2}y/dx^{2 }= [-2x/(1 + x^{2})^{2}](1 + x

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

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

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

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