Differentiation Class 11 OP Malhotra Exe-19B ISC Maths Ch-19 Solutions. In this article you would learn about General Theorems on 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-11 Mathematics.

Differentiation Class 11 OP Malhotra Exe-19B ISC Maths Solutions Ch-19

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-19	Differentiation
Writer	O.P. Malhotra
Exe-19(B)	General Theorems on Differentiation.

General Theorems on Differentiation.

Differentiation Class 11 OP Malhotra Exe-19B ISC Maths Ch-19 Solutions.

Que-1: (ax)^m + b^m

Sol: Let y = (ax)^m + b^m = a^m x^m + b^m
Diff. both sides w.r.t. x, we have
dy/dx = (d/dx)(a^m x^m) + (d/dx) (b^m) = a^m (d/dx) x^m + 0 = ma^m x^m-1

Que-2: x³ + 4x² + 7x + 2

Sol: Let y = x³ + 3x² + 7x + 2
Diff both sides w.r.t. x, we have
dy/dx = (d/dx) (x^m) + (d/dx) 4x² + (d/dx) (7x) + (d/dx) (2)
= 3x² + 4(d/dx) x² + 7(d/dx)(x)  = 0 = 3x² + 8x + 7

Que-3: 7x⁶ + 8x⁵ – 3x⁴ + 11x² + 6x + 7

Sol: Let y = 7×6 + 8×5 – 3×4 + 11×2 + 6x + 7
diff. both sides w.r.t. x, we have
dy/dx = 7 (d/dx) (x^6) + 8(d/dx) (x^5) – 3 (d/dx) (x^4) + 11 (d/dx) x² + 6(d/dx) x + (d/dx) (7)
= 7 × 6x⁵ + 8 × 5x⁴ – 3 × 4x³ + 11 × 2x + 6 + 0
= 42x⁵ + 40x⁴ – 12x³ + 22x + 6

Que-4: 3 + 4x – 7x² – √2x³ + πx⁴ – (2/5)x⁵ + (4/3)

Sol: Diff. both sides w.r.t. x, we have
dy/dx = (d/dx) [3 + 4x – 7x² – √2x³ + πx⁴ – (2/5)x⁵ + (4/3)]
= (d/dx) (3) + 4(d/dx) (x) – 7(d/dx) x² – √2 (d/dx) x³ + π (d/dx) (x^4) – 25 (d/dx) (x^5) + (d/dx) (4/3)
= 0 + 4 – 14x – 3√2 x² + 4πx³ – 2x⁴ + 0
dy/dx = 4 – 14x – 3√2x² + 4πx³ – 2x⁴

Que-5: 3/x^5

Sol: Let y = 3/x^5 = 3x^-5; diff. both sides w.r.t. x; we have
dy/dx = (d/dx)(3x^-5) = (d/dx) x^-5 = 3 (- 5) x^-5-1 = -15/x^6

Que-6: x^(5/3)

Sol: Let y = x^5/3; diff. both sides w.r.t. x ; we have
dy/dx = (d/dx) x^(5/3) = 5/3
= x^{(5/3)−1} = (5/3) x^(2/3)

Que-7: 7 / x^(2/3)

Sol: Let y = 7 / x^(2/3) = 7x^-2/3; diff. both sides w.r.t. x. we have
dy/dx = (d/dx) (7x^(-2/3)) = 7 (d/dx) x^(-2/3)
= 7(−2/3) x^{(−2/3)−1}
= −14/3 x^(−5/3).

Que-8: (√x + (1/√x))², x ≠ 0

Sol: Let y = (√x+(1/√x))² = x + (1/x) + 2√x . 1/√x
⇒ y = x + (1/x) + 2; Diff. both sides w.r.t. x, we have
dy/dx = d/dx (x) + d/dx (1/x) + d/dx (2)
= 1 – (1/x)²

Que-9: √x − (1/√x), x ≠ 0

Sol: Let y = √x − (1/√x); diff. both sides w.r.t. x, we have
∴ dy/dx = d/dx √x – (d/dx) x^(−1/2)
= (1/2) x^{(1/2)−1} – (−1/2) x^{(−1/2)−1}
[∵ d/dx x^n = nx^(n-1)]
= (1/2√x) + 1/2x^(3/2)

Que-10: 1/x + 3/x² + 2/x³

Sol: Let y = 1/x + 3/x² + 2/x³; diff. both sides w.r.t. x, we have
∴ dy/dx = (d/dx) (1/x) + 3 (d/dx) x¯² + 2(d/dx) (x¯³)
= -1 x^-1-1 + 3 (-2) x^-2-1 + 2 (- 3) x^-3-1
= – 1/x² – 6/x³ – 6/x^4

Que-11: {2x^(1/2)} + {6x^(1/3)} − {2x^(3/2)}

Sol: Let y = {2x^(1/2)} + {6x^(1/3)} − {2x^(3/2)}; diff. both sides w.r.t. x, we have
dy/dx = 2(d/dx) x^(1/2) + 6(d/dx) x^(1/3) – 2(d/dx) x^(3/2)
= 2 × (1/2) x^{(1/2)−1} + 6 × (1/3) x^{(1/3)−1} – 2 × (3/2) x^{(3/2)−1}
= 1/√x + 2/x^(2/3) – 3√x

Que-12: 8x³ – x² + 5 – (2/x) + (4/x³)

Sol: Let y = 8x³ – x² + 5 – (2/x) + (4/x³)
Diff both sides w.r.t. x, we have
dy/dx = 8(d/dx) (x^3) – (d/dx) (x^2) + (d/dx) (5) – 2(d/dx) (x^-1) + 4(d/dx) (x^-3)
= 8 × 3x² – 2x + 0 – 2 (- 1) x^-1-1 + 4 (- 3) x^-3-1
= 24x² – 2x + (2/x²) – (12/x^4)

Que-13: {(3x^7)+(x^5)−(2x^4)+x−3}/x^4

Sol: Let y = {(3x^7)+(x^5)−(2x^4)+x−3}/x^4= 3x³ + x – 2 + (1/x³) – (3/x^4)
Diff. both sides w.r.t. x, we have
dy/dx = 3(d/dx) (x^3) + (d/dx) x – (d/dx) (2) + (d/dx) (x^-3) – 3(d/dx) (x^-4)
= 3 × 3x² + 1 – 0 – 3x^-3-1 – 3 ( – 4)x^-4-1 = 9x² + 1 – (3/x^4) + (12/x^5)

Que-14: (i) (2x – 3)²   (ii) (2x – 3)¹⁰⁰

Sol: (i) Let y = (2x – 3)² = 4x² – 12x + 9 ; diff. both sides w.r.t. x, we have
dy/dx = 4(d/dx) (x^2) – (1/2) (d/dx) x + (d/dx) (9)
= 8x – 12

(ii) Let y = (2x – 3)¹⁰⁰ ; diff. both sides w.r.t. x, we get
dy/dx = (d/dx) (2x – 3)¹⁰⁰ = 100 (2x – 3)^100-1  (d/dx) (2x – 3)
= 100 (2x – 3)⁹⁹ [2 (1) – 0] = 200 (2x – 3)⁹⁹

Que-15: √(3x+2)

Sol: Let y = √(3x+2) ; diff. both sides w.r.t. x, we get
dy/dx = (d/dx)(3x+2)^(1/2)
= (1/2) (3x+2)^{(1/2)−1} (d/dx)(3x+2)
= (1/2)(3x+2) − (1/2)(3+0) = (3/2) {1/√(3x+2)}

Que-16: Given f(x) = (7/4)x², find f ‘ (1/7)

Sol: Given f(x) = (7/4)x² ; diff. both sides w.r.t. x, we have
f ‘ (x) = (7/4) × 2x = (7/2)x
⇒ f ‘ (1/7) = (7/2) × (1/7) = 1/2

Que-17: Find the derivative with respect to x of the following:
(i) x – (1/x)
(ii) √x + 1/√x
(iii) 3x² + 3/x²
(iv) (x²+1)/x
(v) (2x+x^4)/x²
(vi) (1+x²)/x³

Sol: (i) Let y = x – (1/x); diff. both sides, w.r.t. x, we get
dy/dx = (d/dx)(x) – (d/dx) x^-1 = 1 – (-1) x^(-1-1)
= 1 + (1/x²)

(ii) Let y = √x + 1/√x; diff. both sides w.r.t. x, we have
dy/dx = (d/dx) x^(1/2) + (d/dx) x^(-1/2)
= (1/2) (1/2)x^{(1/2)−1} + (−1/2)x^{(−1/2)−1}
= 1/2√x – 1/{2x^(3/2)}

(iii) Let y = 3x²+(3/x²); diff. both sides w.r.t. x, we have
dy/dx = 3(d/dx) x^-2
= 3 × 2x + 3 (-2) x^(-2-1)
= 6x – (6/x³)

(iv) Let y = (x²+1)/x; diff. both sides w.r.t. x, we have
dy/dx = (d/dx) x + (d/dx) x^-1
= 1 + (-1)x^(-1-1)
= 1 – (1/x²)

(v) Let y = (2x+x^4)/x² = (2/x) + x²; diff. both sides w.r.t. x,
∴ dy/dx = 2(d/dx) x^-1 + (d/dx) (x^-2)
= 2 (- 1)x^(-1-1) + 2x
= (−2/x²) + 2x

(vi) Let = (1+x²)/x³ = (1/x³) + (1/x); diff. both sides w.r.t. x
∴ dy/dx = (d/dx) (x^-3) + (d/dx) (x^-1)
= 3x^(-3-1) + (-1)^(-1-1)
= (–3/x^4) – (1/x²)

Que-18: If y = x + (1/x), prove that x²(dy/dx) – xy + 2 = 0

Sol: Given y = x + (1/x); diff, both sides w.r.t. x, we have
dy/dx = (d/dx) x + (d/dx) x^-1
= 1 + (-1)x^-1-1 = 1 – (1/x²)
⇒ x²(dy/dx) = x²(dy/dx) – xy + 2
= x² – 1 – x {x+(1/x)} + 2
= x² – 1 x² – 1 + 2 = 0 = R.H.S.

Que-19: If y= √x – 1/√x, prove that 2x (dy/dx) + y = 2√x.

Sol: Given y= √x – 1/√x
Diff. eqn (1) both sides w.r.t x, we have

Que-20: If y = 1/(a−z), show that dz/dy = (z – a)²

Sol: Given y = 1/(a−z)
diff. both sides w.r.t. z, we have
dy/dz = (d/dz) {1/(a−z)} = (d/dz)(a – z)^-1
= (-1)(a – z)^-1-1 (d/dz) (a – z)
= 1 (a – z)^-2 (0 – 1) =  1/(a−z)²
∴ dz/dy = 1/(dy/dz) = (a – z)²

Que-21: If y = 1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + …… to ∞, show that dy/dx = y.

Sol: Given y = 1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + …. ∞ …(1)
Diff. both sides w.r.t. x; we have
dy/dx = 0 + 1 + (x^2)/2! + (x^3)/3! + (x^4)/4! + …. ∞ …(1)
= 1 + x + (x²/2) + (x³/6) …. ∞
= 1 + x + (x²/2!) + (x/³3!) + ….. ∞ = y
Thus dy/dx = y

–: End of Differentiation Class 11 OP Malhotra Exe-19B ISC Maths Ch-19 Solutions. :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions

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