Last updated at Dec. 16, 2024 by Teachoo
Ex 5.2, 7 Differentiate the functions with respect to π₯ 2β(cot ( π₯2 ))Let π¦ = " " 2β(cot ( π₯2 )) We need to find derivative of π¦ π€.π.π‘.π₯ π¦β² = (2β(cot ( π₯^2 ) ) )^β² = 2 Γ 1/(2β(cot ( π₯^2 ) )) . (cot ( π₯^2 ))^β² = 1/β(cot ( π₯^2 ) ) . (βπππ ππ^2 (π₯^2 )) . ( π₯^2 )^β² = 1/β(cot ( π₯^2 ) ) . (βπππ ππ^2 (π₯^2)) . 2π₯ = (β2π₯" " πππ ππ^2 π₯^2)/β(cot ( π₯^2 ) ) = (β2π₯)/(sin^2β‘(π₯^2 ) . β( γcos γβ‘γ(π₯^2)γ/γsin γβ‘γ(π₯^2)γ ) ) "= " (β2π₯)/(γπ ππ γβ‘(π₯^2 ) Γ γπππ γβ‘(π^π ) β( γcos γβ‘γ(π₯^2)γ/γsin γβ‘γ(π₯^2)γ ) ) " " = (β 2π₯ )/(γsin γβ‘γ(π₯)^2 γ Γβ(γγπππγ^π γβ‘γ(π^π)γ Γ γcos γβ‘γ(π₯^2)γ/sinβ‘γ (π₯^2)γ )) = (β 2π₯ )/(γsin γβ‘γ(π₯)^2 γ Γβ(γππ¨π¬ γβ‘γ(π^π)γ πππβ‘γ (π^π)γ )) = (β 2π₯ )/(γsin γβ‘γ(π₯)^2 γ Γβ(2/2 Γ γcos γβ‘γ(π₯^2)γ sinβ‘γ (π₯^2)γ )) = (β 2βπ π₯ )/(γsin γβ‘γ(π₯)^2 γ Γβ(2 γcos γβ‘γ(π₯^2)γ sinβ‘γ (π₯^2)γ )) = (β πβπ π )/(γπ¬π’π§ γβ‘γ(π)^π γ Γβ(γπππ γβ‘γ(ππ^π)γ )) (As sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ)
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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