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) You are here

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)

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, 9 (Method 1) Find the derivative of (iii) x–3 (5 + 3x) Let f(x) = x –3 (5 + 3x) Let u = x–3 & v = 5 + 3x ∴ f(x) = uv So, f’(x) = (uv)’ f’(x) = u’v + v’u Finding u’ & v’ u = x – 3 u’ = –3x–3 – 1 = –3x–4 v = 5 + 3x v’ = 0 + 3 = 3 Now, f’(x) = (uv)’ = u’v + v’u = –3x– 4 (5 + 3x) + 3 (x–3) = –15x–4 – 9x–4 + 1 + 3x –3 (xn)’ = nxn – 1 & (a)’ = 0 where a is constant = –15x–4 – 9x–3 + 3x –3 = –15x–4 – 6x–3 = (−15)/𝑥^4 – 6/𝑥^3 = −3 [5/𝑥^4 +2/𝑥^3 ] = −3 [(5 + 2𝑥)/𝑥^4 ] = (−𝟑)/𝒙^𝟒 (5 + 2x) Hence, f’(x) = (−3)/𝑥^4 (5 + 2x) Ex 12.2, 9 (Method 2) Find the derivative of (iii) x–3 (5 + 3x) 𝑥^(−3) (5+3𝑥) = (5 + 3𝑥)/𝑥^3 = 5/𝑥^3 + 3𝑥/𝑥^3 = 5/𝑥^3 + 3/𝑥^2 = 〖5𝑥〗^(−3) + 〖3𝑥〗^(−2) Differentiating w.r.t.x (〖5𝑥〗^(−3) " + " 〖3𝑥〗^(−2) )^′ = 5 [−3𝑥^(−4) ] + 3 [−2𝑥^(−3) ] = −3 [5/𝑥^4 +2/𝑥^3 ] = −3 [(5 + 2𝑥)/𝑥^4 ] = (−𝟑)/𝒙^𝟒 (5 + 2x)