EXERCISE 13.2
Page No 312:
Question 1:
Find the derivative of x2 – 2 at x = 10.Answer:
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.Answer:
Let f(x) = 99x. Accordingly,
Thus, the derivative of 99x at x = 100 is 99.
Question 3:
Find the derivative of x at x = 1.Answer:
Let f(x) = x. Accordingly,
Thus, the derivative of x at x = 1 is 1.
Question 4:
Find the derivative of the following functions from first principle.(i) x3 – 27 (ii) (x – 1) (x – 2)
(ii)


Answer:
(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


(iv) Let



Question 5:
For the function
Prove that

Answer:
The given function is
Thus,

Page No 313:
Question 6:
Find the derivative of
Answer:
Let

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

Answer:
(i) Let f (x) = (x – a) (x – b)
(ii) Let


(iii)


By quotient rule,

Question 8:
Find the derivative of
Answer:

By quotient rule,

Question 9:
Find the derivative of(i)

(iii) x–3 (5 + 3x) (iv) x5 (3 – 6x–9)
(v) x–4 (3 – 4x–5) (vi)

Answer:
(i) Let

(ii) Let f (x) = (5x3 + 3x – 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 f (x) = x–4 (3 – 4x–5)
By Leibnitz product rule,

(vi) Let f (x) =


By quotient rule,

Question 10:
Find the derivative of cos x from first principle.Answer:
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
Answer:
(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 f (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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