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 π (π₯)=γπ ππγ^(β1) π₯ Let π= γπππγ^(βπ) π secβ‘γπ¦=π₯γ π=π¬ππβ‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (secβ‘π¦ ))/ππ₯ 1 = (π (secβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (secβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = πππβ‘π .πππβ‘π. ππ¦/ππ₯ ππ¦/ππ₯ = 1/(πππβ‘π .γ secγβ‘π¦ ) ππ¦/ππ₯ = 1/((β(γπ¬ππγ^πβ‘π β π)) .γ secγβ‘π¦ ) Putting value of π ππβ‘π¦ = π₯ ππ¦/ππ₯ = 1/((β(π₯^2 β 1 ) ) . π₯) ππ¦/ππ₯ = 1/(π₯ β(π₯^2 β 1 ) ) Hence π(γπππγ^(βπ) π)/ππ = π/(π β(π^π β π ) ) As tan2 ΞΈ = sec2 ΞΈ β 1, tan ΞΈ = β("sec2 ΞΈ β 1" ) As π¦ = γπ ππγ^(β1) π₯ So, πππβ‘π = π