Example 21 (ii) - Chapter 12 Class 11 Limits and Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Examples
Example 1 (i)
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Transcript
Example 21 Compute derivative of (ii) g(x) = cot x g(x) = cot x = cosβ‘π₯/sinβ‘π₯ Let u = cos x & v = sin x β΄ g(x) = π’/π£ So, gβ(x) = (π’/π£)^β² Using quotient rule gβ(x) = (π’^β² π£ βγ π£γ^β² π’)/π£^2 Finding uβ & vβ u = cos x uβ = β sin x & v = sin x vβ = cos x Now, fβ(x) = (π’/π£)^β² = (π’^β² π£ βγ π£γ^β² π’)/π£^2 (π·ππππ£ππ‘ππ£π ππ πππ β‘γπ₯=γβ π ππ γβ‘π₯ γ ) (π·ππππ£ππ‘ππ£π ππ π ππβ‘γπ₯=γπππ γβ‘π₯ γ ) = (βsinβ‘γπ₯ (sinβ‘π₯ ) βγ cosγβ‘γπ₯ (cosβ‘γπ₯)γ γ γ)/(γπ ππγ^2 π₯) = (βsin2β‘γπ₯ βγ cos2γβ‘γπ₯ γ γ)/(γπ ππγ^2 π₯) = (β(π¬π’π§πβ‘γπ + γ ππ¨π¬πγβ‘γπ) γ γ)/(γπ ππγ^2 π₯) = (βπ)/(γπ ππγ^2 π₯) = βcosec2x Hence, fβ(x) = βcosec2x (ππ πππ π ππ2π₯+πππ 2π₯=1)
