Misc  5 - Differentiate cos-1 x/2 / root (2x + 7) - Teachoo

Misc  5 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc  5 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Concept wise

Transcript

Misc 5 Differentiate 𝑀.π‘Ÿ.𝑑. π‘₯ the function, (γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2)/√(2π‘₯+7 ) , – 2 < π‘₯ < 2 Let 𝑦= (γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2)/√(2π‘₯+7 ) Differentiating both sides 𝑀.π‘Ÿ.𝑑. π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑/𝑑π‘₯ ((γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2)/√(2π‘₯+7 )) Using Quotient rule As (𝑒/𝑣)^β€² = (𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣^2 where u = cosβˆ’1 π‘₯/2 & v = √(2π‘₯+7) 𝑑𝑦/𝑑π‘₯ = (𝑑(γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2)/𝑑π‘₯ . √(2π‘₯ + 7 ) βˆ’ 𝑑(√(2π‘₯ + 7 ))/𝑑π‘₯ . γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2 )/(√(2π‘₯ + 7 ))^2 𝑑𝑦/𝑑π‘₯ = ((βˆ’1)/√(1 βˆ’(π‘₯/2)^2 ) . 𝑑(π‘₯/2)/𝑑π‘₯ √(2π‘₯ + 7 ) βˆ’ 1/(2√(2π‘₯ + 7)) . 𝑑(2π‘₯ + 7)/𝑑π‘₯ . γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2 )/(√(2π‘₯ + 7 ))^2 𝑑𝑦/𝑑π‘₯ = ((βˆ’1)/√(γ€–(4 βˆ’ π‘₯)/4γ€—^2 ) . 1/2 √(2π‘₯ + 7 ) βˆ’ 1/(2√(2π‘₯ + 7)) . (2 + 0) . γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2 )/((2π‘₯ + 7) ) 𝑑𝑦/𝑑π‘₯ = ((βˆ’2)/√(γ€–4 βˆ’ π‘₯γ€—^2 ) . 1/2 √(2π‘₯ + 7 ) βˆ’ 1/(2√(2π‘₯ + 7)) . 2 . γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2 )/((2π‘₯ + 7) ) 𝑑𝑦/𝑑π‘₯ = ((βˆ’ √(πŸπ’™ + πŸ• ))/( √(γ€–πŸ’ βˆ’ 𝒙〗^𝟐 )) βˆ’ (. 〖𝒄𝒐𝒔〗^(βˆ’πŸ ) 𝒙/𝟐)/√(πŸπ’™ + πŸ•) )/((2π‘₯ + 7) ) 𝑑𝑦/𝑑π‘₯ = (βˆ’ √(2π‘₯ + 7 ) ( √(2π‘₯ + 7 )) βˆ’ γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2 √(γ€–4 βˆ’ π‘₯γ€—^2 ))/((2π‘₯ + 7) (√(2π‘₯ + 7 )) (√(γ€–4 βˆ’ π‘₯γ€—^2 )) ) 𝑑𝑦/𝑑π‘₯ = γ€–βˆ’(√(2π‘₯ + 7 ))γ€—^2/((2π‘₯ + 7) ( √(2π‘₯ + 7 )) (√(γ€–4 βˆ’ π‘₯γ€—^2 )) ) "+ " (γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2 √(γ€–4 βˆ’ π‘₯γ€—^2 ))/((2π‘₯ + 7) ( √(2π‘₯ + 7 )) (√(γ€–4 βˆ’ π‘₯γ€—^2 )) ) 𝑑𝑦/𝑑π‘₯ = (βˆ’(2π‘₯ + 7))/((2π‘₯ + 7) √(2π‘₯ + 7 ) √(γ€–4 βˆ’ π‘₯γ€—^2 )) "+ " (γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2 )/((2π‘₯ + 7) √(2π‘₯ + 7 ) ) 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/(√(2π‘₯ + 7 ) √(γ€–4 βˆ’ π‘₯γ€—^2 )) "+ " (γ€–π‘π‘œπ‘ γ€—^(βˆ’1 ) π‘₯/2 )/((2π‘₯ + 7) (2π‘₯ + 7)^(1/2) ) π’…π’š/𝒅𝒙 = (βˆ’πŸ)/(√(πŸπ’™ + πŸ•) √(γ€–πŸ’ βˆ’ 𝒙〗^𝟐 ))βˆ’(〖𝒄𝒐𝒔〗^(βˆ’πŸ ) 𝒙/𝟐 )/((πŸπ’™ + πŸ•)^(πŸ‘/𝟐) )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.