Check sibling questions

Ex 5.2, 7 - Differentiate 2 root cot (x^2) - Teachoo - Ex 5.2

Ex 5.2, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.2, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

This video is only available for Teachoo black users

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

Ex 5.2, 7 Differentiate the functions with respect to π‘₯ 2√(cot ( π‘₯2 ))Let 𝑦 = " " 2√(cot ( π‘₯2 )) We need to find derivative of 𝑦 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑦′ = (2√(cot ( π‘₯^2 ) ) )^β€² = 2 Γ— 1/(2√(cot ( π‘₯^2 ) )) . (cot ( π‘₯^2 ))^β€² = 1/√(cot ( π‘₯^2 ) ) . (βˆ’π‘π‘œπ‘ π‘’π‘^2 (π‘₯^2 )) . ( π‘₯^2 )^β€² = 1/√(cot ( π‘₯^2 ) ) . (βˆ’π‘π‘œπ‘ π‘’π‘^2 (π‘₯^2)) . 2π‘₯ = (βˆ’2π‘₯" " π‘π‘œπ‘ π‘’π‘^2 π‘₯^2)/√(cot ( π‘₯^2 ) ) = (βˆ’2π‘₯)/(sin^2⁑(π‘₯^2 ) . √( γ€–cos 〗⁑〖(π‘₯^2)γ€—/γ€–sin 〗⁑〖(π‘₯^2)γ€— ) ) "= " (βˆ’2π‘₯)/(〖𝑠𝑖𝑛 〗⁑(π‘₯^2 ) Γ— γ€–π’”π’Šπ’ 〗⁑(𝒙^𝟐 ) √( γ€–cos 〗⁑〖(π‘₯^2)γ€—/γ€–sin 〗⁑〖(π‘₯^2)γ€— ) ) " " = (βˆ’ 2π‘₯ )/(γ€–sin 〗⁑〖(π‘₯)^2 γ€— Γ—βˆš(γ€–γ€–π’”π’Šπ’γ€—^𝟐 〗⁑〖(𝒙^𝟐)γ€— Γ— γ€–cos 〗⁑〖(π‘₯^2)γ€—/sin⁑〖 (π‘₯^2)γ€— )) = (βˆ’ 2π‘₯ )/(γ€–sin 〗⁑〖(π‘₯)^2 γ€— Γ—βˆš(γ€–πœπ¨π¬ 〗⁑〖(𝒙^𝟐)γ€— π’”π’Šπ’β‘γ€– (𝒙^𝟐)γ€— )) = (βˆ’ 2π‘₯ )/(γ€–sin 〗⁑〖(π‘₯)^2 γ€— Γ—βˆš(2/2 Γ— γ€–cos 〗⁑〖(π‘₯^2)γ€— sin⁑〖 (π‘₯^2)γ€— )) = (βˆ’ 2√𝟐 π‘₯ )/(γ€–sin 〗⁑〖(π‘₯)^2 γ€— Γ—βˆš(2 γ€–cos 〗⁑〖(π‘₯^2)γ€— sin⁑〖 (π‘₯^2)γ€— )) = (βˆ’ 𝟐√𝟐 𝒙 )/(〖𝐬𝐒𝐧 〗⁑〖(𝒙)^𝟐 γ€— Γ—βˆš(γ€–π’”π’Šπ’ 〗⁑〖(πŸπ’™^𝟐)γ€— )) (As sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ)

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.