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Ex 13.2, 11 (iv) - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
Last updated at Aug. 28, 2021 by Teachoo
Ex 13.2
Ex 13.2, 1
Ex 13.2, 2
Ex 13.2, 3
Ex 13.2, 4 (i) Important
Ex 13.2, 4 (ii)
Ex 13.2, 4 (iii) Important
Ex 13.2, 4 (iv)
Ex 13.2, 5
Ex 13.2, 6
Ex 13.2, 7 (i) Important
Ex 13.2, 7 (ii)
Ex 13.2, 7 (iii) Important
Ex 13.2, 8
Ex 13.2, 9 (i)
Ex 13.2, 9 (ii) Important
Ex 13.2, 9 (iii)
Ex 13.2, 9 (iv) Important
Ex 13.2, 9 (v)
Ex 13.2, 9 (vi)
Ex 13.2, 10 Important
Ex 13.2, 11 (i)
Ex 13.2, 11 (ii) Important
Ex 13.2, 11 (iii) Important
Ex 13.2, 11 (iv) You are here
Ex 13.2, 11 (v) Important
Ex 13.2, 11 (vi)
Ex 13.2, 11 (vii) Important
Transcript
Ex 13.2, 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
