Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Derivative of ใ๐๐๐๐๐ใ^(โ๐) ๐ ๐ (๐ฅ)=ใ๐๐๐ ๐๐ใ^(โ1) ๐ฅ Let ๐= ใ๐๐๐๐๐ใ^(โ๐) ๐ cosecโกใ๐ฆ=๐ฅใ ๐=๐๐จ๐ฌ๐๐โกใ๐ ใ Differentiating both sides ๐ค.๐.๐ก.๐ฅ ๐๐ฅ/๐๐ฅ = (๐ (cosecโก๐ฆ ))/๐๐ฅ 1 = (๐ (cosecโก๐ฆ ))/๐๐ฅ ร ๐๐ฆ/๐๐ฆ 1 = (๐ (cosecโก๐ฆ ))/๐๐ฆ ร ๐๐ฆ/๐๐ฅ 1 = โcosecโก๐ฆ .cotโก๐ฆ . ๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ = 1/(ใโcosecใโก๐ฆ .๐๐๐โก๐ ) ๐๐ฆ/๐๐ฅ = 1/(ใโcosecใโก๐ฆ . โ(ใ๐๐จ๐๐๐ใ^๐โก๐ โ ๐)) Putting value of ๐๐๐ ๐๐โก๐ฆ = ๐ฅ ๐๐ฆ/๐๐ฅ = (โ1)/(๐ฅ โ(๐ฅ^2 โ 1 ) ) Hence ๐(ใ๐๐๐๐๐ใ^(โ๐) ๐)/๐๐ = (โ๐)/(๐ โ(๐^๐ โ ๐ ) ) As cot2 ฮธ = cosec2 ฮธ โ 1, cot ฮธ = โ("cosec2 ฮธ โ 1" ) As ๐ฆ = coใ๐ ๐๐ใ^(โ1) ๐ฅ So, co๐๐๐โก๐ = ๐