## If u = γsinγ^(-1) (2x/(1 + x^2 )) and v = γtanγ^(-1) (2x/(1 - x^2 )), then du/dv is

## (A) 1/2Β

## (B) x

## (C) (1 - x^2)/(1 + x^2 ) Β Β

## (D) 1

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Chapter 5 Class 12 Continuity and Differentiability (Term 1)

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This question is similar to

Ex 5.3, 9 - Chapter 5 Class 12 -Continuity and Differentiability

Hello! Teachoo has made this answer with days (even weeks!) worth of effort and love. If Teachoo has been of any help to you in your Board exam preparation, then please support us by clicking on this link to make a donation

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)