posted by Anonymous .
Is the derivative of (squarerootof x)/(3x+1) : 3x+1-3squarerootofx/(2squarerootofx)(3x+1)^2
y = √x/(3x+1)
y' = (1/2√x (3x+1) - √x (3)) / (3x+1)^2
= (3x+1) - 2√x√x(3)) / 2√x (3x+1)^2
= (1-3x) / 2√x (3x+1)^2
Looks like you forgot a factor of √x:
y = u/v
y' = (u'v - uv')/v^2
also ,my own typing was sloppy.
y' = (1/(2√x) (3x+1) - √x (3)) / (3x+1)^2
Looks like you forgot to put the numerator over a common denominator of 2√x