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γ^(βπ) πDerivative of γπππγ^(βπ) π π (π₯)=γπππ γ^(β1) π₯ Let π= γπππγ^(βπ) π cosβ‘γπ¦=π₯γ π=ππ¨π¬β‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (cosβ‘π¦ ))/ππ₯ 1 = (π (cosβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (cosβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = (βsinβ‘π¦) ππ¦/ππ₯ (β1)/sinβ‘π¦ =ππ¦/ππ₯ ππ¦/ππ₯ = (β1)/πππβ‘π ππ¦/ππ₯= (β1)/β(π β γπππγ^π π) Putting πππ β‘γπ¦=π₯γ ππ¦/ππ₯= (β1)/β(1 β π^π ) Hence, (π(γπππγ^(βπ) π" " ))/ππ = (βπ)/β(π β π^π ) "We know that" γπ ππγ^2 π+γπππ γ^2 π=1 γπ ππγ^2 π=1βγπππ γ^2 π πππβ‘π½=β(πβγπππγ^π π½) " " As π¦ = γπππ γ^(β1) π₯ So, πππβ‘π = π