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Last updated at Nov. 30, 2019 by Teachoo

Transcript

Example 22 Find the derivative of (i) (x^5 − cosx)/sinx Let f(x) = (x^5 − cosx)/sinx Let u = x5 – cos x & v = sin 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 = x5 – cos x u’ = 5. x5 – 1 – ( – sin x) = 5x4 + sin x v = sin x v’ = cos x Now, f’(x) = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 Derivative of xn is nxn – 1 & Derivative of cos x = – sin x (Derivative of sin x = cos x) = ((5x4 + sin〖x) sin x −(cos x)(x5 − cos x) 〗)/sin2x = (5x4 sin〖x + sin 2x − cos x . x5 + cos2 x〗)/(sin2 x) = (−x5 cos〖x + 5x4 sinx + 𝐬𝐢𝐧𝟐 𝐱 + 𝐜𝐨𝐬𝟐 𝐱〗)/(sinx )2 = (−x5 cos〖x + 5x4 sinx + 𝟏〗)/(sinx )2 Thus, f’(x) = (−𝐱𝟓 𝐜𝐨𝐬〖𝐱 + 𝟓𝐱𝟒 𝐬𝐢𝐧𝒙 + 𝟏〗)/(𝐬𝐢𝐧𝐱 )𝟐 (Using sin2x + cos2x = 1) Example 22 Find the derivative of (ii) (𝑥 + 𝑐𝑜𝑠𝑥)/𝑡𝑎𝑛𝑥 Let f(x) = (𝑥 + 𝑐𝑜𝑠𝑥)/𝑡𝑎𝑛𝑥 Let u = x + cos x & v = tan 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 = x + cos x u’ = (x + cos x)’ = 1 – sin x v = tan x v’ = sec2x Now, f’(x) = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 = ((𝟏 −〖 𝐬𝐢𝐧〗〖𝒙) (𝐭𝐚𝐧〖𝒙) − 𝒔𝒆𝒄𝟐𝒙 (𝒙 + 〖 𝐜𝐨𝐬〗〖𝒙)〗 〗 〗)/〖(𝐭𝐚𝐧〖𝒙)〗〗^𝟐 (xn)’ = n xn – 1 Derivative of cos x = –sin x Derivative of tan x = sec2x (Calculated in Example 17)

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About the Author

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.