Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Misc 4 Differentiate π€.π.π‘. π₯ the function, γπ ππγ^(β1) (π₯ βπ₯), 0 β€ π₯ β€ 1 Let π¦=γπ ππγ^(β1) (π₯ βπ₯) π¦=γπ ππγ^(β1) (π₯ . π₯^(1/2)) π¦=γπ ππγ^(β1) (π₯^(1 + 1/2)) π=γπππγ^(βπ) (π^( π/π) ) Differentiating π€.π.π‘. π₯ ππ¦/ππ₯ = π(γπ ππγ^(β1) (π₯^( 3/2)))/ππ₯ ππ¦/ππ₯ = 1/β(1 β (π₯^(3/2) )^2 ) Γ (π(π₯)^(3/2))/ππ₯ ("As " π(γπ ππγ^(β1)β‘π₯ )/ππ₯=1/β(1 β π₯^2 )) ππ¦/ππ₯ = 1/β(1 β π₯^3 ) Γ 3/2 (π₯)^(3/2 β1) = 1/β(1 β π₯^3 ) Γ 3/2 γπ₯ γ^(1/2 ) = 1/β(1 β π₯^3 ) Γ 3/2 βπ₯ = π/π β(π/(π βπ^π ))