Example 27 - Find derivative of f(x) = tan-1 x - Class 12 - Examples

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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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 tan⁡𝑦 = 𝑥 ∴ 𝑥 = tan⁡𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑(𝑥)﷮𝑑𝑥﷯ = 𝑑 tan﷮𝑦﷯﷯﷮𝑑𝑥﷯ 1 = 𝑑 tan﷮𝑦﷯﷯﷮𝑑𝑥﷯ We need d𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = 𝑑 tan﷮𝑦﷯﷯﷮𝑑𝑥﷯ × 𝑑𝑦﷮𝑑𝑦﷯ 1 = 𝐬𝐞𝐜﷮𝟐﷯ 𝒚 . 𝑑𝑦﷮𝑑𝑥﷯ 1 = (𝟏 + 𝒕𝒂𝒏𝟐𝒚) 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮1 + 𝐭𝐚𝐧﷮𝟐﷯﷮𝒚﷯﷯ Putting 𝑡𝑎𝑛⁡𝑦 = 𝑥 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮1 + 𝒙﷮𝟐﷯﷯ Hence 𝒅( 𝐭𝐚𝐧﷮−𝟏﷯﷮𝒙)﷯﷮𝒅𝒙﷯ = 𝟏﷮𝟏 + 𝒙﷮𝟐﷯﷯ Find the derivative of f given by f (x) = cos–1 𝑥 assuming it exists. Let y = cos–1 𝑥 cos﷮𝑦﷯= 𝑥 𝑥 = cos﷮𝑦﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑(𝑥)﷮𝑑𝑥﷯ = 𝑑 cos﷮𝑦﷯﷯﷮𝑑𝑥﷯ 1 = 𝑑 cos﷮𝑦﷯﷯﷮𝑑𝑥﷯ We need 𝑑𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = 𝑑 cos﷮𝑦﷯﷯﷮𝑑𝑥﷯ × 𝑑𝑦﷮𝑑𝑦﷯ 1 = − sin﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮− 𝒔𝒊𝒏⁡𝒚﷯ 𝑑𝑦﷮𝑑𝑥﷯ = −1﷮ ﷮𝟏 − 𝒄𝒐𝒔﷮𝟐﷯﷮𝒚﷯﷯ ﷯ Putting value of 𝑐𝑜𝑠⁡𝑦 = 𝑥 𝑑𝑦﷮𝑑𝑥﷯ = −1﷮ ﷮1 − 𝒙﷮𝟐﷯﷯ ﷯ Hence, 𝒅( 𝒄𝒐𝒔﷮−𝟏﷯﷮𝒙)﷯﷮𝒅𝒙﷯ = −1﷮ ﷮1 − 𝒙﷮𝟐﷯﷯ ﷯ Find the derivative of f given by f (x) = cot–1 𝑥 assuming it exists. Let y = cot–1 𝑥 co𝑡﷮𝑦﷯= 𝑥 𝑥 = co𝑡﷮𝑦﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑(𝑥)﷮𝑑𝑥﷯ = 𝑑 co𝑡﷮𝑦﷯﷯﷮𝑑﷯ 1 = 𝑑 co𝑡﷮𝑦﷯﷯﷮𝑑𝑥﷯ We need 𝑑𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = 𝑑 co𝑡﷮𝑦﷯﷯﷮𝑑𝑥﷯ × 𝑑𝑦﷮𝑑𝑦﷯ 1 = − cosec﷮2﷯﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ (− cosec﷮2﷯⁡𝑦 ) = 1 𝑑𝑦﷮𝑑𝑥﷯ = −1﷮ 𝒄𝒐𝒔𝒆𝒄﷮𝟐﷯ 𝒚﷯ 𝑑𝑦﷮𝑑𝑥﷯ = −1﷮𝟏 + 𝐜𝐨𝐭﷮𝟐﷯⁡𝒚﷯ Putting 𝑐𝑜𝑡⁡𝑦 = 𝑥 𝑑𝑦﷮𝑑𝑥﷯ = −1﷮ 𝒙﷮𝟐﷯ + 1﷯ Hence, 𝒅( 𝐜𝐨𝒕﷮−𝟏﷯﷮𝒙)﷯﷮𝒅𝒙﷯ = −𝟏﷮ 𝒙﷮𝟐﷯ + 𝟏﷯ Find the derivative of f given by f (x) = sec–1 𝑥 assuming it exists. Let 𝑦 = sec﷮–1﷯ 𝑥 𝑠𝑒𝑐﷮𝑦﷯= 𝑥 𝑥 = 𝑠𝑒𝑐﷮𝑦﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥﷮𝑑𝑥﷯ = 𝑑 𝑠𝑒𝑐﷮𝑦﷯﷯﷮𝑑𝑥﷯ 1 = 𝑑 𝑠𝑒𝑐﷮𝑦﷯﷯﷮𝑑𝑥﷯ We need 𝑑𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = 𝑑 sec﷮𝑦﷯﷯﷮𝑑𝑥﷯ × 𝑑𝑦﷮𝑑𝑦﷯ 1 = tan﷮𝑦﷯ . sec﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ tan﷮𝑦﷯ . sec﷮𝑦﷯= 1 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮ 𝒕𝒂𝒏﷮𝒚﷯ . sec﷮𝑦﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮ ﷮ 𝐬𝐞𝐜﷮𝟐﷯﷮𝒚﷯ − 𝟏﷯﷯ . sec﷮𝑦﷯﷯ Putting value of 𝑠𝑒𝑐⁡𝑦 = 𝑥 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮ ﷮ 𝑥﷮2﷯ − 1 ﷯ ﷯ . 𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮𝑥 ﷮ 𝑥﷮2﷯ − 1 ﷯ ﷯ Hence 𝒅 𝒔𝒆𝒄﷮–𝟏﷯ 𝒙﷯﷮𝒅𝒙﷯ = 𝟏﷮𝒙 ﷮ 𝒙﷮𝟐﷯ − 𝟏 ﷯ ﷯ Find the derivative of f given by f (x) = cosec–1 𝑥 assuming it exists. Let 𝑦 = cosec﷮–1﷯ 𝑥 cosec﷮𝑦﷯= 𝑥 𝑥 = cosec﷮𝑦﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑(𝑥)﷮𝑑𝑥﷯ = 𝑑 cosec﷮𝑦﷯﷯﷮𝑑𝑥﷯ 1 = 𝑑 cosec﷮𝑦﷯﷯﷮𝑑𝑥﷯ We need 𝑑𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = 𝑑 cosec﷮𝑦﷯﷯﷮𝑑𝑥﷯ × 𝑑𝑦﷮𝑑𝑦﷯ 1 = cosec﷮𝑦﷯ . cot﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ −cosec﷮𝑦﷯ . cot﷮𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮ −cosec﷮𝑦﷯ . 𝒄𝒐𝒕﷮𝒚﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮ −cosec﷮𝑦﷯ . ﷮ 𝐜𝐨𝒔𝒆𝒄﷮𝟐﷯﷮𝒚﷯−𝟏﷯﷯ Putting value of 𝑐𝑜𝑠𝑒𝑐⁡𝑦 = 𝑥 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮− 𝑥 ﷮ 𝑥﷮2﷯ − 1 ﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = −1﷮𝑥 ﷮ 𝑥﷮2﷯ − 1 ﷯ ﷯ Hence 𝒅 𝒄𝒐𝒔𝒆𝒄﷮–𝟏﷯ 𝒙﷯﷮𝒅𝒙﷯ = −𝟏﷮𝒙 ﷮ 𝒙﷮𝟐﷯ − 𝟏 ﷯ ﷯

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