Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Derivative of γπππππγ^(βπ) π π (π₯)=γπππ ππγ^(β1) π₯ Let π= γπππππγ^(βπ) π cosecβ‘γπ¦=π₯γ π=ππ¨π¬ππβ‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (cosecβ‘π¦ ))/ππ₯ 1 = (π (cosecβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (cosecβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = βcosecβ‘π¦ .cotβ‘π¦ . ππ¦/ππ₯ ππ¦/ππ₯ = 1/(γβcosecγβ‘π¦ .πππβ‘π ) ππ¦/ππ₯ = 1/(γβcosecγβ‘π¦ . β(γππ¨πππγ^πβ‘π β π)) Putting value of πππ ππβ‘π¦ = π₯ ππ¦/ππ₯ = (β1)/(π₯ β(π₯^2 β 1 ) ) Hence π(γπππππγ^(βπ) π)/ππ = (βπ)/(π β(π^π β π ) ) As cot2 ΞΈ = cosec2 ΞΈ β 1, cot ΞΈ = β("cosec2 ΞΈ β 1" ) As π¦ = coγπ ππγ^(β1) π₯ So, coπππβ‘π = π