OP Malhotra Class 11 Maths Solutions Chapter 19 Differentiation Ex 19(c)

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S Chand Class 11 ICSE Maths Solutions Chapter 19 Differentiation Ex 19(c)

Question 1.
(ax + b) (cx + d)
Solution:
Let y = (ax + b) (cx + d)
Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = (ax + b)\(\frac { d }{ dx }\)(cx + d) + (cx + d)\(\frac { d }{ dx }\)(ax + b)
[∵ \(\frac { d }{ dx }\) (uv) = u\(\frac { dv }{ dx }\) + v\(\frac { du }{ dx }\)]
= (ax + b) (c1 + 0) + (cx + d) (a1 + 0)
= c (ax + b) + a (ax + d)

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

Question 3.
x (2x – 1)(x + 2)
Solution:
Let y = x (2x – 1) (x + 2); Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = x(2x – 1)\(\frac { d }{ dx }\)(x + 2) + x (x + 2) \(\frac { d }{ dx }\)(2x – 1) (x + 2)\(\frac { d }{ dx }\)x
[∵ \(\frac { d }{ dx }\) (uvw) = uw\(\frac { dw }{ dx }\) + uw\(\frac { dv }{ dx }\) + uw\(\frac { du }{ dx }\)]
= x (2x – 1) + 2x (x + 2) + (2x – 1) (x + 2)

Question 4.
(x – 2) (x + 3) (2x + 5)
Solution:
Let y = (x – 2) (x + 3) (2x + 5)
Diff. both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = (x – 2) (x + 3) \(\frac { d }{ dx }\) (2x + 5) + (x – 2) (2x + 5) \(\frac { d }{ dx }\)(x + 3) + (x + 3) (2x + 5) \(\frac { d }{ dx }\)(x – 2)
= 2 (x – 2) (x + 3) + (x – 2) (2x + 5) + (x + 3) (2x + 5) [using \(\frac { d }{ dx }\) (uvw) = uv\(\frac { dw }{ dx }\) + uw\(\frac { du }{ dx }\) + uw\(\frac { du }{ dx }\)]

Question 5.
y = \(\frac{2 x+5}{3 x-2}\)
Solution:
Let y = \(\frac{2 x+5}{3 x-2}\); Diff. both sides w.r.t. x; we have

Question 6.
y = \(\frac{x^2-3}{x+4}\)
Solution:
Let y = \(\frac{x^2-3}{x+4}\); diff. both sides w.r.t. x; we have

Question 7.
y = \(\frac{2 x-3}{3 x+4}\)
Solution:
Let y = \(\frac{2 x-3}{3 x+4}\); diff. both sides w.r.t. x; we have

Question 8.
y = \(\frac{x^5-x+2}{x^3+7}\)
Solution:
Let y = \(\frac{x^5-x+2}{x^3+7}\); diff. both sides w.r.t. x; we have

Question 9.
s = t² (t + 1)^-1
Solution:
Let s = t² (t + 1)^-1 = \(\frac{t^2}{t+1}\); diff. both sides w.r.t. x; we have

Question 10.
z = \(\frac{u}{u^2+1}\)
Solution:
given z = \(\frac{u}{u^2+1}\); diff. both sides w.r.t. x; we have

Question 11.
y = \(\frac{x^2+2 x+5}{x^3+2 x+4}\)
Solution:
Given y = \(\frac{x^2+2 x+5}{x^3+2 x+4}\); diff. both sides w.r.t. x; we have

Question 12.
f(x) = \(\frac{x^3+2 x}{x^2+4}\)
Solution:
Given f(x) = \(\frac{x^3+2 x}{x^2+4}\); diff. both sides w.r.t. x, we have

Question 13.
If f(x) = \(\frac{x+2}{x-2}\) for all x ≠ 2, find f ‘ ( – 2).
Solution:
Given f(x) = \(\frac{x+2}{x-2}\), x ≠ 2; Diff. both sides w.r.t. x, we have

Question 14.
Differentiate \(\frac{x+2}{x^2-3}\) and find the value of the derivative at x = 0.
Solution:
Given f(x) = \(\frac{x+2}{x^2-3}\); Diff. both sides w.r.t. x, we have

Question 15.
If y = \(\frac{x}{x+a}\), prove that x\(\frac{dy}{dx}\) = y (1 – y).
Solution:
Given y = \(\frac{x}{x+a}\); Diff. both sides w.r.t. x, we have

Question 16.
If \(x \sqrt{1+y}+y \sqrt{1+x}=0\) = 0, prove that \(\frac{d y}{d x}\) = –\(\frac{1}{(1+x)^2}\).
Solution:
Given \(x \sqrt{1+y}+y \sqrt{1+x}=0\) = 0 ⇒ \(x \sqrt{1+y}=-y \sqrt{1+x}\) …(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 = \(\frac{-x}{1+x}\)
Diff. both sides w.r.t. x, we have

Question 17.
Given that y = \(\sqrt{\frac{1-x}{1+x}}\) show that (1 – x²)\(\frac{dy}{dx}\) + y = 0
Solution:
Given y = \(\sqrt{\frac{1-x}{1+x}}\)
Diff. both sides w.r.t. x, we have

Question 18.
Given that y = (3x – 1)² + (2x – 1)³, find \(\frac{dy}{dx}\) and the points on the curve for which \(\frac{dy}{dx}\) = 0.
Solution:
Given y = (3x – 1)² + (2x – 1)³
Diff. eqn. (1) both sides w.r.t. x, we have
\(\frac{dy}{dx}\) = 2(3x -1)\(\frac{d}{dx}\) (3x -1) + 3 (2x -1)² \(\frac{d}{dx}\) (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 \(\frac{dy}{dx}\) = 0 ⇒ 6x (4x – 1) = 0 ⇒ x = 0, \(\frac{1}{4}\)
when x = 0 ∴ from (1); y = (0 – 1)² + (0 – 1)³ = 1 – 1 = 0
When x = \(\frac{1}{4}\) ∴ from (1); y = \(\left(\frac{3}{4}-1\right)^2\) + \(\left(\frac{1}{2}-1\right)^3\) = \(\frac{1}{16}\) – \(\frac{1}{8}\) = \(\frac{-1}{16}\)
Hence the required points on curve are (0, 0) and \(\left(\frac{1}{4},-\frac{1}{16}\right) .\)

Question 19.
(i) If y = \(\frac{x-1}{2 x^2-7 x+5}\), find \(\frac{dy}{dx}\) at x = 2.
(ii) If y = \(\frac{x^2+3}{x^3+2 x}\), find \(\frac{dy}{dx}\) at x = 1.
Solution:
(i) Given y = \(\frac{x-1}{2 x^2-7 x+5}\); Diff. both sides w.r.t. x, we have

(ii) Given y = \(\frac{x^2+3}{x^3+2 x}\)
Diff. both sides w.r.t. x, we have

Question 20.
Find the coordinates of the points on the curve y = \(\frac{x}{1-x^2}\) for which \(\frac{dy}{dx}\) = 1.
Solution:
Given eqn. of curve be y = \(\frac{x}{1-x^2}\)
Diff. both sides w.r.t. x, we have

Now \(\frac{dy}{dx}\) = 1
⇒ \(\frac{1+x^2}{\left(1-x^2\right)^2}\) = 1
⇒ 1 + x² = (1 – x²)²
⇒ 1 + x² = 1 + x⁴ – 2x²
⇒ x⁴ – 3x² = 0
⇒ x² (x² – 3) = 0 ⇒ x = 0, ± \(\sqrt{3}\)
When x = 0 ∴ from (1); y = 0
When x = \(\sqrt{3}\)
∴ from (1); y = \(\frac{\sqrt{3}}{1-3}\) = –\(\frac{\sqrt{3}}{2}\)
When x = –\(\sqrt{3}\)
∴ from (1); y = \(\frac{-\sqrt{3}}{1-3}\) = –\(\frac{\sqrt{3}}{2}\)
Hence the coordinates of required points on given curve are (0, 0);
\(\left(\sqrt{3}, \frac{-\sqrt{3}}{2}\right)\) and \(\left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right)\).