Ex 5.3, 11 - Chapter 5 Class 12 Continuity and Differentiability

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Ex 5.3, 11 Find 𝑑𝑦/𝑑𝑥 in, 𝑦 = cos–1 ((1− 𝑥^2)/( 1+ 𝑥2 )) , 0 < x < 1 𝑦 = cos–1 ((1− 𝑥^2)/( 1+ 𝑥2 )) Putting x = tan θ y = 〖𝑐𝑜𝑠〗^(−1) ((1−tan⁡2 𝜃)/(1+ tan⁡2 𝜃)) y = cos−1 (cos 2𝜃) 𝑦 =2θ Putting value of θ = 〖𝑡𝑎𝑛〗^(−1) 𝑥 𝑦=2 (〖𝑡𝑎𝑛〗^(−1) 𝑥) (cos⁡2θ " =" (1 − tan⁡2 𝜃)/(1+ tan⁡2 𝜃)) Since x = tan θ ∴ 〖𝑡𝑎𝑛〗^(−1) x = θ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 . (𝑑(𝑦))/𝑑𝑥 = (𝑑 (2 〖𝑡𝑎𝑛〗^(−1) 𝑥" ) " )/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 2 (𝑑 (〖𝑡𝑎𝑛〗^(−1) 𝑥" ) " )/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 2 (1/(1+ 𝑥^2 )) 𝒅𝒚/𝒅𝒙 = 𝟐/(𝟏+〖 𝒙〗^𝟐 ) ((〖𝑡𝑎𝑛〗^(−1) 𝑥")‘ = " 1/(1 + 𝑥^2 ))