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

Chapter 12 of RD Sharma's Class 12 Mathematics textbook focuses on higher-order derivatives, a crucial concept in calculus. Exercise 12.1 specifically deals with finding higher-order derivatives of various functions, providing a comprehensive set of practice problems to reinforce understanding.

Find the second-order derivative of the 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). e^xsin5x

Solution:

Let us considered

y = e^xsin5x

On differentiating both sides w.r.t x,

(dy/dx) = e^xsin5x + 5e^xcos5x

Again differentiating both sides w.r.t x,

d²y/dx² = e^xsin5x + 5e^xcos5x + 5(e^xcos5x - 5e^xsin5x)

d²y/dx² = -24e^xsin5x + 10e^xcos5x

d²y/dx² = 2e^x(5cos5x - 12sinx)

Question 1(v). e^6xcos3x

Solution:

Let us considered

y = e^6xcos3x

On differentiating both sides w.r.t x,

(dy/dx) = 6e^6xcos3x - 3e^6xsin3x

Again differentiating both sides w.r.t x,

d²y/dx² = 6(6e^6xcos3x - 3e^6xsin3x) - 3(6e^6xsin3x + 3e^6xcos3x)

d²y/dx² = 36e^6xcos3x - 18e^6xsin3x - 18e^6xsin3x - 9e^6xcos3x

d²y/dx² = 27e^6xcos3x - 36e^6xsin3x

d²y/dx² = 9e^6x(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 = e^xcosx, prove that d²y/dx² = 2e^xcos(x + π/2).

Solution:

We have,

y = e^xcosx

On differentiating both sides w.r.t x,

(dy/dx) = e^xcosx - e^xsinx

Again differentiating both sides w.r.t x,

d²y/dx² = (e^xcosx - e^xsinx) - (e^xsinx + e^xcosx)

d²y/dx² = e^xcosx - e^xcosx - e^xsinx - e^xsinx

d²y/dx² = -2e^xsinx

d²y/dx² = 2e^xcos(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 = e^2x(ax + b), show that y₂ - 4y₁ + 4y = 0.

Solution:

We have,

y = e^2x(ax + b)

On differentiating both sides w.r.t θ,

y₁ = 2e^2x(ax + b) + a.e^2x

Again differentiating both sides w.r.t x,

y₂ = 4e^2x(ax + b) + 2ae^2x + 2a.e^2x

y₂ = 4e^2x(ax + b) + 4a.e^2x

Lets take L.H,S,

= y₂ - 4y₁ + 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 - 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,

y₂ =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

Summary

Exercise 12.1 in Chapter 12 of RD Sharma's Class 12 Mathematics provides a comprehensive set of problems on higher order derivatives. These questions cover a wide range of functions, including polynomial, exponential, trigonometric, and composite functions. By practicing these problems, students can enhance their skills in applying differentiation rules recursively and understand the patterns that emerge in higher order derivatives.