Example 40 (iii) - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
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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 )
