Finding derivative of Inverse trigonometric functions
Example 24 Important You are here
Question 3 Deleted for CBSE Board 2025 Exams
Derivative of cot-1 x (cot inverse x)
Derivative of sec-1 x (Sec inverse x)
Derivative of cosec-1 x (Cosec inverse x)
Ex 5.3, 14
Ex 5.3, 9 Important
Ex 5.3, 13 Important
Ex 5.3, 12 Important
Ex 5.3, 11 Important
Ex 5.3, 10 Important
Ex 5.3, 15 Important
Misc 5 Important
Misc 4
Misc 13 Important
Finding derivative of Inverse trigonometric functions
Last updated at April 16, 2024 by Teachoo
Example 24 Find the derivative of f given by π (π₯)=γπ ππγ^(β1) π₯ assuming it exists. π (π₯)=γπ ππγ^(β1) π₯ Let π= γπππγ^(βπ) π sinβ‘γπ¦=π₯γ π=π¬π’π§β‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (sinβ‘π¦ ))/ππ₯ 1 = (π (sinβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (sinβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = cosβ‘π¦ ππ¦/ππ₯ 1/cosβ‘π¦ =ππ¦/ππ₯ ππ¦/ππ₯ = 1/πππβ‘π ππ¦/ππ₯= 1/β(π β γπππγ^π π) Putting π ππβ‘γπ¦=π₯γ ππ¦/ππ₯= 1/β(1 β π^π ) Hence, (π (γπππγ^(βπ) π" " ))/π π = π/β(π β π^π ) "We know that" γπ ππγ^2 π+γπππ γ^2 π=1 γπππ γ^2 π=1βγπ ππγ^2 π πππβ‘π½=β(πβγπππγ^π π½) " " As π¦ = γπ ππγ^(β1) π₯ So, πππβ‘π = π