Example 40 (iii) - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 40 Differentiate the following w.r.t. x. (iii) sin^(β1) ((2^( π₯+1) )/( 1 +γ 4 γ^π₯ )) Let π(π₯) = sin^(β1) ((2^( π₯+1) )/( 1 +γ 4 γ^π₯ )) π(π₯) = sin^(β1) ((2^( π₯). 2)/( 1 + (2^π₯ )^2 )) Let π^π = tan ΞΈ π(π₯) = sin^(β1) ((tanβ‘γπ γ. 2)/( 1 + tan^2β‘π )) = sin^(β1) ((2 tanβ‘γπ γ )/( 1 +γ tan^2γβ‘π )) = sin^(β1) (sin 2π) = 2π (sinβ‘2π "= " (2 tanβ‘π)/(1 +γ tan^2γβ‘π )) (As γπ ππγ^(β1)β‘γ(π ππβ‘π)γ =π) Since 2^π₯= tanβ‘π tan^(β1) (2^π₯ )=π β΄ π(π) = π (γπππγ^(βπ) (π^π )) Differentiating π€.π.π‘.π₯ πβ(π₯) = 2 (π (tan^(β1) 2^π₯ )" " )/ππ₯ πβ(π₯) = 2 . 1/(1 + (2^π₯ )^2 ) . (π (π^π )" " )/π π πβ(π₯) = (2 )/(1 + (2^π₯ )^2 ) . π^π . πππβ‘π (π΄π π/ππ₯(γπ‘ππγ^(β1))=1/(1+π₯^2 )) (π΄π π/ππ₯ (π^π₯ )=π^π₯. πππβ‘π₯ ) πβ(π₯) = (γ2. 2γ^π₯.γ logγβ‘2)/(1 + (2^π₯ )^2 ) πβ(π₯) = (2^(π₯ + 1).γ logγβ‘2)/(1 + (2^π₯ )^2 ) πβ(π₯) = (2^(π₯ + 1).γ logγβ‘2)/(1 + (2^2 )^π₯ ) πβ(π) = (π^(π + π).γ πππγβ‘π)/(π + π^π )
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo