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Example 21 - Chapter 13 Class 11 Limits and Derivatives - Part 3

Example 21 - Chapter 13 Class 11 Limits and Derivatives - Part 4
Example 21 - Chapter 13 Class 11 Limits and Derivatives - Part 5

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Example 21 Compute derivative of (ii) g(x) = cot x g(x) = cot x = cos⁑π‘₯/sin⁑π‘₯ Let u = cos x & v = sin x ∴ g(x) = 𝑒/𝑣 So, g’(x) = (𝑒/𝑣)^β€² Using quotient rule g’(x) = (𝑒^β€² 𝑣 βˆ’γ€– 𝑣〗^β€² 𝑒)/𝑣^2 Finding u’ & v’ u = cos x u’ = – sin x & v = sin x v’ = cos x Now, f’(x) = (𝑒/𝑣)^β€² = (𝑒^β€² 𝑣 βˆ’γ€– 𝑣〗^β€² 𝑒)/𝑣^2 (π·π‘’π‘Ÿπ‘–π‘£π‘Žπ‘‘π‘–π‘£π‘’ π‘œπ‘“ π‘π‘œπ‘ β‘γ€–π‘₯=γ€–βˆ’ 𝑠𝑖𝑛 〗⁑π‘₯ γ€— ) (π·π‘’π‘Ÿπ‘–π‘£π‘Žπ‘‘π‘–π‘£π‘’ π‘œπ‘“ 𝑠𝑖𝑛⁑〖π‘₯=γ€–π‘π‘œπ‘  〗⁑π‘₯ γ€— ) = (βˆ’sin⁑〖π‘₯ (sin⁑π‘₯ ) βˆ’γ€– cos〗⁑〖π‘₯ (cos⁑〖π‘₯)γ€— γ€— γ€—)/(〖𝑠𝑖𝑛〗^2 π‘₯) = (βˆ’sin2⁑〖π‘₯ βˆ’γ€– cos2〗⁑〖π‘₯ γ€— γ€—)/(〖𝑠𝑖𝑛〗^2 π‘₯) = (βˆ’(π¬π’π§πŸβ‘γ€–π’™ + γ€– πœπ¨π¬πŸγ€—β‘γ€–π’™) γ€— γ€—)/(〖𝑠𝑖𝑛〗^2 π‘₯) = (βˆ’πŸ)/(〖𝑠𝑖𝑛〗^2 π‘₯) = –cosec2x Hence, f’(x) = –cosec2x (π‘ˆπ‘ π‘–π‘›π‘” 𝑠𝑖𝑛2π‘₯+π‘π‘œπ‘ 2π‘₯=1)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.