Last updated at Dec. 16, 2024 by Teachoo
Derivative of γπππγ^(βπ) πDerivative of γπππγ^(βπ) π π (π₯)=γπππ γ^(β1) π₯ Let π= γπππγ^(βπ) π cosβ‘γπ¦=π₯γ π=ππ¨π¬β‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (cosβ‘π¦ ))/ππ₯ 1 = (π (cosβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (cosβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = (βsinβ‘π¦) ππ¦/ππ₯ (β1)/sinβ‘π¦ =ππ¦/ππ₯ ππ¦/ππ₯ = (β1)/πππβ‘π ππ¦/ππ₯= (β1)/β(π β γπππγ^π π) Putting πππ β‘γπ¦=π₯γ ππ¦/ππ₯= (β1)/β(1 β π^π ) Hence, (π (γπππγ^(βπ) π" " ))/π π = (βπ)/β(π β π^π ) "We know that" γπ ππγ^2 π+γπππ γ^2 π=1 γπ ππγ^2 π=1βγπππ γ^2 π πππβ‘π½=β(πβγπππγ^π π½) " " As π¦ = γπππ γ^(β1) π₯ So, πππβ‘π = π
Finding derivative of Inverse trigonometric functions
Example 24 Important
Question 3
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
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo