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ใ^(โ1) ๐ฅ Let ๐= ใ๐๐๐ใ^(โ๐) ๐ secโกใ๐ฆ=๐ฅใ ๐=๐ฌ๐๐โกใ๐ ใ Differentiating both sides ๐ค.๐.๐ก.๐ฅ ๐๐ฅ/๐๐ฅ = (๐ (secโก๐ฆ ))/๐๐ฅ 1 = (๐ (secโก๐ฆ ))/๐๐ฅ ร ๐๐ฆ/๐๐ฆ 1 = (๐ (secโก๐ฆ ))/๐๐ฆ ร ๐๐ฆ/๐๐ฅ 1 = ๐๐๐โก๐ .๐๐๐โก๐. ๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ = 1/(๐๐๐โก๐ .ใ secใโก๐ฆ ) ๐๐ฆ/๐๐ฅ = 1/((โ(ใ๐ฌ๐๐ใ^๐โก๐ โ ๐)) .ใ secใโก๐ฆ ) Putting value of ๐ ๐๐โก๐ฆ = ๐ฅ ๐๐ฆ/๐๐ฅ = 1/((โ(๐ฅ^2 โ 1 ) ) . ๐ฅ) ๐๐ฆ/๐๐ฅ = 1/(๐ฅ โ(๐ฅ^2 โ 1 ) ) Hence ๐(ใ๐๐๐ใ^(โ๐) ๐)/๐๐ = ๐/(๐ โ(๐^๐ โ ๐ ) ) As tan2 ฮธ = sec2 ฮธ โ 1, tan ฮธ = โ("sec2 ฮธ โ 1" ) As ๐ฆ = ใ๐ ๐๐ใ^(โ1) ๐ฅ So, ๐๐๐โก๐ = ๐