# 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 offor 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 offor 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,

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