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

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

Question 1.
(ax)^m + b^m
Solution:
Let y = (ax)^m + b^m = a^m x^m + b^m
Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (a^m x^m) + \(\frac { d }{ dx }\) (b^m) = a^m \(\frac { d }{ dx }\) x^m + 0 = ma^m x^m-1

Question 2.
x³ + 4x² + 7x + 2
Solution:
Let y = x³ + 3x² + 7x + 2
Diff both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) x^m + \(\frac { d }{ dx }\) 4x² + \(\frac { d }{ dx }\) (7x) + \(\frac { d }{ dx }\) (2) = 3x² + 4\(\frac { d }{ dx }\) x² + 7\(\frac { d }{ dx }\)(x) = 0 = 3x² + 8x + 7

Question 3.
7x⁶ + 8x⁵ – 3x⁴ + 11x² + 6x + 7
Solution:
Let y = 7x⁶ + 8x⁵ – 3x⁴ + 11x² + 6x + 7
diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = 7 \(\frac { d }{ dx }\) x⁶ + 8\(\frac { d }{ dx }\)x⁵ – 3 \(\frac { d }{ dx }\) x⁴ + 11 \(\frac { d }{ dx }\) x² + 6\(\frac { d }{ dx }\) x + \(\frac { d }{ dx }\) (7)
= 7 × 6x⁵ + 8 × 5x⁴ – 3 × 4x³ + 11 × 2x + 6 + 0
= 42x⁵ + 40x⁴ – 12x³ + 22x + 6

Question 4.
3 + 4x – 7x² – √2x³ + πx⁴ – \(\frac { 2 }{ 5 }\)x⁵ + \(\frac { 4 }{ 3 }\)
Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) [3 + 4x – 7x² – √2x³ + πx⁴ – \(\frac { 2 }{ 5 }\)x⁵ + \(\frac { 4 }{ 3 }\)]
= \(\frac { d }{ dx }\) (3) + 4\(\frac { d }{ dx }\) (x) – 7\(\frac { d }{ dx }\) x² – √2 \(\frac { d }{ dx }\) x³ + π \(\frac { d }{ dx }\) x⁴ – \(\frac { 2 }{ 5 }\) \(\frac { d }{ dx }\) x⁵ + \(\frac { d }{ dx }\) \(\left(\frac{4}{3}\right)\)
= 0 + 4 – 14x – 3√2 x² + 4πx³ – 2x⁴ + 0
∴ \(\frac { dy }{ dx }\) = 4 – 14x – 3√2x² + 4πx³ – 2x⁴

Question 5.
\(\frac{3}{x^5}\)
Solution:
Let y = \(\frac{3}{x^5}\) = 3x^-5; diff. both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (3x^-5) = \(\frac { d }{ dx }\) x^-5 = 3 (- 5) x^-5-1 = \(\frac{-15}{x^6}\) [∵ \(\frac { d }{ dx }\) xⁿ = nx^n-1]

Question 6.
\(x^{\frac{5}{3}}\)
Solution:
Let y = x^5/3; diff. both sides w.r.t. x ; we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) x^5/3 = \(\frac { 5 }{ 3 }\) = \(x^{\frac{5}{3}-1}\) = \(\frac{5}{3} x^{2 / 3}\)

Question 7.
\(\frac{7}{x^{\frac{2}{3}}}\)
Solution:
Let y = \(\frac{7}{x^{2 / 3}}\) = 7x^-2/3; diff. both sides w.r.t. x. we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (7x^-2/3) = 7 \(\frac { d }{ dx }\) x^-2/3 = 7\(\left(-\frac{2}{3}\right) x^{-\frac{2}{3}-1}\) = \(\frac{-14}{3} x^{-5 / 3}\)

Question 8.
\(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2, x \neq 0\)
Solution:
Let y = \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2\) = x + \(\frac { 1 }{ x }\) + 2√x . \(\frac{1}{\sqrt{x}}\)
⇒ y = x + \(\frac { 1 }{ x }\) + 2; Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (x) + \(\frac { d }{ dx }\) \(\left(\frac{1}{x}\right)\) + \(\frac { d }{ dx }\) (2) = 1 – \(\frac{1}{x^2}\)

Question 9.
\(\sqrt{x}-\frac{1}{\sqrt{x}}, x \neq 0\)
Solution:
Let y = \(\sqrt{x}-\frac{1}{\sqrt{x}}\); diff. both sides w.r.t. x, we have
∴ \(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) \(\sqrt{x}\) – \(\frac{d}{d x} x^{-1 / 2}\) = \(\frac{1}{2} x^{\frac{1}{2}-1}\) – \(\left(\frac{-1}{2}\right) x^{-\frac{1}{2}-1}\)
[∵ \(\frac { d }{ dx }\) xⁿ =nx^n-1]
= \(\frac{1}{2 \sqrt{x}}\) + \(\frac{1}{2 x^{3 / 2}}\)

Question 10.
\(\frac { 1 }{ x }\) + \(\frac{3}{x^2}\) + \(\frac{2}{x^3}\)
Solution:
Let y = \(\frac { 1 }{ x }\) + \(\frac{3}{x^2}\) + \(\frac{2}{x^3}\); diff. both sides w.r.t. x, we have
∴ \(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) \(\left(\frac{1}{x}\right)\) + 3 \(\frac { d }{ dx }\) x^-2 + 2\(\frac { d }{ dx }\) (x^-3) = -1 x^-1-1 + 3 (-2) x^-2-1 + 2 (- 3) x^-3-1 = – \(\frac{1}{x^2}\) – \(\frac{6}{x^3}\) – \(\frac{6}{x^4}\)

Question 11.
\(2 x^{\frac{1}{2}}+6 x^{\frac{1}{3}}-2 x^{\frac{3}{2}}\)
Solution:
Let y = \(2 x^{\frac{1}{2}}+6 x^{\frac{1}{3}}-2 x^{\frac{3}{2}}\); diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = 2\(\frac { d }{ dx }\) x^1/2 + 6\(\frac { d }{ dx }\) x^1/3 – 2\(\frac { d }{ dx }\) x^3/2
= 2 × \(\frac{1}{2} x^{\frac{1}{2}-1}\) + 6 × \(\frac{1}{3} x^{\frac{1}{3}-1}\) – 2 × \(\frac{3}{2} x^{\frac{3}{2}-1}\)
= \(\frac{1}{\sqrt{x}}\) + \(\frac{2}{x^{2 / 3}}\) – \(3 \sqrt{x}\)

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

Question 13.
\(\frac{3 x^7+x^5-2 x^4+x-3}{x^4}\)
Solution:
Let y = \(\frac{3 x^7+x^5-2 x^4+x-3}{x^4}\) = 3x³ + x – 2 + \(\frac{1}{x^3}\) – \(\frac{3}{x^4}\)
Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = 3\(\frac { d }{ dx }\) x³ + \(\frac { d }{ dx }\) x – \(\frac { d }{ dx }\) (2) + \(\frac { d }{ dx }\) x^-3 – 3\(\frac { d }{ dx }\) x^-4
= 3 × 3x² + 1 – 0 – 3x^-3-1 – 3 ( – 4)x^-4-1 = 9x² + 1 – \(\frac{3}{x^4}\) + \(\frac{12}{x^5}\)

Question 14.
(i) (2x – 3)²
(ii) (2x – 3)¹⁰⁰
Solution:
(i) Let y = (2x – 3)² = 4x² – 12x + 9 ; diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = 4\(\frac { d }{ dx }\) x² – 12 \(\frac { d }{ dx }\) x + \(\frac { d }{ dx }\) (9) = 8x – 12

(ii) Let y = (2x – 3)¹⁰⁰ ; diff. both sides w.r.t. x, we get
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (2x – 3)¹⁰⁰ = 100 (2x – 3)^{100 – 1} \(\frac { d }{ dx }\) (2x – 3) [∵ \(\frac { d }{ dx }\) xⁿ = nx^n-1]
= 100 (2x – 3)⁹⁹ [2 (1) – 0] = 200 (2x – 3)⁹⁹

Question 15.
\(\sqrt{3 x+2}\)
Solution:
Let y = \(\sqrt{3 x+2}\) ; diff. both sides w.r.t. x, we get
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\)\((3 x+2)^{\frac{1}{2}}\) = \(\frac{1}{2}(3 x+2)^{\frac{1}{2}-1} \frac{d}{d x}(3 x+2)\)
= \(\frac{1}{2}(3 x+2)^{-\frac{1}{2}}(3+0)\) = \(\frac{3}{2} \frac{1}{\sqrt{3 x+2}}\)
[∵ \(\frac { d }{ dx }\) (ax + b)ⁿ = n (ax + b)^n-1 \(\frac { d }{ dx }\) (ax + b)]

Question 16.
Given f(x) = \(\frac { 7 }{ 4 }\)x², find f ‘ \(\left(\frac{1}{7}\right)\)
Solution:
Given f(x) = \(\frac { 7 }{ 4 }\)x²;diff. both sides w.r.t. x, we have
f ‘ (x) = \(\frac { 7 }{ 4 }\) × 2x = \(\frac { 7 }{ 2 }\)x
⇒ f ‘ \(\left(\frac{1}{7}\right)\) = \(\frac{7}{2}\) × \(\frac{1}{7}\) = \(\frac{1}{2}\)

Question 17.
Find the derivative with respect to x of the following:
(i) x – \(\frac { 1 }{ x }\)
(ii) √x + \(\frac{1}{\sqrt{x}}\)
(iii) 3x² + \(\frac{3}{x^2}\)
(iv) \(\frac{x^2+1}{x}\)
(v) \(\frac{2 x+x^4}{x^2}\)
(vi) \(\frac{1+x^2}{x^3}\)
Solution:
(i) Let y = x – \(\frac { 1 }{ x }\); diff. both sides, w.r.t. x, we get
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\)(x) – \(\frac { d }{ dx }\) x^-1 = 1 – (-1) x^-1-1 = 1 + \(\frac{1}{x^2}\)

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

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

(iv) Let y = \(\frac{x^2+1}{x}\); diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) x + \(\frac { d }{ dx }\) x^-1 = 1 + (-1)x^-1-1 = 1 – \(\frac{1}{x^2}\)

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

(vi) Let = \(\frac{1+x^2}{x^3}\) = \(\frac{1}{x^3}\) + \(\frac{1}{x}\); diff. both sides w.r.t. x
∴ \(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (x^-3) + \(\frac { d }{ dx }\) x^-1 = 3x^-3-1 + (-1)^-1-1 = –\(\frac{3}{x^4}\) – \(\frac{1}{x^2}\)

Question 18.
If y = x + \(\frac { 1 }{ x }\), prove that x²\(\frac { dy }{ dx }\) – xy + 2 = 0
Solution:
Given y = x + \(\frac { 1 }{ x }\); diff, both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) x + \(\frac { d }{ dx }\) x^-1 = 1 + (-1)x^-1-1 = 1 – \(\frac{1}{x^2}\)
⇒ x²\(\frac { dy }{ dx }\) = x²\(\frac { dy }{ dx }\) – xy + 2 = x² – 1 – x\(\left(x+\frac{1}{x}\right)\) + 2
= x² – 1 x² – 1 + 2 = 0 = R.H.x.

Question 19.
If y= \(\sqrt{x}\) – \(\frac{1}{\sqrt{x}}\), prove that 2x\(\frac{d y}{d x}\) + y = \(2 \sqrt{x}\).
Solution:

Question 20.
If y = \(\frac{1}{a-z}\), show that \(\frac{d z}{d y}\) = (z – a)²
Solution:
Given y = \(\frac{1}{a-z}\)
diff. both sides w.r.t. z, we have
\(\frac { dy }{ dz }\) = \(\frac { d }{ dz }\)\(\left(\frac{1}{a-z}\right)\) = \(\frac { d }{ dz }\)(a – z)^-1 = (-1)(a – z)^-1-1 \(\frac { d }{ dz }\) (a – z)
= 1 (a – z)^-2 (0 – 1) = \(\frac{1}{(a-z)^2}\)
∴ \(\frac { dz }{ dy }\) = \(\frac{1}{\frac{d y}{d z}}\) = (a – z)²

Question 21.
If y = 1 + x + \(\frac{x^2}{2 !}\) + \(\frac{x^3}{3 !}\) + \(\frac{x^4}{4 !}\) + …… to ∞, show that \(\frac { dy }{ dx }\) = y.
Solution:
Given y = 1 + x + \(\frac{x^2}{2 !}\) + \(\frac{x^3}{3 !}\) + \(\frac{x^4}{4 !}\) + …. ∞ …(1)
Diff. both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = 0 + 1 + \(\frac{x^2}{2 !}\) + \(\frac{x^3}{3 !}\) + \(\frac{x^4}{4 !}\) + …. ∞ …(1)
= 1 + x + \(\frac{x^2}{2}\) + \(\frac{x^3}{6}\) …. ∞
= 1 + x + \(\frac{x^2}{2 !}\) + \(\frac{x^3}{3 !}\) + ….. ∞ = y
Thus \(\frac { dy }{ dx }\) = y
Aliter = Given y = 1 + x + \(\frac{x^2}{2 !}\) + \(\frac{x^3}{3 !}\) + \(\frac{x^4}{4 !}\) + …. ∞ = e^x
Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dy }\) e^x = e^x = y