Ex 12.2, 11 (iv) - Chapter 12 Class 11 Limits and Derivatives
Last updated at April 16, 2024 by Teachoo
Ex 12.2
Ex 12.2, 1
Ex 12.2, 2
Ex 12.2, 3
Ex 12.2, 4 (i) Important
Ex 12.2, 4 (ii)
Ex 12.2, 4 (iii) Important
Ex 12.2, 4 (iv)
Ex 12.2, 5
Ex 12.2, 6
Ex 12.2, 7 (i) Important
Ex 12.2, 7 (ii)
Ex 12.2, 7 (iii) Important
Ex 12.2, 8
Ex 12.2, 9 (i)
Ex 12.2, 9 (ii) Important
Ex 12.2, 9 (iii)
Ex 12.2, 9 (iv) Important
Ex 12.2, 9 (v)
Ex 12.2, 9 (vi)
Ex 12.2, 10 Important
Ex 12.2, 11 (i)
Ex 12.2, 11 (ii) Important
Ex 12.2, 11 (iii) Important
Ex 12.2, 11 (iv) You are here
Ex 12.2, 11 (v) Important
Ex 12.2, 11 (vi)
Ex 12.2, 11 (vii) Important
Transcript
Ex 12.2, 11 Find the derivative of the following functions: (iv) cosec x Let f (x) = cosec x f(x) = 1/sinβ‘π₯ Let u = 1 & v = sin x β΄ f(x) = π’/π£ So, fβ(x) = (π’/π£)^β² Using quotient rule fβ(x) = (π’^β² π£ βγ π£γ^β² π’)/π£^2 Finding uβ & vβ u = 1 uβ = 0 & v = sin x vβ = cos x Now, fβ(x) = (π’^β² π£ βγ π£γ^β² π’)/π£^2 = (0 (sinβ‘γπ₯) βγ cosγβ‘γπ₯ (1)γ γ)/(γπ ππγ^2 π₯) (Derivative of constant function = 0) (Derivative of sin x = cos x) = (0 β πππ π₯)/(γπ ππγ^2 π₯) = (β πππ π₯)/(γπ ππγ^2 π₯) = (β πππ π₯)/sinβ‘π₯ . 1/sinβ‘π₯ = β cot x cosec x = β cosec x cot x Hence fβ(x) = β cosec x cot x Using cot x = πππ /sinβ‘π₯ & 1/sinβ‘π₯ = cosec x
