Ex 5.6, 11 - Show that dy/dx = - y/x, if x = root a sin-1 t - Ex 5.6

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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Ex 5.6, 11 If 𝑥= ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯ , y= ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯, show that 𝑑𝑦﷮𝑑𝑥﷯ = − 𝑦﷮𝑥﷯ 𝑥= ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯ , y= ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯ We need to show 𝑑𝑦﷮𝑑𝑥﷯ = − 𝑦﷮𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑥﷯ × 𝑑𝑡﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑡﷯ × 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑡﷯﷮ 𝑑𝑥﷮𝑑𝑡﷯﷯ Calculating 𝒅𝒚﷮𝒅𝒕﷯ 𝑦 = ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑑 ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯﷯﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 1﷮2 ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯﷯ . 𝑑 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 1﷮2 ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯﷯ . 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯. log⁡𝑎 × 𝑑 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 1﷮2 ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯﷯ . 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯. log⁡𝑎 × −1﷮ ﷮1 − 𝑡﷮2﷯﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = − 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯. log⁡𝑎 ﷮2 ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯ . ﷮1 − 𝑡﷮2﷯﷯﷯ Calculating 𝒅𝒙﷮𝒅𝒕﷯ 𝑥= ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑑 ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯﷯﷮𝑑𝑡﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 1﷮2 ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯﷯ × 𝑑 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯﷮𝑑𝑡﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 1﷮2 ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯﷯ × 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯. log⁡𝑎 . 𝑑 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷮𝑑𝑡﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 1﷮2 ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯﷯ × 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯. log⁡𝑎 . 1﷮ ﷮1 − 𝑡﷮2﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷮2 ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯﷯ × log﷮𝑎﷯﷮ ﷮1 − 𝑡﷮2﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯ . log﷮𝑎﷯﷮2 ﷮1 − 𝑡﷮2﷯﷯﷯ Therefore, 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑡﷯﷮ 𝑑𝑥﷮𝑑𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯ . log⁡𝑎 ﷮2 ﷮1 − 𝑡﷮2﷯﷯﷯﷮ ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯ . log﷮𝑎﷯﷮2 ﷮1 − 𝑡﷮2﷯﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯ . − log⁡𝑎 ﷮2 ﷮1 − 𝑡﷮2﷯﷯﷯ × 2 ﷮1 − 𝑡﷮2﷯﷯﷮ ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯ . log﷮𝑎﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯⁡﷮ ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯﷯ × log﷮𝑎﷯ × 2 ﷮1 − 𝑡﷮2﷯﷯﷮ log﷮𝑎﷯ × 2 ﷮1 − 𝑡﷮2﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − ﷮ 𝑎﷮ 𝑐𝑜𝑠﷮−1﷯𝑡﷯﷯⁡﷮ ﷮ 𝑎﷮ 𝑠𝑖𝑛﷮−1﷯𝑡﷯﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = −𝒚⁡﷮𝒙﷯ Hence proved.

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