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Misc 30 Important You are here

Last updated at April 16, 2024 by Teachoo

Misc 30 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): π₯/(π πππ π₯) Let f(x) = π₯/(π πππ π₯) Let u = x & v = sinn x β΄ f(x) = π’/π£ So, fβ(x) = (π’/π£)^β² Using quotient rule fβ(x) = (π’^β² π£ βγ π£γ^β² π’)/π£^2 Finding uβ & vβ u = x uβ = 1 Now, v = sinn x Let p = sin x v = pn By Leibnitz product rule vβ = (pn)β pβ = n pn β 1 pβ Putting p = sin x = n sinn β 1 x (sin x)β = n sinn β 1 x cos x Now, fβ(x) = (π’/π£)^β² = (π’^β² π£ βγ π£γ^β² π’)/π£^2 = ( 1 (sinπβ‘γ π₯γ ) β γπ π ππγ^(πβ1) π₯ cosβ‘γπ₯ (π₯)γ)/γγ(π ππγ^π π₯)γ^2 = ( γπ ππγ^π π₯ β π₯ (πγπ ππγ^(πβ1) π₯ cosβ‘γπ₯) γ)/γγ(π ππγ^π π₯)γ^2 = ( γπππγ^(πβπ) π . sinβ‘γπ₯ β π₯ (π γ γπ ππγ^(πβ1) π₯ cosβ‘γπ₯) γ)/γγ(π ππγ^π π₯)γ^2 = ( γπππγ^(πβπ) π γ(sinγβ‘γπ₯ β ππ₯ . γ cosβ‘γπ₯) γ)/(γπ ππγ^2π π₯) = sinβ‘γπ₯ β ππ₯ cosβ‘π₯ γ/(γπ ππγ^2π π . γπππγ^(β(πβπ) ) π) = sinβ‘γπ₯ β ππ₯ cosβ‘π₯ γ/(γπππγ^((ππ β π+π)) π) = sinβ‘γπ₯ β ππ₯ cosβ‘π₯ γ/(γπ ππγ^(π + 1) π₯) Thus, fβ(x) = πππβ‘γπ β ππ πππβ‘π γ/(γπππγ^(π + π) π)