Differentiation Class 11 OP Malhotra Exe-19C ISC Maths Ch-19 Solutions. In this article you would learn about Rule for Differentiating Product of Functions. 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-19C ISC Maths Solutions Ch-19

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-19	Differentiation
Writer	O.P. Malhotra
Exe-19(C)	Rule for Differentiating Product of Functions.

Rule for Differentiating Product of Functions.

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

Differentiate the following w.r.t. x or t or u as the case may be :

Que-1: (ax + b) (cx + d)

Sol: Let y = (ax + b) (cx + d)
Diff. both sides w.r.t. x, we have
dy/dx = (ax + b) (d/dx) (cx + d) + (cx + d) (d/dx) (ax + b)
[∵ d/dx (uv) = u(dv/dx) + v(du/dx)]
= (ax + b) (c1 + 0) + (cx + d) (a1 + 0)
= c (ax + b) + a (ax + d)

Que-2: (x¹⁰⁰ + 2x⁵⁰ – 3) (7x⁸ + 20x + 5)

Sol: Let y = (x¹⁰⁰ + 2x⁵⁰ – 3) (7x⁸ + 20x + 5)
Diff. both sides w.r.t. x, we have
dy/dx = (x¹⁰⁰ + 2x⁵⁰ – 3) (d/dx) (7x⁸ + 20x + 5) + (7x⁸ + 20x + 5) (d/dx) (x¹⁰⁰ + 2x⁵⁰ – 3)
= (x¹⁰⁰ + 2x⁵⁰ – 3) (56x⁷ + 20) + (7x⁸ + 20x + 5) (100x⁹⁹ + 100x⁴⁹)

Que-3: x (2x – 1)(x + 2)

Sol: Let y = x (2x – 1) (x + 2); Diff. both sides w.r.t. x, we have
dy/dx = x(2x – 1) (d/dx) (x + 2) + x (x + 2) (d/dx) (2x – 1) (x + 2) (d/dx)x
[∵ d/dx (uvw) = uw (dw/dx) + uw(dv/dx) + uw(du/dx)]
= x (2x – 1) + 2x (x + 2) + (2x – 1) (x + 2)

Que-4: (x – 2) (x + 3) (2x + 5)

Sol: Let y = (x – 2) (x + 3) (2x + 5)
Diff. both sides w.r.t. x; we have
dy/dx = (x – 2) (x + 3) (d/dx) (2x + 5) + (x – 2) (2x + 5) (d/dx)(x + 3) + (x + 3) (2x + 5) (d/dx) (x – 2)
= 2 (x – 2) (x + 3) + (x – 2) (2x + 5) + (x + 3) (2x + 5)

Que-5: y = (2x+5)/(3x-2)

Sol: Let y = (2x+5)/(3x−2); Diff. both sides w.r.t. x; we have

Que-6: y = (x²−3)/(x+4)

Sol: Let y = (x²−3)/(x+4); diff. both sides w.r.t. x; we have

Que-7: y = (2x−3)/(3x+4)

Sol: Let y = (2x−3)/(3x+4); diff. both sides w.r.t. x; we have

Que-8: y = [(x^5)−x+2]/(x³+7)

Sol: Let y = [(x^5)−x+2]/(x³+7) ; diff. both sides w.r.t. x; we have

Que-9: s = t² (t + 1)^-1

Sol: Let s = t² (t + 1)^-1 = t²/(t+1); diff. both sides w.r.t. x; we have

Que-10: z = u/(u²+1)

Sol: given z = u/(u²+1); diff. both sides w.r.t. x; we have

Que-11: y = (x²+2x+5)/(x³+2x+4)

Sol: Given y = y = (x²+2x+5)/(x³+2x+4); diff. both sides w.r.t. x; we have

Que-12: f(x) = (x³+2x)/(x²+4)

Sol: Given f(x) = (x³+2x)/(x²+4); diff. both sides w.r.t. x, we have

Que-13: If f(x) = (x+2)/(x−2) for all x ≠ 2, find f ‘ ( – 2).

Sol: Given f(x) = (x+2)/(x−2), x ≠ 2; Diff. both sides w.r.t. x, we have

Que-14: Differentiate (x+2)/(x²−3) and find the value of the derivative at x = 0.

Sol: Given f(x) = (x+2)/(x²−3); Diff. both sides w.r.t. x, we have

Que-15: If y = x/(x+a), prove that x(dy/dx) = y (1 – y).

Sol: Given y = x/(x+a); Diff. both sides w.r.t. x, we have

Que-16: If x√(1+y) + y√(1+x) = 0, prove that dy/dx = –1/(1+x)².

Sol: Given x√(1+y) + y√(1+x) = 0  ⇒ x√(1+y)=−y√(1+x) …(1)
On squaring both sides, we have
x² (1 + y) = y² (1 + x)
⇒ x² + x²y – y² – xy² = 0
⇒(x – y)(x + y) + xy(x – y) = 0
⇒ (x – y) (x + y + xy) = 0
⇒ x + y + xy = 0 [∵ x ≠ y if x = y then given eqn. is meaningless]
⇒ x + y (1 + x) = 0
⇒ y = −x/(1+x)
Diff. both sides w.r.t. x, we have

Que-17: Given that y = √{(1-x)/(1+x)} show that (1 – x²)(dy/dx) + y = 0

Sol: Given y = √{(1-x)/(1+x)}
Diff. both sides w.r.t. x, we have

Que-18: Given that y = (3x – 1)² + (2x – 1)³, find (dy/dx) and the points on the curve for which (dy/dx) = 0.

Sol: Given y = (3x – 1)² + (2x – 1)³
Diff. eqn. (1) both sides w.r.t. x, we have
dy/dx = 2(3x -1)(d/dx) (3x -1) + 3 (2x -1)² (d/dx) (2x -1)
= 2 (3x – 1) (3.1 -0) + 3(2x – 1)² (2 × 1 – 0) = 6 (3x – 1) + 6 (2x – 1)²
= 6 [3x – 1 + 4x² – 4x + 1] = 6 [4x² – x] = 6x (4x- 1)
Now dy/dx = 0 ⇒ 6x (4x – 1) = 0 ⇒ x = 0, 1/4
when x = 0 ∴ from (1); y = (0 – 1)² + (0 – 1)³ = 1 – 1 = 0
When x = 1/4 ∴ from (1); y = (3/4−1)² + (1/2−1)² = (1/16) – (1/8) = −1/16
Hence the required points on curve are (0, 0) and (1/4,−1/16).

Que-19: (i) If y = (x−1)/{2x²−7x+5}, find dy/dx at x = 2.
(ii) If y = (x²+3)/(x³+2x), find dy/dx at x = 1

Sol: (i) Given y =(x−1)/{2x²−7x+5}; Diff. both sides w.r.t. x, we have

(ii) Given y = (x²+3)/(x³+2x)
Diff. both sides w.r.t. x, we have

Que-20: Find the coordinates of the points on the curve y = x/(1−x²) for which dy/dx = 1.

Sol: Given eqn. of curve be y = x/(1−x²)
Diff. both sides w.r.t. x, we have

Now dy/dx = 1
⇒ (1+x²){(1−x²)²} = 1
⇒ 1 + x² = (1 – x²)²
⇒ 1 + x² = 1 + x⁴ – 2x²
⇒ x⁴ – 3x² = 0
⇒ x² (x² – 3) = 0 ⇒ x = 0, ± √3
When x = 0 ∴ from (1); y = 0
When x = √3
∴ from (1); y = √3/(1−3) = –√3/2
When x = –√3
∴ from (1); y = −√3/(1−3) = –√3/2
Hence the coordinates of required points on given curve are (0, 0);
(√3,−√3/2) and (−√3,√3/2).

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

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

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