Asked by Anonymous
                Is the derivative of (squarerootof x)/(3x+1) : 3x+1-3squarerootofx/(2squarerootofx)(3x+1)^2
            
            
        Answers
                    Answered by
            Steve
            
    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 - <b>u</b>v')/v^2
    
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 - <b>u</b>v')/v^2
                    Answered by
            Steve
            
    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
    
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
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