Last updated at Nov. 30, 2019 by
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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)
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