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). 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) – 2x $e^{tan^{-1}x}$ /(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

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Last Updated : 28 Mar, 2021

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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). x³+ tanx

Question 1(ii). sin(logx)

Question 1(iii). log(sinx)

Question 1(iv). e^xsin5x

Question 1(v). e^6xcos3x

Question 1(vi). x³logx

Question 1(vii). tan^-1x

Question 1(viii). x.cosx

Question 1(ix). log(logx)

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

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

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

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

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

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

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

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

Question 10. If y = e^xcosx, prove that d²y/dx²= 2e^xcos(x + π/2).

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

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

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

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

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

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

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

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

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

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

Question 21. If y = $e^{tan^{-1}x}$ , prove that (1 + x²)y₂+ (2x – 1)y₁= 0.

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

Question 23. If y = e^2x(ax + b), show that y₂– 4y₁+ 4y = 0.

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

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

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

CBSE Class 12 Computer Science

GATE CS & IT 2024

Data Structures & Algorithms in JavaScript - Self Paced

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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). x3 + tanx

Question 1(ii). sin(logx)

Question 1(iii). log(sinx)

Question 1(iv). exsin5x

Question 1(v). e6xcos3x

Question 1(vi). x3logx

Question 1(vii). tan-1x

Question 1(viii). x.cosx

Question 1(ix). log(logx)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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CBSE Class 12 Computer Science

GATE CS & IT 2024

Data Structures & Algorithms in JavaScript - Self Paced

Question 1(i). x³+ tanx

Question 1(iv). e^xsin5x

Question 1(v). e^6xcos3x

Question 1(vi). x³logx

Question 1(vii). tan^-1x

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

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

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

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

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

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

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

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

Question 10. If y = e^xcosx, prove that d²y/dx²= 2e^xcos(x + π/2).

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

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

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

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

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

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

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

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

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

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

Question 21. If y = $e^{tan^{-1}x}$ , prove that (1 + x²)y₂+ (2x – 1)y₁= 0.

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

Question 23. If y = e^2x(ax + b), show that y₂– 4y₁+ 4y = 0.

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

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

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