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Example 45 (ii) - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at Aug. 19, 2021 by Teachoo
Transcript
Example 45 Differentiate the following w.r.t. x. (ii) tan β1 (sinβ‘π₯/( 1 +γ cosγβ‘γπ₯ γ )) Let π(π₯) = tan β1 (πππβ‘π/( 1 +γ πππγβ‘γπ γ )) π(π₯) = tan β1 ((π γπ¬π’π§ γβ‘γπ/πγ γ ππ¨π¬ γβ‘γπ/πγ)/( 1+ (π γππ¨π¬γ^πβ‘γ π/πγ β π))) = tan β1 ((2 γsin γβ‘γπ₯/2γ γ cos γβ‘γπ₯/2 γ)/( 2 cos^2β‘γ π₯/2γ )) = tan β1 ((2 γsin γβ‘γπ₯/2γ)/( 2 cosβ‘γπ₯/2γ )) We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by π/2 sin ΞΈ = 2 πππβ‘γπ½/πγ πππβ‘γπ½/πγ and cos 2ΞΈ = 2cos2 ΞΈ β 1 Replacing ΞΈ by π/2 cos ΞΈ = 2cos2 π½/π β 1 = tan β1 (γsin γβ‘γπ₯/2γ/( cosβ‘γ π₯/2γ )) = tan β1 (γtan γβ‘γπ₯/2γ ) = π/π π(π) = π/π Differentiating π€.π.π‘.π₯ πβ(π₯) = 1/2 (π(π₯))/ππ₯ πβ(π) = π/π (As tan^(β1)β‘γ(tanβ‘π)γ =π) Example 45 (Method 2) Differentiate the following w.r.t. x. (ii) tan β1 (sinβ‘π₯/( 1 +γ cosγβ‘γπ₯ γ )) Let π(π₯) = tan β1 (πππβ‘π/( 1 +γ πππγβ‘γπ γ )) Differentiating w.r.t x π^β² (π₯) = 1/(1 + (πππβ‘π/( 1 +γ πππγβ‘γπ γ ))^2 ) (πππβ‘π/( 1 +γ πππγβ‘γπ γ ))^β² = 1/(((1 + cosβ‘π₯ )^2 + sin^2β‘π₯)/(1 + cosβ‘π₯ )^2 ) (((πππβ‘π )^β² (π + πππ π)β(π + πππ π)^β² πππ π)/( (1 +γ πππγβ‘π )^π )) = (1 + cosβ‘π₯ )^2/((1 + cosβ‘π₯ )^2 + sin^2β‘π₯ ) ((πππ π(π + πππ π). β (β πππ π)πππ π)/( (1 +γ πππγβ‘π )^π )) = (πππ π₯ + πππ^π π + πππ^π π)/(1 + γππ¨π¬γ^πβ‘π + 2 cosβ‘π₯ + γπππγ^πβ‘π ) = (πππ π +π)/(1 + 1 + 2 cosβ‘π₯ ) = (πππ π + π)/(2 + 2 cosβ‘π₯ ) = (1 + πππ π)/(2(1 + cosβ‘π₯) ) = 1/2
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Example 42 Important Deleted for CBSE Board 2022 Exams
Example 43 Deleted for CBSE Board 2022 Exams
Example 44 (i)
Example 44 (ii) Important
Example 44 (iii) Important
Example 45 (i)
Example 45 (ii) Important You are here
Example 45 (iii) Important
Example 46
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Chapter 5 Class 12 Continuity and Differentiability (Term 1)
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.