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Last updated at March 22, 2023 by Teachoo

Misc 3 (Method 1) Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (px + q) (r/x+s) Let f(x) = (px + q) (r/x+s) = (px + q) (rx –1 + s) Let u = (px + q) & v = (rx-1 + s) ∴ f(x) = uv So, f’(x) = (uv)’ f’(x) = u’v + v’u Using product rule, (uv)’ = u’v + v’u Finding u’ & v’ u = px + q u’ = p × 1 + 0 = p v = rx – 1 + s v’ = r( – 1) x–1 – 1 + 0 = – rx – 2 = (− 𝑟)/𝑥^2 Now , f’(x) = (uv)’ = u’v + v’u = p(rx – 1 + s) + ((− 𝑟)/𝑥^2 ) (px + q) = p (𝑟/𝑥+𝑠) − ( 𝑟)/𝑥^2 (px + q) = 𝑝𝑟/𝑥 + ps − 𝑝𝑟𝑥/𝑥^2 – 𝑟𝑞/𝑥^2 = 𝑝𝑟/𝑥 – 𝑝𝑟/𝑥 + ps – 𝑟𝑞/𝑥^2 = 0 + ps – 𝑟𝑞/𝑥^2 = ps – 𝑟𝑞/𝑥^2 Hence, f’(x) = ps – 𝒒𝒓/𝒙^𝟐 Misc 3 (Method 2) Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (px + q) (r/x+s) Let f(x) = (px + q) (r/x+s) = px (r/x+s) + q (r/x+s) = px (r/x) + px (s) + q (r/x) + qs = pr + pxs + qr/x + qs = pxs + qrx-1 + qs + ps Now, So, f’(x) = ("pxs + qrx−1 + qs + ps" )^′ = (pxs)’ + (qrx-1)’ + (qs)’ + (ps)’ = ps × 1 + qr (–1 x – 1 –1) + 0 + 0 = ps – qr (x-2) = ps – 𝑞𝑟/𝑥^2 Hence, f’(x) = ps – 𝒒𝒓/𝒙^𝟐