# NCERT solution class 11 chapter 13 Limits and Derivatives exercise 13.2 mathematics

## EXERCISE 13.2

#### Question 1:

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

Let f(x) = x2 – 2. Accordingly, Thus, the derivative of x2 – 2 at x = 10 is 20.

#### Question 2:

Find the derivative of 99x at x = 100.

Let f(x) = 99x. Accordingly, Thus, the derivative of 99x at x = 100 is 99.

#### Question 3:

Find the derivative of at = 1.

Let f(x) = x. Accordingly, Thus, the derivative of at = 1 is 1.

#### Question 4:

Find the derivative of the following functions from first principle.

(i) x3 – 27 (ii) (x – 1) (– 2)

(ii) (iv) (i) Let f(x) = x3 – 27. Accordingly, from the first principle, (ii) Let f(x) = (x – 1) (x – 2). Accordingly, from the first principle, (iii) Let . Accordingly, from the first principle, (iv) Let . Accordingly, from the first principle,  #### Question 5:

For the function Prove that The given function is Thus, #### Question 6:

Find the derivative of for some fixed real number a.

Let  #### Question 7:

For some constants a and b, find the derivative of

(i) (– a) (x – b) (ii) (ax2 + b)2 (iii) (i) Let f (x) = (– a) (x – b) (ii) Let  (iii)  By quotient rule, #### Question 8:

Find the derivative of for some constant a. By quotient rule, #### Question 9:

Find the derivative of

(i) (ii) (5x3 + 3– 1) (x – 1)

(iii) x–3 (5 + 3x) (iv) x5 (3 – 6x–9)

(v) x–4 (3 – 4x–5) (vi) (i) Let  (ii) Let f (x) = (5x3 + 3– 1) (x – 1)

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

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

By Leibnitz product rule, (v) Let (x) = x–4 (3 – 4x–5)

By Leibnitz product rule, (vi) Let (x) =  By quotient rule, #### Question 10:

Find the derivative of cos x from first principle.

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

(i) Let f (x) = sin x cos x. Accordingly, from the first principle, (ii) Let f (x) = sec x. Accordingly, from the first principle, (iii) Let f (x) = 5 sec x + 4 cos x. Accordingly, from the first principle, (iv) Let f (x) = cosec x. Accordingly, from the first principle, (v) Let (x) = 3cot x + 5cosec x. Accordingly, from the first principle,   From (1), (2), and (3), we obtain (vi) Let f (x) = 5sin x – 6cos x + 7. Accordingly, from the first principle, (vii) Let f (x) = 2 tan x – 7 sec x. Accordingly, from the first principle, error: Content is protected !! 