1. Chapter 5 Class 12 Continuity and Differentiability
2. Serial order wise
3. Ex 5.7

Transcript

Ex 5.7, 17 If 𝑦= ( 𝑡𝑎𝑛﷮−1﷯ 𝑥)﷮2 ﷯, show that ( 𝑥﷮2﷯+1)﷮2 ﷯ 𝑦2 + 2𝑥 ( 𝑥﷮2﷯+1)﷮ ﷯ 𝑦1 = 2 We have y = ( 𝑡𝑎𝑛﷮−1﷯ 𝑥)﷮2 ﷯ Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 ( ( 𝑡𝑎𝑛﷮−1﷯ 𝑥)﷮2 ﷯)﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 2 𝑡𝑎𝑛﷮−1﷯ 𝑥 . 𝑑 𝑡𝑎𝑛﷮−1﷯ 𝑥﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 2 𝑡𝑎𝑛﷮−1﷯ 𝑥 . 1﷮1 + 𝑥﷮2﷯﷯ Hence, 𝑦1 = 𝑑𝑦﷮𝑑𝑥﷯ = 2 𝑡𝑎𝑛﷮−1﷯ 𝑥﷮1 + 𝑥﷮2﷯﷯ Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ = 𝑑﷮𝑑𝑥﷯ 2 𝑡𝑎𝑛﷮−1﷯ 𝑥﷮1 + 𝑥﷮2﷯﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = − 2 𝑑﷮𝑑𝑥﷯ 2 𝑡𝑎𝑛﷮−1﷯ 𝑥﷮1 + 𝑥﷮2﷯﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 2 𝑑 2 𝑡𝑎𝑛﷮−1﷯ 𝑥﷯﷮𝑑𝑥﷯ . 1 + 𝑥﷮2﷯﷯ − 𝑑 1 + 𝑥﷮2﷯﷯﷮𝑑𝑥﷯ . 𝑡𝑎𝑛﷮−1﷯ 𝑥﷮ 1 + 𝑥﷮2﷯﷯﷮2﷯﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 2 1﷮1 + 𝑥﷮2﷯﷯ . 1 + 𝑥﷮2﷯﷯ − 𝑑 1﷯﷮𝑑𝑥﷯ + 𝑑 𝑥﷮2﷯﷯﷮𝑑𝑥﷯﷯ . 𝑡𝑎𝑛﷮−1﷯ 𝑥﷮ 1 + 𝑥﷮2﷯﷯﷮2﷯﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 2 1− 0 + 2𝑥﷯ 𝑡𝑎𝑛﷮−1﷯ 𝑥﷮ 1 + 𝑥﷮2﷯﷯﷮2﷯﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 2 1−2𝑥 . 𝑡𝑎𝑛﷮−1﷯ 𝑥﷮ 1 + 𝑥﷮2﷯﷯﷮2﷯﷯﷯ Thus, 𝒚2 = 2 𝟏−𝟐𝒙 . 𝒕𝒂𝒏﷮−𝟏﷯ 𝒙﷮ 𝟏 + 𝒙﷮𝟐﷯﷯﷮𝟐﷯﷯﷯ We need to show ( 𝑥﷮2﷯+1)﷮2 ﷯ 𝑦2 + 2𝑥 ( 𝑥﷮2﷯+1)﷮ ﷯ 𝑦1 = 2 Solving LHS ( 𝑥﷮2﷯+1)﷮2 ﷯ 𝑦2 + 2𝑥 ( 𝑥﷮2﷯+1)﷮ ﷯ 𝑦1 = ( 𝑥﷮2﷯+1)﷮2 ﷯. 2 1−2𝑥 . 𝑡𝑎𝑛﷮−1﷯ 𝑥﷮ 1 + 𝑥﷮2﷯﷯﷮2﷯﷯﷯ + 2𝑥 (𝑥2+1) . 2 𝑡𝑎𝑛﷮−1﷯ 𝑥﷮1 + 𝑥﷮2﷯﷯﷯ = 2 (1−2𝑥 𝑡𝑎𝑛﷮−1﷯ 𝑥) + 4𝑥 𝑡𝑎𝑛﷮−1﷯ 𝑥 = 2 – 4x 𝑡𝑎𝑛﷮−1﷯ x + 4x 𝑡𝑎𝑛﷮−1﷯ x = 2 Hence proved .

Ex 5.7