Ex 5.3, 15 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at March 11, 2021 by Teachoo
Last updated at March 11, 2021 by Teachoo
Transcript
Ex 5.3, 15 Find ๐๐ฆ/๐๐ฅ in, y = secโ1 (1/( 2๐ฅ2โ1 )), 0 < x < 1/โ2 y = secโ1 (1/( 2๐ฅ^2 โ 1 )) ๐๐๐โก๐ = 1/(2๐ฅ^2 โ 1) ๐/๐๐จ๐ฌโก๐ = 1/(2๐ฅ^2 โ 1) cosโก๐ฆ = 2๐ฅ2โ1 y = cos โ1 (2๐ฅ2โ1) Putting ๐ฅ = cosโกฮธ ๐ฆ = cos โ1 (2๐๐๐ 2๐โ1) ๐ฆ = cos โ1 (cosโก2 ๐) ๐ฆ = 2๐ Putting value of ฮธ = cosโ1 x ๐ฆ = 2 ใ๐๐๐ ใ^(โ1) ๐ฅ Differentiating both sides ๐ค.๐.๐ก.๐ฅ . (๐(๐ฆ))/๐๐ฅ = (๐ (2ใ๐๐๐ ใ^(โ1) ๐ฅ" " ))/๐๐ฅ ๐๐ฆ/๐๐ฅ = 2 (๐ (ใ๐๐๐ ใ^(โ1) ๐ฅ" " ))/๐๐ฅ ๐๐ฆ/๐๐ฅ = 2 . ((โ1)/โ(1 โ ๐ฅ^2 )) ๐ ๐/๐ ๐ = (โ๐)/โ(๐ โ ๐^๐ ) (๐๐๐ โก2๐ " = 2 " ใ๐๐๐ ใ^2 ๐โ1) Since x = cos ฮธ โด ใ๐๐๐ ใ^(โ1) x = ฮธ ((ใ๐๐๐ ใ^(โ1) ๐ฅ")โ = " (โ1)/โ(1 โ ๐ฅ^2 ))
Finding derivative of Inverse trigonometric functions
Example 26 Important
Example 27
Derivative of cot-1 x (cot inverse x)
Derivative of sec-1 x (Sec inverse x)
Derivative of cosec-1 x (Cosec inverse x)
Ex 5.3, 14
Ex 5.3, 9 Important
Ex 5.3, 13 Important
Ex 5.3, 12 Important
Ex 5.3, 11 Important
Ex 5.3, 10 Important
Ex 5.3, 15 Important You are here
Misc 5 Important
Misc 4
Misc 13 Important
Finding derivative of Inverse trigonometric functions
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