Find the derivative of following functions:
(i) sin x cos x (ii) sec x (iii) 5sec x + 4cos x
(iv) cosec x (v) 3cot x + 5 cosec x (vi) 5sin x - 6cos x + 7
(vii) 2 tan x - 7sec x
Solution:
(i) Let f (x) = sin x cos x.
Accordingly, from the first principle,
f' (x) = limₕ→₀ [f (x + h) - f (x)]/h
= limₕ→₀ [sin (x + h)cos(x + h) - sin x cos x]/h
Multiply and divide by 2,
= limₕ→₀ 1/(2h)[2 sin (x + h)cos(x + h) - 2 sin x cos x]
Using double angle formulas,
= limₕ→₀ 1/(2h)[sin 2 (x + h) - sin 2x]
Using the product to sum formulas,
= limₕ→₀ 1/2h[2 cos (2x + 2h + 2x)/2 sin (2x + 2h - 2x)/h]
= limₕ→₀ 1/2h[2 cos (4x + 2h)/2 .sin (2h)/2]
= limₕ→₀ 1/h[cos (2x + h) sin h]
f' (x) = limₕ→₀ cos(2x + h).limₕ→₀ sin h/h
= cos(2x + 0).1
= cos 2x
(ii) Let f (x) = sec x
Accordingly, from the first principle,
f' (x) = limₕ→₀ [f (x + h) - f (x)]/h
f' (x) = limₕ→₀ [sec (x + h) - sec x]/h
= limₕ→₀ 1/h[1/(cos (x + h) - 1/cos x)]
= limₕ→₀ 1/h [cos x - cos(x + h)] / [cos x cos(x + h)]
Using the product to sum formulas,
= 1/cos x limₕ→₀ 1/h [- 2 sin (x + x + h)/2 sin (x - x - h)/2] / [cos(x + h)]
= 1/cos x limₕ→₀ 1/h [- 2 sin (2x + h)/2 sin (- h)/2] / [cos(x + h)]
Multiply and divide by h/2,
= 1/cos x limₕ→₀ (1/h) (h/2) [- 2 sin (2x + h)/2 sin (- h/2)/(h/2)]/[cos(x + h)]
= 1/cos x limₕ/₂→₀ sin (h/2)/(h/2). limₕ→₀ (sin(2x + h)/2)/cos(x + h)
= 1/cos x. 1. sin x/cos x
= sec x . tan x
(iii) Let f (x) = 5sec x + 4cos x
Accordingly, from the first principle,
f' (x) = limₕ→₀ [f (x + h) - f (x)]/h
f' (x) = limₕ→₀ [5sec(x + h) + 4 cos(x + h) - (5sec x + 4 cos x)]/h
= 5 limₕ→₀ [sec (x + h) - sec x]/h + 4 limₕ→₀ [cos (x + h) - cos x]/h
= 5 limₕ→₀ 1/h[1/cos(x + h) - 1/cos x] + 4 limₕ→₀ 1/h[cos x cos h - sin x sin h - cos x]
= 5/cos x limₕ→₀ 1/h[- 2 sin (2x + h)/2 sin (- h)/2]/[cos(x + h)] + 4 [- cos limₕ→₀ (1 - cos x)/h - sin x limₕ→₀ sin h/h]
= 5/cos x limₕ→₀ [sin (2x + h)/2 sin (- h/2)/(h/2)]/[cos(x + h)] + 4 [- cos x(0) - sin x(1)]
= 5/cos x. limₕ→₀ [sin (h/2)/(h/2). limₕ→₀ (sin(2x + h)/2)/cos(x + h) - 4 sin x
= 5/cos x. sin x/cos x. 1 - 4 sin x
= 5sec x tan x - 4 sin x
(iv) Let f (x) = cosec x
Accordingly, from the first principle,
f' (x) = limₕ→₀ [f (x + h) - f (x)]/h
f' (x) = limₕ→₀ 1/h[cosec (x + h) - cosec x]
= limₕ→₀ 1/h[1/(sin (x + h) - 1/sin x)]
= limₕ→₀ 1/h[sin x - sin(x + h)]/[sin x sin(x + h)]
= limₕ→₀ 1/h[2 cos (x + x + h)/2. sin (x - x - h)/2]/[sin x sin(x + h)]
= limₕ→₀ 1/h[2 cos (2x + h)/2. sin (- h)/2]/[sin x sin(x + h)]
= limₕ→₀ [- cos (2x + h)/2. sin (- h/2)/(h/2)]/[sin x sin(x + h)]
= limₕ→₀ [- cos (2x + h)/2. sin (- h/2)/(h/2)]/[sin x sin(x + h)]. limₕ/₂→₀ sin (h/2)/(h/2)
= (- cos x/sinx sin x). 1
= - cosec x cot x
(v) Let f (x) = 3cot x + 5 cosec x
Accordingly, from the first principle,
f' (x) = limₕ→₀ [f (x + h) - f (x)]/h
= limₕ→₀ 1/h[3cot (x + h) + 5 cosec(x + h) - 3cot x - 5 cosec x]
= 3 limₕ→₀ 1/h[cot (x + h) - cot x] + 5 limₕ→₀ 1/h [cosec (x + h) - cosec x] ....(1)
Now,
limₕ→₀ 1/h[cot (x + h) - cot x] = limₕ→₀ 1/h [cos (x + h)/sin (x + h) - cos x/sin x]
= limₕ→₀ 1/h [sin (x - x - h)/sin x sin(x + h)]
= limₕ→₀ 1/h [sin (- h)/sin x sin(x + h)]
= limₕ→₀ - sin h/h. limₕ→₀ [1//sin x sin(x + h)]
= 1. 1//sin x sin(x + 0) = - 1/sin² x = - cosec² x ....(2)
limₕ→₀ 1/h[cosec (x + h) - cosec x] = limₕ→₀ 1/h [1/sin (x + h) - 1/sin x]
= limₕ→₀ 1/h [sin x - sin (x + h)/sin x sin (x + h)]
= limₕ→₀ 1/h [2 cos (x + x + h)/2. sin (- h/2)]/[sin x sin (x + h)]
= x limₕ→₀ 1/h[2 cos (2x + h)/2 sin (- h)/2]/[sin x sin (x + h)]
= x limₕ→₀ [cos (2x + h)/2. sin (- h/2)/(h/2)]/[sin x sin (x + h)]
= limₕ→₀ [- cos (2x + h)/2]/[sin x sin (x + h)]. limₕ→₀ (sin h/2)/(h/2)
= - cos x./(sin x/sin x). 1
= - cosec x cot x ....(3)
From (1), (2) and (3), we obtain
f' (x) = - 3 cosec2x - 5 cosec x cot x
(vi) Let f (x) = 5sin x - 6cos x + 7
Accordingly, from the first principle,
f' (x) = limₕ→₀ [f (x + h) - f (x)]/h
limₕ→₀ 1/h[5sin (x + h) - 6 cos(x + h) + 7 - 5sin x + 6 cos x - 7]
= 5 limₕ→₀ 1/h [sin (x + h) - sin x] - 6 limₕ→₀ 1/h [cos (x + h) - cos x]
= 5 limₕ→₀ 1/h [2 cos (x + h + x)/2. sin (x + h - x)/2] - 6 limₕ→₀ 1/h [cos x cos h - sin x sin h - cos x]
= 5 limₕ→₀ 1/h [2 cos (2x + h)/2. sin (h)/2] - 6 limₕ→₀ 1/h [- cos x (1 - cos h) - sin x sin h]
f' (x) = 5 limₕ→₀ 1/h [cos (2x + h)/2. (sin (h)/2)/(h/2)] - 6 limₕ→₀ 1/h [- cos x (1 - cos h)/h - (sin x sin h)/h)]
= 5 [limₕ→₀ cos (2x + h)/2] [limₕ→₀ (sin (h)/2)/(h/2)] - 6 [- cos x (limₕ→₀ (1 - cos h)/h) - sin x (limₕ→₀ sin h)/h)]
= 5 cos x.1 - 6 [(- cos x).(0) - sin x.1]
= 5 cos x + 6 sin x
(vii) Let f (x) = 2 tan x - 7sec x
Accordingly, from the first principle,
f' (x) = limₕ→₀ [f (x + h) - f (x)]/h
= limₕ→₀ 1/h [2 tan (x + h) - 7 sec(x + h) - 2 tan x + 7 sec x]/h
= 2 limₕ→₀ 1/h [tan (x + h) - tan x] - 7 limₕ→₀ 1/h [sec (x + h) - sec x]
= 2 limₕ→₀ 1/h [cos (x + h)/sin (x + h) - cos x/sin x] - 7 limₕ→₀ 1/h [1/cos (x + h) - 1/cos x]
= 2 limₕ→₀ 1/h [cos x sin (x + h) - sin x cos(x + h)]/[cos x cos(x + h)] - 7 limₕ→₀ 1/h [cos x - cos (x + h)]/[cos x cos(x + h)]
= 2 limₕ→₀ 1/h [sin (x + h - x)/cos x cos(x + h)] - 7 limₕ→₀ 1/h [- 2 sin (x + x + h)/2. sin (x - x - h)/2]/[cos x cos(x + h)]
= 2 [limₕ→₀ (sin h/h). 1/cos x cos(x + h)] - 7 limₕ→₀ 1/h [- 2 sin (2x + h)/2. sin (- h)/2]/[cos x cos(x + h)]
= 2 (limₕ→₀ (sin h/h) [limₕ→₀ 1/cos x cos(x + h)] - 7 [limₕ→₀ sin (h)/2/(h)/2] [limₕ→₀ sin (2x + h)/cos x cos(x + h)]
= 2 × 1 × 1(1/cos x cos x) - 7 x 1 (sin x/cos x cos x)
= 2 sec2 x - 7 sec x tan x
NCERT Solutions Class 11 Maths Chapter 13 Exercise 13.2 Question 11
Find the derivative of following functions: (i) sin x cos x (ii) sec x (iii) 5sec x + 4cos x (iv) cosec x (v) 3cot x + 5 cosec x (vi) 5sin x - 6cos x + 7 (vii) 2 tan x - 7sec x
Summary:
The derivatives of the following functions are (i) cos 2x (ii) sec x . tan x (iii) 5sec x tan x - 4 sin x (iv) - cosec x cot x (v) - 3 cosec2x - 5 cosec x cot x (vi) 5 cos x + 6 sin x (vii) 2 sec2 x - 7 sec x tan x
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