NCERT Solutions For Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.2

Class 11 Maths NCERT Solutions for Chapter 13 Limits and Derivatives Exercise 13.2

Limits and Derivatives Exercise 13.2 Solutions

1. Find the derivative of x² – 2 at x = 10.

Solution

Let f(x) = x² – 2. Accordingly, 

Thus, the derivative of x² – 2 at x = 10 is 20 .

2. Find the derivative of 99x at x = 100.

Solution

Let f(x) = 99x . Accordingly, 

Thus, the derivative of x at x = 1 is  1.

3. Find the derivative of x at x = 1.

Solution

Let f(x) = 99x, Accordingly,  

Thus, the derivative of 99x at x = 100 is 99.

4. Find the derivative of the following functions from first principle.
(i) x³ – 27
(ii) (x – 1) (x – 2) 
(iii) 1/x² 
(iv) (x + 1)/(x - 1)

Solution

(i) Let f(x) = x³ - 27, Accordingly, from the first principle,  

(ii) Let f(x) = (x - 1)/(x - 2)  Accordingly, from the first principle,  

= (2x + 0 - 3) 
= 2x - 3

(iii) Let f(x) = 1/x², Accordingly, from the first principle,  

(iv) let f(x) = (x + 1)/(x - 1). Accordingly, from the first principle,  

5. For the function
f(x) = x¹⁰⁰ /100 + x⁹⁹ /99 + ....+ x² /2 + x + 1
Prove that f '(1) = 100 f '(0)

Solution

The given function is  

At x = 0, 
f '(0) = 1 
At x = 1, 
f '(1) = 1⁹⁹ + 1⁹⁸ + ... + 1 +1 = [1 + 1 +....+ 1 + 1] 100 terms  = 1 × 100 = 100 
Thus, f '(1) = 100 × f¹ (0)

6. Find the derivative of xⁿ + ax^n-1 + a² x^n-2 + .....+ a^n-1 x + aⁿ for some fixed real number a 

Solution

Let f(x) = xⁿ + ax^n-1 + a² x^n-2 + ......+ a^n-1 x + aⁿ  

7. For some constants a and b, find the derivative of
(i) (x – a) (x – b)
(ii) (ax² + b)² 
(iii) (x- a)/(x - b)  

Solution

(i) Let f(x)  = (x - a)(x -b) 
⇒ f(x) = x² - (a +b)x + ab 

f '(x) = 2x - (a + b) + 0 = 2x - a -b

(ii) Let f(x) = (ax² + b²)
⇒ f(x) = a² x⁴ + 2abx² + b² 

f '(x) = a²(4x³) + 2ab (2x) + b² (0) 
= 4a² x³ + 4abx 
= 4ax(ax² + b)

8. Find the derivative of  (xⁿ - aⁿ )/(x-a) for some constant a. 

Solution

9. Find the derivative of
(i) 2x - 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) 2/(x + 1) - x²/(3x - 1)

Solution

= 2 - 0 
= 2

(ii) Let f(x) = (5x³ + 3x - 1)(x- 1) 
By Leibnitz product rule,  

= (5x³ + 3x - 1)(1) + (x - 1) (5.3x² + 3 - 0) 
- (5x³ + 3x - 1) + (x - 1)(15x³ + 3) 
= 5x³ + 3x - 1 + 15x³ + 3x - 15x² - 3 
= 20x³ - 15x² + 6x - 4

(iii) Let f(x) = x^-3 (5 + 3x)
By Leibnitz product rule, 

(iv) Let f(x) = x⁵ (3 - 6x^-9) 
By Leibnitz product rule,  

= x⁵ [0 - 6(-9)x^-9-1] + (3 - 6x^-9)(5x⁴) 
= x⁵ (54x^-10) + 15x⁴ - 30x^-5
= 54x^-5 + 15x⁴ - 30x^-5
= 24x^-5 + 15x⁴
= 15x⁴ + 24/x⁵

(v) Let f(x) = x^-4 (3 - 4x^-5) 
By Leibnitz product rule,  

= x^-4 [0 - 4(-5)x^-5-1] + (3 - 4x^-5)(-4)x^-4-1 

= x^-4 (20x^-6)+ (3 -4x^-5)(-4x^-5) 
= 20x^-10 - 12x^-5 + 16x^-10 
= 36x^-10 - 12x^-5 
= -12/x⁵ + 36/x¹⁰ 

10. Find the derivative of cos x from first principle. 

Solution

Let f(x) = cos x, Accordingly, from the first principle,  

= - sin x
∴ f '(x) = -sin x

11. Find the derivative of the following functions:
(i) sin x cos x 
(ii) sec x 
(iii) 5 sec x + 4 cos x
(iv) cosec x 
(v) 3cot x + 5cosec x
(vi) 5sin x – 6cos x + 7
(vii) 2tan x – 7sec x 

Solution

(i) Let f(x) = sin x cos x. Accordingly, from the first principle,  

= cos (2x + 0).1 
= cos 2x

(ii) Let f(x) = sec x. Accordingly, from the first principle,  

= sec x tan x

(iii) Let f(x) = 5 sec x + 4cos x. Accordingly, from the first principle,  

(iv) Let f(x) = cosec x, Accordingly, from the first principle,  

(v) Let f(x) = 3cot x + 5cosec x. Accordingly, from the first principle,  

= - cosecx cot x  ...(3)
From (1), (2) and (3), we obtain  
f '(x) = -3cosec² x - 5cosec x cot x

(vi) Let f(x) = 5sin x - 6 cosx + 7. Accordingly, from the first principle,  

= 5cosx. 1 -6[(-cos x).(0) - sinx. 1] 
= 5 cos x + 6 sin x

(vii) Let f(x) = 2 tan x - 7 sec x. Accordingly, from the first principle,