Last updated at Dec. 8, 2016 by Teachoo

Transcript

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) rx+s Let f(x) = (px + q) rx+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 Finding u’ & v’ u = px + q u’ = p 1 . x1-1 + 0 = px0 = p v = rx – 1 + s v’ = r( – 1) x1 – 1 + 0 = – rx – 2 = − 𝑟 𝑥2 Now , f’(x) = (uv)’ = u’v + v’u = p(rx – 1 + s) + − 𝑟 𝑥2 (px + q) = p 𝑟𝑥+𝑠 − 𝑟 𝑥2 (px + q) = 𝑝𝑟𝑥 + ps − 𝑝𝑟 𝑥 𝑥2 – 𝑟𝑞 𝑥2 = 𝑝𝑟𝑥 – 𝑝𝑟𝑥 + ps – 𝑟𝑞 𝑥2 = 0 + ps – 𝑟𝑞 𝑥2 = ps – 𝑟𝑞 𝑥2 Hence, f’(x) = ps – 𝒒𝒓 𝒙𝟐 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) rx+s Let f(x) = (px + q) rx+s = px rx+s + q rx+s = px rx + px (s) + q rx + qs = pr + pxs + qrx + qs = pxs + qrx-1 + qs + ps Now, So, f’(x) = pxs + qrx−1 + qs + ps′ = (pxs)’ + (qrx-1)’ + (qs)’ + (ps)’ = ps . (1 . x1–1 ) + qr (-1 . x – 1 –1) + 0 + 0 = ps (x0) + qr ( -1 x-2) = ps (1) – qr x –2 = ps – 𝑞 𝑟 𝑥2 Hence f’(x) = ps – 𝑞𝑟 𝑥2

Misc 1
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Misc 2

Misc 3 You are here

Misc 4

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Misc 6 Important

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Misc 9 Important

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Misc 24 Important

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Misc 27 Important

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.