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

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

(ii)

(iii)

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

Putting the above obtained values in the formula :-

(uv)′ = u′v + uv′

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

= 12x² – 9 + 3e^x+ 24x² + 3xe^x – 40x - 5e^x

= 36x² + x(3e^x – 40) – 9 - 2e^x

Ans) 36x² + x(3e^x – 40) – 9 - 2e^x