GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Question 1.
Find the derivative of x2 – 2 at x = 10.
Solution:
Derivative of f(x) at x = a is given by
f(a) = \(\lim _{h \rightarrow 0}\) \(\frac{f(a+h)-f(a)}{h}\)
Here, f(x) = x2 – 2
a = 10, ∴ f(10 + h) = (10 + h)2 – 2
and f(10) = 102 – 2.
∴ f(10 + h) – f(10) = [(10 + h)2 – 2] – (102 – 2)
= (10 + h)2 – 102 = (20 + h)h
∴ f(10) = \(\lim _{h \rightarrow 0}\) \(\frac{f(10+h)-f(10)}{h}\)
= \(\lim _{h \rightarrow 0}\) \(\frac{h(20+h)}{h}\).
= \(\lim _{h \rightarrow 0}\) (20 + h) = 20.

GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Question 2.
Find the derivative of 99x at x =100.
Solution:
Derivative of f(x) at x = 100 is
\(\lim _{h \rightarrow 0}\) = \(\frac{f(100+h)-f(100)}{h}\)
Now, f(x) = 99x.
∴ f(100 + h) = 99(100 + h).
and f(100) = 99 × 100.
∴ f(100 + h) – f(100) = 99(100 + h) – 99 × 100
= 99[100 + h – 100] = 99 × h
∴ f(100) = \(\lim _{h \rightarrow 0}\) \(\frac{99h}{h}\) = 99.

GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Question 3.
Find the derivative of x at x = 1.
Solution:
Derivative of f(x) = x at x = 1
f'(1) = \(\lim _{h \rightarrow 0}\) = \(\frac{f(x+h)-f(x)}{h}\).
= \(\lim _{h \rightarrow 0}\) \(\frac{(1+h)-1}{h}\)
= \(\lim _{h \rightarrow 0}\) \(\frac{h}{h}\) = 1.

Question 4.
Find the derivative of the following functions from first principles:
(i) x3 – 27
(ii) (x – 1)(x – 2)
(iii) \(\frac{1}{x^{2}}\)
(iv) \(\frac{x+1}{x-1}\)
Solution:
(i) f(x) = x3 – 27, Derivative f'(x) by first principle is given by
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 1
= 3x2

(ii) f(x) = (x – 1)(x – 2) = x2 – 3x + 2
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 2

(iii) f(x) = \(\frac{1}{x^{2}}\).
Derivative of f(x) is given by
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 3

(iv) f(x) = \(\frac{x+1}{x-1}\).
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 4

GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Question 5.
For the function f(x) = \(\frac{x^{100}}{100}\) + \(\frac{x^{99}}{99}\) + …………… + \(\frac{x^{2}}{2}\) + x, prove that f ‘(1) = 100f ‘(0).
Solution:
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 5
Now, f ‘(1) = 1 + 1 + ………… to 100 terms = 100
Also, f ‘(0) = 1.
∴ f ‘(1) = 100 × 1 = 100 f ‘(0)
Hence, f ‘(1) = 100 f ‘(0).

Question 6.
Find the derivative of xn + axn-1 + a2xn-2 + …………. + an-1x + an for some fixed real number.
Solution:
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 6
∴ f(x) = nxn-1 + (n – 1)axn-2 + (n – 2)a2xn-3 + …………. + an-1.

GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Question 7.
For some constants a and b, find the derivative of:
(i) (x – a)(x – b)
(ii) (ax2 + b)2
(iii) \(\frac{x-a}{x-b}\)
Solution:
(i) f(x) = (x – a)(x – b)
We have:
f'(uv) = u’v + uv’.
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 7
∴ f'(x) = x – b + x – a = 2x – a – b.

(ii) f(x) = (ax2 + b)2 = a2x4 + 2abx2 + b2
Now, \(\frac{d}{dx}\) x4 = 4x3 and \(\frac{d}{dx}\) x2 = 2x and \(\frac{d}{dx}\) b2 = 0.
∴ f(x) = a2.4x3 + 2ab.2x + 0
= 4a2x3 + 4abx.

(iii)
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 8

Question 8.
Find the derivative of \(\frac{x^{n}-a^{n}}{x-a}\) for some constant a.
Solution:
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 9

GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Question 9.
Find the derivative of
(i) 2x – \(\frac{3}{4}\)
(ii) (5x3 + 3x – 1)(x – 1)
(iii) x-3(5 + 3x)
(iv) x5(3 – 6x-9)
(v) x-4(3 – 4x-5)
(vi) \(\frac{2}{x+1}\) – \(\frac{x^{2}}{3 x-1}\)
Solution:
(i) Let f(x) = 2x – \(\frac{3}{4}\).
∴ f ‘(x) = 2.1 [∵ \(\frac{d}{dx}\) x = 1, \(\frac{d}{dx}\) (\(\frac{3}{4}\)) = 0]
= 2

(ii) Let f(x) = (5x3 + 3x – 1)(x – 1)
(uv)’ = u’v + uv’
∴ f(5x3 + 3x – 1)(x – 1)
= [\(\frac{d}{dx}\) (5x3 + 3x – 1)] (x – 1) + (5x3 + 3x – 1) \(\frac{d}{dx}\) (x – 1)
= (15x2 + 3)(x – 1) + (5x3 + 3x – 1).1
= x(15x2 + 3) – (15x2 + 3) + (5x3 + 3x – 1)
= 20x3 – 15x2 + 6x – 4.

(iii) Let f(x) = x-3(5 + 3x) = 5x-3 + 3x-2
∴ f ‘(x) = 5(- 3) x-4 + 3(- 2)x-3
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 10

(iv) Let f(x) = x5(3 – 6x-9) = 3x5 – 6x-4
f ‘(x) = 3 × 5x4 – 6 × (- 4)x-5
= 15x4 + 24x-5
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 11

(v) Let f(x) = x-4(3 – 4x-5) = 3x-4 – 4x-9
f ‘(x) = 3.(- 4)x-5 – 4. (- 9)x-10
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 12

(vi) Let f(x) = \(\frac{2}{x+1}\) – \(\frac{x^{2}}{3 x-1}\)
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 13

GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Question 10.
Find the derivative of cos x from the first principle.
Solution:
f(x) = cos x.
∴ By first principle,
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 14

Question 11.
Find the derivative of the following functions:
(i) sin x cos x
(ii) sec x
(iii) 5secx + 4cos x
(iv) cosec x
(v) 3cot x + 5cosec x
(vi) 5sinx – 6cosx + 7
(vii) 2tanx – 7 secx
Solution:
(i) Let f(x) = sin x.cos x
So, f ‘(x) = u’v + uv’
= (\(\frac{d}{dx}\) sin x)cos x + sin x \(\frac{d}{dx}\)(cos x)
= cos x. cos x + sin x(- sin x)
= cos2x – sin2x = cos 2x.

(ii)
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 15
= sec x. tan x

Alternative method:
GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2 img 16

(vi) \(\frac{d}{dx}\) sin x = cos x and \(\frac{d}{dx}\) cos x = – sin x.
∴ \(\frac{d}{dx}\) (5sin x – 6cos x + 7) = 5cos x + 6sin x = 0
= 5cos x + 6sin x.

(vii) \(\frac{d}{dx}\) (sec x) = sec x tan x
and \(\frac{d}{dx}\)(tan x) = sec2x
∴ \(\frac{d}{dx}\)(2tan x – 7 secx) = 2sec2x – 2 sec x tan x
= sec x(2 sec x – 7 tan x).

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