Example 21 (ii) - Chapter 12 Class 11 Limits and Derivatives
Last updated at April 16, 2024 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
Example 1 (iii)
Example 2 (i)
Example 2 (ii) Important
Example 2 (iii) Important
Example 2 (iv)
Example 2 (v)
Example 3 (i) Important
Example 3 (ii) Important
Example 4 (i)
Example 4 (ii) Important
Example 5
Example 6
Example 7 Important
Example 8
Example 9
Example 10 Important
Example 11
Example 12
Example 13 Important
Example 14
Example 15 Important
Example 16
Example 17 Important
Example 18
Example 19 (i) Important
Example 19 (ii)
Example 20 (i)
Example 20 (ii) Important
Example 21 (i)
Example 21 (ii) Important You are here
Example 22 (i)
Example 22 (ii) Important
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)
