differential
in Calculus Answers by Level 1 User (140 points)

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2 Answers

y = 1\2log((x+3)\(x-3))

2y = log((x+3)\(x-3))

log(m/n) = logm-logn

2y = log x+3 - log x-3

2 dy/dx = 1/x+3 - 1/x-3

2 dy/dx = (x-3) - (x+3) / [x^2-3^2]

2 dy/dx = x-3-x-3 / (x^2-9)

2 dy/dx = -6 / (x^2-9)

dy/dx = -3 / (x^2-9)
by Level 1 User (340 points)
y = 1\2log((x+3)\(x-3))

2y = log((x+3)\(x-3))

2 dy/dx = x-3/x+3 . d/dx ((x+3)\(x-3))

use quotient rule

2 dy/dx = (x-3/x+3) .[(x-3) . 1 - (x+3).1 / (x-3)^2]

2 dy/dx = (x-3/x+3). [-6 / (x-3)^2]

2 dy/dx = -6 / [(x-3) (x+3)]

dy/dx = -3 / (x^2-9)
by Level 1 User (340 points)

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