Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

Transcript

Example 24 Find the derivative of f given by π (π₯)=γπ ππγ^(β1) π₯ assuming it exists. π (π₯)=γπ ππγ^(β1) π₯ Let π= γπππγ^(βπ) π sinβ‘γπ¦=π₯γ π=π¬π’π§β‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (sinβ‘π¦ ))/ππ₯ 1 = (π (sinβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (sinβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = cosβ‘π¦ ππ¦/ππ₯ 1/cosβ‘π¦ =ππ¦/ππ₯ ππ¦/ππ₯ = 1/πππβ‘π ππ¦/ππ₯= 1/β(π β γπππγ^π π) Putting π ππβ‘γπ¦=π₯γ ππ¦/ππ₯= 1/β(1 β π^π ) Hence, (π(γπππγ^(βπ) π" " ))/ππ = 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