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 cotβ‘γπ¦=π₯γ π=ππ¨π­β‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (cotβ‘π¦ ))/ππ₯ 1 = (π (cotβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (cotβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = βππ¨γπ¬ππγ^π π . ππ¦/ππ₯ 1 = β(π +πππππ) ππ¦/ππ₯ ππ¦/ππ₯ = (β1)/(1 + γππ¨π­γ^πβ‘π ) Putting πππ‘β‘π¦ = π₯ ππ¦/ππ₯ = (β1)/(π^π + π) Hence (π(γππ¨π­γ^(βπ)β‘γπ)γ)/ππ = (βπ)/(π^π + π) (π΄π  γ πππ ππγ^2β‘γπ¦= γ1+γβ‘γπππ‘γ^2β‘π¦ γ) As π¦ = γπππ‘γ^(β1) π₯ So, πππβ‘π = π