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To find: Differentiation of (3x – 5) (4x2 – 3 + ex)

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

Let us take u = (3x – 5) and v = (4x2 – 3 + ex)

Putting the above obtained values in the formula :-

(uv)′ = u′v + uv′

[(3x – 5)(4x2 – 3 + ex)]’ = 3×(4x2 – 3 + ex) + (3x – 5)×(8x + ex)

= 12x2 – 9 + 3ex+ 24x2 + 3xex – 40x - 5ex

= 36x2 + x(3ex – 40) – 9 - 2ex

Ans) 36x2 + x(3ex – 40) – 9 - 2ex

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