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

Last updated at Nov. 30, 2019 by Teachoo

Last updated at Nov. 30, 2019 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 (𝐷𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑠𝑖𝑛〖𝑥=𝑐𝑜𝑠𝑥 〗) & 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 (𝐷𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑐𝑜𝑠〖𝑥=〖− 𝑠𝑖𝑛〗𝑥 〗) (𝐷𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑐𝑜𝑠〖𝑥=〖− 𝑠𝑖𝑛〗𝑥 〗) Ex 13.2, 11 Find the derivative of the following functions: (ii) sec x Let f (x) = sec x f(x) = 1/cos𝑥 Let u = 1 & v = cos x So, f(x) = 𝑢/𝑣 ∴ f’(x) = (𝑢/𝑣)^′ Using quotient rule f’(x) = (𝑢^′ 𝑣 − 𝑣^′ 𝑢)/𝑣^2 Finding u’ & v’ u = 1 u’ = 0 & v = cos x v’ = – sin x Now, f’(x) = (𝑢^′ 𝑣 − 𝑣^′ 𝑢)/𝑣^2 = (0(cos〖𝑥) − (−sin〖𝑥) (1)〗 〗)/(〖𝑐𝑜𝑠〗^2 𝑥) (Derivative of constant is 0) (Derivative of cos x = – sin x) = (0 +〖 sin〗𝑥)/(〖𝑐𝑜𝑠〗^2 𝑥) = 〖 sin〗𝑥/(〖𝑐𝑜𝑠〗^2 𝑥) = 〖 sin〗𝑥/cos𝑥 . 1/cos𝑥 = tan x . sec x Hence f’(x) = tan x . sec x Using tan θ = sin𝜃/𝑐𝑜𝑠𝜃 & 1/cos𝜃 = sec θ Ex 13.2, 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 Derivative of sec x = sec x tan x (calculated in Ex 13.2.11 (ii)) & Derivative of cos x = – sin x Ex 13.2, 11 Find the derivative of the following functions: (iv) cosec x Let f (x) = cosec x f(x) = 1/sin𝑥 Let u = 1 & v = sin x ∴ f(x) = 𝑢/𝑣 So, f’(x) = (𝑢/𝑣)^′ Using quotient rule f’(x) = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 Finding u’ & v’ u = 1 u’ = 0 & v = sin x v’ = cos x Now, f’(x) = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 = (0 (sin〖𝑥) −〖 cos〗〖𝑥 (1)〗 〗)/(〖𝑠𝑖𝑛〗^2 𝑥) (Derivative of constant function = 0) (Derivative of sin x = cos x) = (0 − 𝑐𝑜𝑠 𝑥)/(〖𝑠𝑖𝑛〗^2 𝑥) = (− 𝑐𝑜𝑠 𝑥)/(〖𝑠𝑖𝑛〗^2 𝑥) = (− 𝑐𝑜𝑠 𝑥)/sin𝑥 . 1/sin𝑥 = – cot x cosec x = – cosec x cot x Hence f’(x) = – cosec x cot x Using cot x = 𝑐𝑜𝑠/sin𝑥 & 1/sin𝑥 = cosec x Ex 13.2, 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)’ = – 3 cosec2x – 5 cosec x cot x Derivative of cot x = – cosec2x Calculated in Example 21(ii) Derivative of cosec x = – cot x cosec x Calculated in 13.2,11 part (iv) Ex 13.2, 11 Find the derivative of the following functions: (vi) f (x) = 5sin x – 6 cos x + 7. Let f (x) = 5 sin x – 6 cos x + 7 f’(x) = (5 sin x – 6 cos x + 7)’ = (5 sin x)’ – (6 cos x)’ + (7)’ Using (sinx)’ = cos x , (cos x)’ = – sinx , & Derivative of constant = 0 = 5 cos x – 6 (–sin x) + 0 = 5 cos x + 6 sin x Ex 13.2, 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 Derivative of tan x = sec2x Calculated in Example 17 Derivative of sec x = sec x tan x Calculated in Ex 13.2, 11 part (ii)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.