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Ex 13.2
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
Last updated at Aug. 28, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 13.2, 11 Find the derivative of the following functions: (iv) cosec x Let f (x) = cosec x f(x) = 1/sin𝑥 Let u = 1 & v = sin x ∴ f(x) = 𝑢/𝑣 So, f’(x) = (𝑢/𝑣)^′ Using quotient rule f’(x) = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 Finding u’ & v’ u = 1 u’ = 0 & v = sin x v’ = cos x Now, f’(x) = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 = (0 (sin〖𝑥) −〖 cos〗〖𝑥 (1)〗 〗)/(〖𝑠𝑖𝑛〗^2 𝑥) (Derivative of constant function = 0) (Derivative of sin x = cos x) = (0 − 𝑐𝑜𝑠 𝑥)/(〖𝑠𝑖𝑛〗^2 𝑥) = (− 𝑐𝑜𝑠 𝑥)/(〖𝑠𝑖𝑛〗^2 𝑥) = (− 𝑐𝑜𝑠 𝑥)/sin𝑥 . 1/sin𝑥 = – cot x cosec x = – cosec x cot x Hence f’(x) = – cosec x cot x Using cot x = 𝑐𝑜𝑠/sin𝑥 & 1/sin𝑥 = cosec x