# Example 27

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 27 Find the derivative of f given by f (x) = tan–1 𝑥 assuming it exists. f (x) = tan–1 𝑥 Let y = tan–1 𝑥 tan𝑦 = 𝑥 ∴ 𝑥 = tan𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑(𝑥)𝑑𝑥 = 𝑑 tan𝑦𝑑𝑥 1 = 𝑑 tan𝑦𝑑𝑥 We need d𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = 𝑑 tan𝑦𝑑𝑥 × 𝑑𝑦𝑑𝑦 1 = 𝐬𝐞𝐜𝟐 𝒚 . 𝑑𝑦𝑑𝑥 1 = (𝟏 + 𝒕𝒂𝒏𝟐𝒚) 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = 11 + 𝐭𝐚𝐧𝟐𝒚 Putting 𝑡𝑎𝑛𝑦 = 𝑥 𝑑𝑦𝑑𝑥 = 11 + 𝒙𝟐 Hence 𝒅( 𝐭𝐚𝐧−𝟏𝒙)𝒅𝒙 = 𝟏𝟏 + 𝒙𝟐 Find the derivative of f given by f (x) = cos–1 𝑥 assuming it exists. Let y = cos–1 𝑥 cos𝑦= 𝑥 𝑥 = cos𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑(𝑥)𝑑𝑥 = 𝑑 cos𝑦𝑑𝑥 1 = 𝑑 cos𝑦𝑑𝑥 We need 𝑑𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = 𝑑 cos𝑦𝑑𝑥 × 𝑑𝑦𝑑𝑦 1 = − sin𝑦 . 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = 1− 𝒔𝒊𝒏𝒚 𝑑𝑦𝑑𝑥 = −1 𝟏 − 𝒄𝒐𝒔𝟐𝒚 Putting value of 𝑐𝑜𝑠𝑦 = 𝑥 𝑑𝑦𝑑𝑥 = −1 1 − 𝒙𝟐 Hence, 𝒅( 𝒄𝒐𝒔−𝟏𝒙)𝒅𝒙 = −1 1 − 𝒙𝟐 Find the derivative of f given by f (x) = cot–1 𝑥 assuming it exists. Let y = cot–1 𝑥 co𝑡𝑦= 𝑥 𝑥 = co𝑡𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑(𝑥)𝑑𝑥 = 𝑑 co𝑡𝑦𝑑 1 = 𝑑 co𝑡𝑦𝑑𝑥 We need 𝑑𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = 𝑑 co𝑡𝑦𝑑𝑥 × 𝑑𝑦𝑑𝑦 1 = − cosec2𝑦 . 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 (− cosec2𝑦 ) = 1 𝑑𝑦𝑑𝑥 = −1 𝒄𝒐𝒔𝒆𝒄𝟐 𝒚 𝑑𝑦𝑑𝑥 = −1𝟏 + 𝐜𝐨𝐭𝟐𝒚 Putting 𝑐𝑜𝑡𝑦 = 𝑥 𝑑𝑦𝑑𝑥 = −1 𝒙𝟐 + 1 Hence, 𝒅( 𝐜𝐨𝒕−𝟏𝒙)𝒅𝒙 = −𝟏 𝒙𝟐 + 𝟏 Find the derivative of f given by f (x) = sec–1 𝑥 assuming it exists. Let 𝑦 = sec–1 𝑥 𝑠𝑒𝑐𝑦= 𝑥 𝑥 = 𝑠𝑒𝑐𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥𝑑𝑥 = 𝑑 𝑠𝑒𝑐𝑦𝑑𝑥 1 = 𝑑 𝑠𝑒𝑐𝑦𝑑𝑥 We need 𝑑𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = 𝑑 sec𝑦𝑑𝑥 × 𝑑𝑦𝑑𝑦 1 = tan𝑦 . sec𝑦 . 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 tan𝑦 . sec𝑦= 1 𝑑𝑦𝑑𝑥 = 1 𝒕𝒂𝒏𝒚 . sec𝑦 𝑑𝑦𝑑𝑥 = 1 𝐬𝐞𝐜𝟐𝒚 − 𝟏 . sec𝑦 Putting value of 𝑠𝑒𝑐𝑦 = 𝑥 𝑑𝑦𝑑𝑥 = 1 𝑥2 − 1 . 𝑥 𝑑𝑦𝑑𝑥 = 1𝑥 𝑥2 − 1 Hence 𝒅 𝒔𝒆𝒄–𝟏 𝒙𝒅𝒙 = 𝟏𝒙 𝒙𝟐 − 𝟏 Find the derivative of f given by f (x) = cosec–1 𝑥 assuming it exists. Let 𝑦 = cosec–1 𝑥 cosec𝑦= 𝑥 𝑥 = cosec𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑(𝑥)𝑑𝑥 = 𝑑 cosec𝑦𝑑𝑥 1 = 𝑑 cosec𝑦𝑑𝑥 We need 𝑑𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = 𝑑 cosec𝑦𝑑𝑥 × 𝑑𝑦𝑑𝑦 1 = cosec𝑦 . cot𝑦 . 𝑑𝑦𝑑𝑥 −cosec𝑦 . cot𝑦 𝑑𝑦𝑑𝑥 = 1 𝑑𝑦𝑑𝑥 = 1 −cosec𝑦 . 𝒄𝒐𝒕𝒚 𝑑𝑦𝑑𝑥 = 1 −cosec𝑦 . 𝐜𝐨𝒔𝒆𝒄𝟐𝒚−𝟏 Putting value of 𝑐𝑜𝑠𝑒𝑐𝑦 = 𝑥 𝑑𝑦𝑑𝑥 = 1− 𝑥 𝑥2 − 1 𝑑𝑦𝑑𝑥 = −1𝑥 𝑥2 − 1 Hence 𝒅 𝒄𝒐𝒔𝒆𝒄–𝟏 𝒙𝒅𝒙 = −𝟏𝒙 𝒙𝟐 − 𝟏

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.