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use logarithm differentiation to find dy/dx for the following y=(8x + 1)^3 e^x/ sqrt(4x^2 + 1)
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take ln of both sides \[lny = \ln[(8x+1)^3]+lne^x-\ln[\sqrt{4x^2+1}]\]
\[lny = 3\ln[8x+1]+xlne-1/2\ln[4x^2+1]\]
\[lny = 3\ln[8x+1]+x-1/2\ln[4x^2+1]\]
lne = 1 so the lne goes away in that last one
\[1/y*dy/dx = 24/(8x+1)+1-4/(4x^2+1)\]
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\[1/y*dy/dx = 24/[4*(2x+1)]+1-4/[(2x+1)(2x-1)]\]
etc multiply by y and replace y with the original equation
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