Ex 5.7

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

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### Transcript

Ex 5.7, 17 (Method 1) If π¦= γ(γπ‘ππγ^(β1) π₯)γ^(2 ), show that γ(π₯^2+1)γ^(2 ) π¦2 + 2π₯ γ(π₯^2+1)γ^ π¦1 = 2 We have y = γ(γπ‘ππγ^(β1) π₯)γ^(2 ) Differentiating π€.π.π‘.π₯ yβ = 2 tanβ1 π₯ Γ 1/(1 + π₯^2 ) (1 + π₯^2) yβ = 2 tanβ1 π₯ Again differentiating π€.π.π‘.π₯ [π¦^β² (1+π₯^2 )]^β² = 2 Γ 1/(1 +γ π₯γ^2 ) [π¦^β² (1+π₯^2 )]^β² = 2/(1 + π₯^2 ) (tan^(β1)β‘π₯ )^β²=1/(1+π₯^2 ) Using product rule (1+π₯^2 )^β² + π¦^β²β² (1 + π₯^2) = 2/(1 + π₯^2 ) 2π₯ π¦^β²+π¦^β²β² (1 +π₯^2 ) = 2/(1 + π₯^2 ) 2π₯ π¦^β² (1 +π₯^2 )+π¦^β²β² (1 +π₯^2 )Γ(1 +π₯^2 ) = 2 2π₯(1 +π₯^2 ) π¦^β²+π¦^β²β² (1 +π₯^2 )^2 = 2 π^β²β² (π +π^π )^π+ππ(π +π^π ) π^β² = 2 Hence Proved Ex 5.7, 17 (Method 2) If π¦= γ(γπ‘ππγ^(β1) π₯)γ^(2 ), show that γ(π₯^2+1)γ^(2 ) π¦2 + 2π₯ γ(π₯^2+1)γ^ π¦1 = 2 We have y = γ(γπ‘ππγ^(β1) π₯)γ^(2 ) Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = (π (γ(γπ‘ππγ^(β1) π₯)γ^(2 )))/ππ₯ ππ¦/ππ₯ = 2 γπ‘ππγ^(β1) π₯ . π(γπ‘ππγ^(β1) π₯)/ππ₯ ππ¦/ππ₯ = 2 γπ‘ππγ^(β1) π₯ . 1/(1 +γ π₯γ^2 ) Hence, π¦1 = ππ¦/ππ₯ = (2 γπ‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 ) Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = π/ππ₯ ((2 γπ‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )) (π^2 π¦)/(ππ₯^2 ) = 2 π/ππ₯ ((γπ‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )) (π^2 π¦)/(ππ₯^2 ) = 2 [(π(γπ‘ππγ^(β1) π₯)/ππ₯ . (1 +γ π₯γ^2 ) β π(1 +γ π₯γ^2 )/ππ₯ .γ π‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )^2 ] (π^2 π¦)/(ππ₯^2 ) = 2 [(1/(1 +γ π₯γ^2 ) . (1 +γ π₯γ^2 ) β (π(1)/ππ₯ + π(π₯^2 )/ππ₯) . γ π‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )^2 ] Using quotient Rule As, (π’/π£)^β²= (π’βπ£ β π£βπ’)/π£^2 where u = tan-1 x & v = 1 + x2 (π^2 π¦)/(ππ₯^2 ) = 2 [( 1 β (0 + 2π₯) γ π‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )^2 ] (π^2 π¦)/(ππ₯^2 ) = 2 [( 1 β 2π₯ .γ π‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )^2 ] Thus, π2 = 2 [( π β ππ .γ πππγ^(βπ) π)/(π +γ πγ^π )^π ] We need to show γ(π₯^2+1)γ^(2 ) π¦2 + 2π₯ γ(π₯^2+1)γ^ π¦1 = 2 Solving LHS γ(π₯^2+1)γ^(2 ) π¦2 + 2π₯ γ(π₯^2+1)γ^ π¦1 = γ(π₯^2+1)γ^(2 ). 2 [( 1 β 2π₯ .γ π‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )^2 ] + 2π₯ (π₯2+1) . ((2 γπ‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )) = 2 (1β2π₯ γπ‘ππγ^(β1) π₯) + 4π₯ γπ‘ππγ^(β1) π₯ = 2 β 4x γπππγ^(βπ) x + 4x γπππγ^(βπ) x = 2 Hence proved .

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.