NCERT Solutions
Class 11 Maths
Limits and Derivatives
Ex.13.2 Q.11
Find the derivative of the following functions:
1. sin x cos x
2. sec x
3. 5 sec x + 4 cos x
4. cosec x
5. 3cot x + 5cosec x
6. 5sin x – 6cos x + 7
7. 2tan x – 7sec x
(1) Let y = sin x cos x
Now, 
= d (sin x cos x) ÷ dx
=> = {d (sin x) ÷ dx} × cos x + {sin x × d (cos x) ÷ dx}
=> = cos x × cos x + sin x × (-sin x)
[Since {d (sin x) ÷ dx} = cos x, {d (cos x) ÷ dx} = -sin x] => = cos2 x – sin2 x
=> = cos 2x
[Since cos 2x = cos2 x – sin2 x]
(2) Let f(x) = sec x
Now, from the first principle, f’(x) = limh->0 [{f (x+ h) – f(x)} ÷ h] = limh->0 [{sec (x + h) – sec x} ÷ h]
= limh->0 [{(1 ÷ cos (x + h)) – (1 ÷ cos x)} ÷ h]
= limh->0 {cos x - cos (x + h)} ÷ [{cos x × cos (x + h) × h]
= (1 ÷ cos x) × limh->0 [{(-2 × sin (x + x + h) ÷ 2) × (sin (x - x - h) ÷ 2)} ÷ {h × cos (x + h)}]
= (1 ÷ cos x) × limh->0 [{(-2 × sin (2x + h) ÷ 2) × (sin (-h) ÷ 2)} ÷ {h × cos (x + h)}]
= (1 ÷ cos x) × limh->0 [{(2 × sin (2x + h) ÷ 2) × (sin (h) ÷ 2)} ÷ {h × cos (x + h)}]
= (1 ÷ cos x) × limh->0 [{sin (h) ÷ 2] ÷ (
) × limh->0 [{sin (2x + h) ÷ 2} ÷ cos (x + h)]
= (1 ÷ cos x) × 1 × (sin x ÷ cos x)
= sec x × tan x
So, f’(x) = sec x × tan x
(3) Let y = 5 sec x + 4 cos x
Now, = d (5 sec x + 4 cos x) ÷ dx
=> = {d (5 sec x) ÷ dx} + {d (4 cos x) ÷ dx}
=> = 5 × sec x × tan x – 4 × sin x
(4) Let f(x) = cosec x
Now, from the first principle, f’(x) = limh->0 [{f (x + h) – f(x)} ÷ h]
= limh->0 [{cosec (x + h) – cosec x} ÷ h]
= limh->0 [{(1 ÷ sin (x + h)) – (1 ÷ sin x)} ÷ h]
= limh->0 {sin x - sin (x + h)} ÷ [{sin x × sin (x + h) × h]
= (1 ÷ sin x) × limh->0 [{(2 × cos (x + x + h) ÷ 2) × (sin (x - x - h) ÷ 2)} ÷ {h × sin (x + h)}]
= (1 ÷ sin x) × limh->0 [{(2 × cos (2x + h) ÷ 2) × (sin (-h) ÷ 2)} ÷ {h × sin (x + h)}]
= -(1 ÷ sin x) × limh->0 [{(2 × cos (2x + h) ÷ 2) × (sin(h) ÷ 2)} ÷ {h × sin (x + h)}]
= -(1 ÷ sin x) × limh->0 [{sin (h) ÷ 2] ÷ (h ÷ 2) × limh->0 [{cos (2x + h) ÷ 2} ÷ sin (x + h)]
= -(1 ÷ sin x) × 1 × (cos x ÷ sin x)
= -cosec x × cot x
So, f’(x) = -cosec x × cot x
5.Let f(x) = 3cot x + 5cosec x
Now, 
= d (3cot x + 5cosec x) ÷ dx
=> f’(x) = {d (3cot x) ÷ dx} + {d (5cosec x) ÷ dx}
=> f’(x) = -3cosec2 x – 5cosec x × cot x
[Since {d (cot x) ÷ dx} = -cosec2 x]
6.Let f(x) = 5sin x – 6cos x + 7
Now, = d (5sin x – 6cos x + 7) ÷ dx
=> f’(x) = (d (5sin x) ÷ dx) – (d (6cos x) ÷ dx) + (d (7) ÷ dx)
=> f’(x) = 5cos x – 6(-sin x) + 0
=> f’(x) = 5cos x + 6sin x
7. Let f(x) = 2tan x – 7sec x
Now, = d (2tan x – 7sec x) ÷ dx
=> f’(x) = (d (2tan x) ÷ dx) – (d (7sec x) ÷ dx)
=> f’(x) = 2sec2 x – 7sec x tan x
[Since d (tan x) ÷ dx = sec2 x, d (sec x) ÷ dx = sec x tan x]