Ex 5.2

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

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¦β² = (2β(cot ( π₯^2 ) ) )^β² = 2 Γ 1/(2β(cot ( π₯^2 ) )) . (cot ( π₯^2 ))^β² = 1/β(cot ( π₯^2 ) ) . (βπππ ππ^2 (π₯^2 )) . ( π₯^2 )^β² = 1/β(cot ( π₯^2 ) ) . (βπππ ππ^2 (π₯^2)) . 2π₯ = (β2π₯" " πππ ππ^2 π₯^2)/β(cot ( π₯^2 ) ) = (β2π₯)/(sin^2β‘(π₯^2 ) . β( γcos γβ‘γ(π₯^2)γ/γsin γβ‘γ(π₯^2)γ ) ) "= " (β2π₯)/(γπ ππ γβ‘(π₯^2 ) Γ γπππ γβ‘(π^π ) β( γcos γβ‘γ(π₯^2)γ/γsin γβ‘γ(π₯^2)γ ) ) " " = (β 2π₯ )/(γsin γβ‘γ(π₯)^2 γ Γβ(γγπππγ^π γβ‘γ(π^π)γ Γ γcos γβ‘γ(π₯^2)γ/sinβ‘γ (π₯^2)γ )) = (β 2π₯ )/(γsin γβ‘γ(π₯)^2 γ Γβ(γππ¨π¬ γβ‘γ(π^π)γ πππβ‘γ (π^π)γ )) = (β 2π₯ )/(γsin γβ‘γ(π₯)^2 γ Γβ(2/2 Γ γcos γβ‘γ(π₯^2)γ sinβ‘γ (π₯^2)γ )) = (β 2βπ π₯ )/(γsin γβ‘γ(π₯)^2 γ Γβ(2 γcos γβ‘γ(π₯^2)γ sinβ‘γ (π₯^2)γ )) = (β πβπ π )/(γπ¬π’π§ γβ‘γ(π)^π γ Γβ(γπππ γβ‘γ(ππ^π)γ )) (As sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ)