jake-patel_609560f68691b
jake-patel_609560f68691b
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y = ln (1/x + 1/x²) Vou chamar (1/x + 1/x^2) de g y = ln g y' = (1/g) * g' g = (1/x + 1/x²) g' = -1/x² - 2/x³ g' = ( - x -2) / x³ y' = [1/(1/x + 1/x²)] * [( - x -2) / x³] y' = 1/ [( x + 1)/ x²] * [( - x -2) / x³] y' = x² / (x+1) * (-x…