# Ex 13.2, 11 - Chapter 13 Class 11 Limits and Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 13.2, 11 Find the derivative of the following functions: (i) sin x cos x Let f (x) = sin x cos x. Let u = sin x & v = cos x f(x) = uv So, f (x) = (uv) = u v + v u Here, u = sin x So, u = cos x & v = cos x So, v = sin x Now, f (x) = (uv) = u v + v u = cos x . cos x + ( sin x) sin x = cos2x sin2x = cos 2x Hence f (x) = cos 2x Ex13.2, 11 Find the derivative of the following functions: (ii) sec x Let f (x) = sec x f(x) = 1 cos Let u = 1 & v = cos x So, f(x) = f (x) = using quotient rule f (x)= 2 Finding u & v u = 1 u = 0 & v = cos x v = sin x Now, f (x) = = 2 = 0( cos ) ( sin ) (1) 2 = 0 + sin 2 = sin 2 = sin cos . 1 cos = tan x . sec x Hence f (x) = tan x . sec x Ex13.2, 11 Find the derivative of the following functions: (iii) 5 sec x + 4 cos x Let f (x) = 5 sec x + 4 cos x. Now, f (x) = ( 5 sec x + 4 cos x) = (5 sec x) + (4 cos x) = 5 (sec x . tan x) + 4 ( sin x) = 5 sec x . tan x 4 sin x Ex 13.2, 11 Find the derivative of the following functions: (iv) cosec x Let f (x) = cosec x f(x) = 1 sin Let u = 1 & v = sin x f(x) = So, f (x) = Using quotient rule f (x) = 2 Finding u & v u = 1 u = 0 & v = sin x v = cos x Now, f (x) = = 2 = 0 ( sin ) cos (1) 2 = 0 2 = 2 = sin . 1 sin = cot x cosec x = cosec x cot x Hence f (x) = cosec x cot x Ex13.2, 11 Find the derivative of the following functions: (v) f (x) = 3cot x + 5cosec x. Given f (x) = 3cot x + 5cosec x Now, f (x) = (3cot x + 5cosec x) = 3(cot x) + 5( cosec x) = 3cosec2x 5 cot x cosec x = 3 cosec2x 5 cosec x cot x Ex13.2, 11 Find the derivative of the following functions: (vi) f (x) = 5sin x 6 cos x + 7. Let f (x) = 5sin x 6cos x + 7 f (x) = (5 sin x 6 cos x + 7) = (5 sin x) (6 cos x) + (7) = 5 cos x 6 ( sin x ) + 0 = 5 cos x + 6 sin x Ex13.2, 11 Find the derivative of the following functions: (vii) f (x) = 2 tan x 7 sec x f (x) = 2 tan x 7 sec x. Now, f (x) = (2 tan x 7 sec x) = (2tan x) (7sec x) = 2 (tan x) 7 (sec x) = 2 sec2 x 7 (sec x tan x) = 2 sec2 x 7 sec x tan x

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.