

Ex 12.2
Last updated at Dec. 16, 2024 by Teachoo
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