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Class 12 RD Sharma Solutions – Chapter 12 Higher Order Derivatives – Exercise 12.1 | Set 1
  • Last Updated : 28 Mar, 2021

Find the second order derivative of following function

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) × (1/x)

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

Again differentiating both sides w.r.t x,

d2y/dx2 = d/dx[cos(logx)/x]

=\frac{x\frac{d}{dx}cos(logx)-cos(logx)\frac{d}{dx}x}{x^2}

\frac{-x[\frac{sin(logx)}{x}]-cos(logx).1}{x^2}

= -[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/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,

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

d2y/dx2 = (-1)(1 + x2)-2.2x

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

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) × (1/x)

(dy/dx) = 1/xlogx

Again differentiating both sides w.r.t x,

\frac{d^2y}{dx^2}=\frac{d}{dx}(xlogx)^{-1}

d2y/dx2 = (-1)(xlogx)-2.[(d/dx)xlogx]

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

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

\frac{d^2y}{dx^2}=\frac{-(1+logx)}{(xlogx)^2}

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) × (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,

\frac{dy}{dx}=\frac{x\frac{d}{dx}logx-logx\frac{dx}{dx}}{x^2}

(dy/dx) = (1 – logx)/x2

Again differentiating both sides w.r.t x,

\frac{d^2y}{dx^2}=\frac{x^2\frac{d}{dx}(1-logx)-(1-logx)\frac{dx}{dx}}{x^4}

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 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/dx2 = -(b/a2).(1/tan3θ)

d2y/dx2 = -(b/a2tan3θ).(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 < 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/dx2 = (excosx – exsinx) – (exsinx + excosx)

d2y/dx2 = excosx – excosx – exsinx – exsinx

d2y/dx2 = -2exsinx

d2y/dx2 = 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/dx2 = -(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/dx2 = sec4(π/6)/3asin(π/6)

\frac{d^2y}{dx^2}=\frac{(\frac{2}{\sqrt{3}})^4}{3a\frac{1}{2}}

d2y/dx2 = 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,

\frac{d^2y}{dx^2}=-[\frac{(1+cosθ)\frac{d}{dx}sinθ+sinθ\frac{d}{dx}(1+cosθ)}{(1+cosθ)^2}]\frac{dθ}{dx}

\frac{d^2y}{dx^2}=-[\frac{(1+cosθ)cosθ-sinθsinθ)}{(1+cosθ)^2}]\frac{1}{a(1+cosθ)}

\frac{d^2y}{dx^2}=[\frac{-cosθ-cos^2θ-sin^2θ}{(1+cosθ)^2}]\frac{1}{a(1+cosθ)}

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

(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,

\frac{d^2y}{dx^2}=-[\frac{(1-cosθ)\frac{d}{dx}sinθ+sinθ\frac{d}{dx}(1-cosθ)}{(1-cosθ)^2}]\frac{dθ}{dx}

\frac{d^2y}{dx^2}=-[\frac{(1-cosθ)cosθ-sinθsinθ)}{(1-cosθ)^2}]\frac{1}{a(1-cosθ)}

\frac{d^2y}{dx^2}=[\frac{-cosθ+cos^2θ+sin^2θ}{(1-cosθ)^2}]\frac{1}{a(1-cosθ)}

\frac{d^2y}{dx^2}=\frac{1-cosθ}{a(1-cosθ)^3}

(d2y/dx2) = 1/a(1 – cosθ)2

\frac{d^2y}{dx^2}=\frac{1}{a(2sin^2\frac{θ}{2})^2}

d2y/dx2 = (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,

\frac{d^2y}{dx^2}=[\frac{sinθ\frac{d}{dx}(1+cosθ)-(1+cosθ)\frac{d}{dx}sinθ}{(sinθ)^2}]\frac{dθ}{dx}

\frac{d^2y}{dx^2}=[\frac{sinθ.sinθ-(1+cosθ)cosθ}{(sinθ)^2}]\frac{1}{asinθ}

d2y/dx2 = (-sin2θ – cosθ – cos2θ)/asin3θ

d2y/dx2 = -(sin2θ + cosθ + cos2θ)/asin3θ

At θ = π/2,

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

d2y/dx2 = -(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,

\frac{d^2y}{dx^2}=-[\frac{sinθ\frac{d}{dx}(1+cosθ)-(1+cosθ)\frac{d}{dx}sinθ}{(sinθ)^2}]\frac{dθ}{dx}

\frac{d^2y}{dx^2}=-[\frac{sinθ.sinθ-(1+cosθ)cosθ}{(sinθ)^2}]\frac{1}{-asinθ}

d2y/dx2 = (-sin2θ – cosθ – cos2θ)/asin3θ

d2y/dx2 = -(sin2θ + cosθ + cos2θ)/asin3θ

At θ = π/2,

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

d2y/dx2 = -(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/dx2 = -3[sinθ(-sinθ) + cosθ.cosθ](dθ/dx)

d2y/dx2 = (3sin2θ – 3cos2θ)/-sinθ

d2y/dx2 = -(3sin2θ – 3cos2θ)/sinθ

L.H.S,

y(d2y/dx2) + (dy/dx)2 = -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/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) × (dt/dx)

(dy/dx) = pcos pt×[1/cos t]

(dy/dx) = pcos pt/cos t

Again differentiating both sides w.r.t x,

\frac{d^2y}{dx^2}=\frac{-p^2sin pt.cos t+pcos pt.sint}{cos^2t}.\frac{dt}{dx}

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

d2y/dx2 = -(p2sin pt)/cos2t + (pcos pt.sin t)/cos3t

\frac{d^2y}{dx^2}=-\frac{p^2y}{cos^2t}+\frac{x\frac{dy}{dx}}{cos^2y}

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,

\frac{dy}{dx}=2sin^{-1}x.\frac{1}{\sqrt{1-x^2}}

Again differentiating both sides w.r.t x,

\frac{d^2y}{dx^2}=\frac{2xsin^{-1}x}{(1-x^2)^\frac{3}{2}}+\frac{2}{1-x^2}

\frac{d^2y}{dx^2}=\frac{2xsin^{-1}x}{(1-x^2)\sqrt{1-x^2}}+\frac{2}{1-x^2}

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 = e^{tan^{-1}x}, prove that (1 + x2)y2 + (2x – 1)y1 = 0.

Solution:

We have,

y = e^{tan^{-1}x}

On differentiating both sides w.r.t t,

y1 e^{tan^{-1}x} × [1/(1 + x2)]

Again differentiating both sides w.r.t x,

y2 e^{tan^{-1}x}\frac{1}{(1+x^2)^2}+e^{tan^{-1}x}\frac{-2x}{(1+x^2)^2}

(1 + x2)y2 e^{tan^{-1}x}/(1 + x2) – 2xe^{tan^{-1}x}/(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) × (1/x) + 4cos(logx) × (1/x)

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

Again differentiating both sides w.r.t x,

xy2 + y1 = -3cos(logx)×(1/x) – 4sin(logx) × (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 θ,

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/√(1 – x2)

y1 = ay/√(1 – x2)

Again differentiating both sides w.r.t x,

y=a[\frac{\sqrt{1-x^2}\frac{dy}{dx}-\frac{2xy}{2\sqrt{1-x^2}}}{1-x^2}]

(1 – x2)y2 = a√(1 – x2) × y1 + axy/√(1 – x2)

(1 – x2)y2 = a√(1 – x2) × [ay/√(1 – x2)] + x[ay/√(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 θ,

(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] × (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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