Differentiation Class 12 OP Malhotra Exe-8L ISC Maths Solutions Ch-8 Solutions. In this article you would learn about higher derivatives (successive differentiation). Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Differentiation Class 12 OP Malhotra Exe-8L ISC Maths Solutions Ch-8

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-8	Differentiation
Writer	OP Malhotra
Exe-8(L)	higher derivatives (successive differentiation)

Higher Derivatives (Successive Differentiation)

Differentiation Class 12 OP Malhotra Exe-8L ISC Maths Solutions Ch-8 Solutions

Que-1: (i) x²
(ii) a^x
(iii) ax³ + bx² + cx + d
(iv) log x
(v) 1/√x
(vi) x/√x-1
(vii) sin^-1 x

Sol: (i) Let y = x² ; Diff. both sides w.r.t. x,
dy/dx = 2x ; Again diff. both sides w.r.t. x
∴ d²y/dx² = 2

(ii) Let y = a^x ; Diff. both sides w.r.t. x, we have
dy/dx = a^x log a; Again diff. both sides w.r.t. x ; we have
∴ d²y/dx² = a²(log a)²

(iii) Let y = ax³ + bx² + cx + d; Diff. both sides w.r.t. x
dy/dx = 3ax² + 2bx + c; Diff. again w.r.t. x
d²y/dx² = 6ax + 2b

(iv) Let y = log x ; Diff. both sides w.r.t. x
dy/dx = 1/x ; Diff. again w.r.t. x; we have
d²y/dx² = – 1/x²

Que-2: (i) e^x + sin x
(ii) e^-x sin x

Sol: (i) Let y = e^x + sin x ;
Diff. both sides w.r.t. x; we have
dy/dx = e^x + cos x ;
Diff. again both sides w.r.t. x; we have
d²y/dx² = e^x – sin x

(ii) Let y = e^-x sin x ;
Diff. both sides w.r.t. x,

Que-3: (i) If y = 2 sin x + 3 cos x, prove that y + d²y/dx² = 0.
(ii) If y = a + bx², prove that x.d²y/dx² = dy/dx
(iii) If y = tan x + sec x, prove that d²y/dx² = cos x/(1-sin x)².
(iv) If y = 500, e^7x + 600 e^-7x, show that d²y/dx² = 49 y.
(iv) If e^y (1 + x) = 1, show that d²y/dx² = (dy/dx)².

Sol: (i) Given y = 2 sin x + 3 cos x …(1)
Diff. both sides w.r.t. x; we have
dy/dx = 2 cos x – 3 sin x ;
Again diff. both sides w.r.t. x

(iii) Given y = tan x + sec x … (1)
Diff. eqn. (1) both sides w.r.t. x; we have

(iv) Given y = 500 e^7x + 600 e^-7x …(1)
Diff. eqn. (1) both sides w.r.t. x; we have
dy/dx = 3500 e^7x – 4200 e^-7x
Again diff. both sides w.r.t. x
dy/dx = 7 x 3500 e^7x + 4200 x 7 e^-7x
= 49[500 e^7x + 600 e^-7x] = 49 y [using eqn. (1)]

(v) Given e^y (1 + x) = 1 ⇒ e^y = 1/1+x
Taking logorithm both sides w.r.t. x, we have
y = log(1/1+x) = – log(1 + x)
Diff. both sides w.r.t. x ; we have
dy/dx = – (1/1+x)… (1)
Again diff. both sides w.r.t. x
d²y/dx² = 1/(1+x)² = (dy/dx)² [using eqn. (1)]

Que-4: If y = tan x, prove that d²y/dx² = 2y dy/dx .

Sol:

Que-5: If y = logx / x, prove that d²y/dx² = 2 log x – 3 / x³.

Sol: Given y = log x / x
Diff. both sides w.r.t. x; we have

Que-6: (i) If y = tan^-1 x, prove that
(1 + x²) d²y/dx² + 2x dy/dx = 0.
(ii) If y = sin^-1x, then show that
(1 + x²) d²y/dx² – x dy/dx = 0.

Sol: (i) Given y = tan^-1 x;
Diff. both sides w.r.t. x, we have

(ii) Given y = sin^-1 x;
Diff. both sides w.r.t. x, we have

Que-7: If y = e^(tan^-1 x) , prove that
(1+x²)d²y/dx² + (2x-1) dy/dx = 0.

Sol:

Que-8: If y = x^x, prove that
d²y/dx² – 1/y (dy/dx)² -y/x = 0

Sol: Given y = x^x, … (1)
Taking logarithm on eqn. (1); we have
log y = x log x …(2)
Diff. eqn. (2) w.r.t. x, we have

Que-9: If y = sin^-1x / √1-x², prove that
(1-x²) d²y/dx² – 3x dy/dx – y = 0.

Sol:

Que-10: If y = (tan^-1 x)², prove that
(x²+1)² d²y/dx² + 2x (x²+1) dy/dx = 2.

Sol: Given y = (tan^-1x)²,
Diff. both sides w.r.t. x, we have

Que-11: If y = sin (m sin^-1 x) show that (1 – x²) d²y/dx² – x dy/dx + m²y = 0

Sol:

Que-12: If y = (A + Bx)e^3x, prove that y” + 6y’ + 9y + 2 = 2.

Sol: Given y = (A + Bx)e^-3x …(1)
Diff. eqn. (1) w.r.t. x, we have

Que-13: If x^myⁿ = (x + y)^m+n, prove that d²y/dx² = 0.

Sol: Given x^myⁿ = (x + y)^m+n
Taking logaritum on both sides, we have

Que-14: If y = ae^mx + be^-mx, prove that d²y/dx² + m²y = 0.

Sol:

Que-15: If y = a cos (log x) + b sin (log x), prove that x² d²y/dx² + x dy/dx + y = 0.

Sol:

Que-16: Find d²y/dx² when
(i) x = t², y = t³.
(ii) x = at², y = 2at.
(iii) x = a cos θ, y = b sin θ
(iv) x = cos t, y = sin t

Sol: (i) Let x = t² … (1)
& y = t³ … (2)
Diff. eqn. (1) & (2) w.r.t. t; we have

(ii) x = at² … (1)
& y = 2at … (2)

(iii) Let x = a cos θ …(1)
& y = b sin θ …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have

Que-17: Find d²y/dx² when θ = π/2 :
(i) x = a(θ + sin θ), y = a(1 – cos θ)
(ii) x = a(1 – cos θ), y = a(θ + sin θ).

Sol: (i) Let x = a(θ + sin θ) …(1)
& y = a(1 – cos θ) …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have

(ii) Given x = a(1 – cos θ) …(1)
& y = a(θ + sin θ) …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have

Que-18: If x = a sec³θ, y = a tan³θ, find d²y/dx² at θ = π/4 .

Sol: Let x = a sec³θ …(1)
& y = a tan³θ …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have

Que-19: If x = cos θ + θ sin θ, y = sin θ – θ cos θ, 0 < θ < π/2 , prove that d²y/dx² = sec³θ / θ.

Sol:

Que-20: If x = cos θ, y = sin³θ, show that d²y/dx² . (dy/dx)² = 3 sin²θ(5cos²θ-1).

Sol:

Que-21:

(a) 0
(b) – 1
(c) independent of θ
(d) None of these.

Sol:

Que-22:

Sol: 

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