Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

### Transcript

Derivative of ใ๐๐๐ใ^(โ๐) ๐ ๐ (๐ฅ)=ใ๐๐๐กใ^(โ1) ๐ฅ Let ๐= ใ๐๐๐ใ^(โ๐) ๐ cotโกใ๐ฆ=๐ฅใ ๐=๐๐จ๐ญโกใ๐ ใ Differentiating both sides ๐ค.๐.๐ก.๐ฅ ๐๐ฅ/๐๐ฅ = (๐ (cotโก๐ฆ ))/๐๐ฅ 1 = (๐ (cotโก๐ฆ ))/๐๐ฅ ร ๐๐ฆ/๐๐ฆ 1 = (๐ (cotโก๐ฆ ))/๐๐ฆ ร ๐๐ฆ/๐๐ฅ 1 = โ๐๐จใ๐ฌ๐๐ใ^๐ ๐ . ๐๐ฆ/๐๐ฅ 1 = โ(๐ +๐๐๐๐๐) ๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ = (โ1)/(1 + ใ๐๐จ๐ญใ^๐โก๐ ) Putting ๐๐๐กโก๐ฆ = ๐ฅ ๐๐ฆ/๐๐ฅ = (โ1)/(๐^๐ + ๐) Hence (๐(ใ๐๐จ๐ญใ^(โ๐)โกใ๐)ใ)/๐๐ = (โ๐)/(๐^๐ + ๐) (๐ด๐  ใ ๐๐๐ ๐๐ใ^2โกใ๐ฆ= ใ1+ใโกใ๐๐๐กใ^2โก๐ฆ ใ) As ๐ฆ = ใ๐๐๐กใ^(โ1) ๐ฅ So, ๐๐๐โก๐ = ๐