Finding derivative of Inverse trigonometric functions
Example 26 Important
Example 27
Derivative of cot-1 x (cot inverse x)
Derivative of sec-1 x (Sec inverse x) You are here
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 26, 2021 by Teachoo
Derivative of γπππγ^(βπ) π π (π₯)=γπ ππγ^(β1) π₯ Let π= γπππγ^(βπ) π secβ‘γπ¦=π₯γ π=π¬ππβ‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (secβ‘π¦ ))/ππ₯ 1 = (π (secβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (secβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = πππβ‘π .πππβ‘π. ππ¦/ππ₯ ππ¦/ππ₯ = 1/(πππβ‘π .γ secγβ‘π¦ ) ππ¦/ππ₯ = 1/((β(γπ¬ππγ^πβ‘π β π)) .γ secγβ‘π¦ ) Putting value of π ππβ‘π¦ = π₯ ππ¦/ππ₯ = 1/((β(π₯^2 β 1 ) ) . π₯) ππ¦/ππ₯ = 1/(π₯ β(π₯^2 β 1 ) ) Hence π (γπππγ^(βπ) 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¦ = γπ ππγ^(β1) π₯ So, πππβ‘π = π