If u = γsinγ^(-1) (2x/(1 + x^2 )) and v = γtanγ^(-1) (2x/(1 - x^2 )), then du/dv is
(A) 1/2
(B) x
(C) (1 - x^2)/(1 + x^2 )
(D) 1
This question is similar to Ex 5.3, 9 - Chapter 5 Class 12 - Continuity and Differentiability
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Last updated at April 16, 2024 by Teachoo
This question is similar to Ex 5.3, 9 - Chapter 5 Class 12 - Continuity and Differentiability
Question 10 If u = γπ ππγ^(β1) (2π₯/(1 + π₯^2 )) and v = γπ‘ππγ^(β1) (2π₯/(1 β π₯^2 )), then ππ’/ππ£ is (A) 1/2 (B) π₯ (C) (1 β π₯^2)/(1 + π₯^2 ) (D) 1 π π/π π=(π πβπ π)/(π πβπ π) So, we will find ππ’/ππ₯ and ππ£/ππ₯ Simplifying u u = γπ ππγ^(β1) (2π₯/(1 + π₯^2 )) Putting π₯=tanβ‘π u = γπ ππγ^(β1) ((2 πππβ‘π½)/(1 + γπππγ^π π½)) Using sinβ‘γ2π=γ (2 tanβ‘π)/(1 + γπ‘ππγ^2 π) u = γπ ππγ^(β1) (sinβ‘2π ) u = 2π Putting π=γπ‘ππγ^(β1) π₯ u = πγπππγ^(βπ) π Simplifying v v = γπ‘ππγ^(β1) (2π₯/(1 β π₯^2 )) Putting π₯=tanβ‘π v = γπ‘ππγ^(β1) ((2 πππβ‘π½)/(1 β γπππγ^π π½)) Using tanβ‘γ2π=γ (2 tanβ‘π)/(1 β γπ‘ππγ^2 π) v = γπ‘ππγ^(β1) (tanβ‘2π ) v = 2π Putting π=γπ‘ππγ^(β1) π₯ v = πγπππγ^(βπ) π Finding π π/π π u = πγπππγ^(βπ) π Differentiating w.r.t. π₯ π π/π π=(πγ(2π‘ππγ^(β1) π₯))/ππ₯ =2 Γ(πγ(π‘ππγ^(β1) π₯))/ππ₯ =2 Γ1/γ1 + π₯γ^2 = π/γπ + πγ^π Finding π π/π π v = πγπππγ^(βπ) π Differentiating w.r.t. π₯ π π/π π=(πγ(2π‘ππγ^(β1) π₯))/ππ₯ =2 Γ(πγ(π‘ππγ^(β1) π₯))/ππ₯ =2 Γ1/γ1 + π₯γ^2 = π/γπ + πγ^π Now, π π/π π=(ππ’βππ₯)/(ππ£βππ₯) =(2/γ1 + π₯γ^2 )/(1/γ1 + π₯γ^2 ) =π So, the correct answer is (D)