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

Last updated at May 7, 2024 by Teachoo

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