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Misc 4 - Differentiate sin-1 (x root x) - Chapter 5 NCERT

Misc  4 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Misc 4 Differentiate 𝑀.π‘Ÿ.𝑑. π‘₯ the function, 〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯ √π‘₯), 0 ≀ π‘₯ ≀ 1 Let 𝑦=〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯ √π‘₯) 𝑦=〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯ . π‘₯^(1/2)) 𝑦=〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯^(1 + 1/2)) π’š=γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) (𝒙^( πŸ‘/𝟐) ) Differentiating 𝑀.π‘Ÿ.𝑑. π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯^( 3/2)))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 1/√(1 βˆ’ (π‘₯^(3/2) )^2 ) Γ— (𝑑(π‘₯)^(3/2))/𝑑π‘₯ ("As " 𝑑(〖𝑠𝑖𝑛〗^(βˆ’1)⁑π‘₯ )/𝑑π‘₯=1/√(1 βˆ’ π‘₯^2 )) 𝑑𝑦/𝑑π‘₯ = 1/√(1 βˆ’ π‘₯^3 ) Γ— 3/2 (π‘₯)^(3/2 βˆ’1) = 1/√(1 βˆ’ π‘₯^3 ) Γ— 3/2 γ€–π‘₯ γ€—^(1/2 ) = 1/√(1 βˆ’ π‘₯^3 ) Γ— 3/2 √π‘₯ = πŸ‘/𝟐 √(𝒙/(𝟏 βˆ’π’™^πŸ‘ ))

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