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

Transcript

Ex 5.7, 12 If y= 𝑐𝑜𝑠﷮−1﷯ 𝑥 , Find 𝑑2𝑦﷮𝑑𝑥2﷯ in terms of 𝑦 alone. Let y = 𝑐𝑜𝑠﷮−1﷯ 𝑥 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑( 𝑐𝑜𝑠﷮−1﷯ 𝑥)﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = −1﷮ ﷮1 − 𝑥﷮2﷯ ﷯﷯ Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ = 𝑑﷮𝑑𝑥﷯ −1﷮ ﷮1 − 𝑥﷮2﷯ ﷯﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 𝑑﷮𝑑𝑥﷯ −1﷮ ﷮1 − 𝑥﷮2﷯ ﷯﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = − 𝑑 1﷯ ﷮𝑑𝑥﷯ . ﷮1 − 𝑥﷮2﷯ ﷯− 𝑑 ﷮1 − 𝑥﷮2﷯ ﷯ ﷯﷮𝑑𝑥﷯ . 1﷮ ﷮1 − 𝑥﷮2﷯ ﷯ ﷯﷮2﷯﷯ ﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = − 0 . ﷮1 − 𝑥﷮2﷯ ﷯− 1﷮2 ﷮1 − 𝑥﷮2﷯ ﷯﷯ . 𝑑﷮𝑑𝑥﷯ (1 − 𝑥﷮2﷯) ﷮ ﷮1 − 𝑥﷮2﷯ ﷯ ﷯﷮2﷯﷯ ﷯ = − 1﷮2 ﷮1 − 𝑥﷮2﷯ ﷯﷯ . 0−2𝑥﷯﷮ 1 − 𝑥﷮2﷯﷯﷯ ﷯ = − −1 − 2𝑥﷯﷮2 ﷮1 − 𝑥﷮2﷯ ﷯ ﷯ 1 − 𝑥﷮2﷯﷯﷯ = −2𝑥﷮2 ﷮1 − 𝑥﷮2﷯ ﷯ ﷯ 1 − 𝑥﷮2﷯﷯﷯ = − 𝑥﷮ 1 − 𝑥﷮2﷯﷯﷮ 1﷮2﷯﷯ 1 − 𝑥﷮2﷯﷯﷯ = − 𝑥﷮ 1 − 𝑥﷮2﷯﷯﷮ 1﷮2﷯ + 1﷯ ﷯ 𝒅﷮𝟐﷯𝒚﷮𝒅 𝒙﷮𝟐﷯﷯ = − 𝒙﷮ 𝟏 − 𝒙﷮𝟐﷯﷯﷮ 𝟑﷮𝟐﷯ ﷯ ﷯ But we need to calculate 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ in terms of y . y = 𝑐𝑜𝑠﷮−1﷯ 𝑥 cos⁡𝑦 = 𝑥 𝑥 = cos⁡𝑦 Putting 𝑥 = cos⁡𝑦 in equation 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = − 𝑥﷮ 1 − 𝑥﷮2﷯﷯﷮ 3﷮2﷯ ﷯ ﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = −cos⁡𝑦 ﷮ 1 − ( cos﷮𝑦﷯)﷮2﷯﷯﷮ 3﷮2﷯ ﷯ ﷯ = − cos⁡𝑦 ﷮ ( sin2﷮𝑦) ﷯﷮ 3﷮2﷯ ﷯ ﷯ = − cos⁡𝑦 ﷮ ( sin﷮𝑦) ﷯﷮2 × 3﷮2﷯ ﷯ ﷯ = − cos⁡𝑦 ﷮ ( sin﷮𝑦) ﷯﷮3 ﷯ ﷯ = cos⁡𝑦 ﷮ sin 𝑦 ﷯ × 1﷮ sin2﷮𝑦﷯﷯ 𝒅﷮𝟐﷯𝒚﷮𝒅 𝒙﷮𝟐﷯﷯ = 𝒄𝒐𝒕﷮𝒚﷯. 𝒄𝒐𝒔𝒆𝒄𝟐 𝒚