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
To find the derivative of the function (sqrt(x))/(3x+1), we will use the quotient rule.
According to the quotient rule, if we have a function in the form f(x)/g(x), then the derivative is given by:
(f'(x) * g(x) - f(x) * g'(x)) / [g(x)]^2
Let's break down the function and differentiate each part separately:
f(x) = sqrt(x)
g(x) = 3x + 1
Now let's find the derivative of f(x) and g(x):
f'(x) = (1/2) * (x)^(-1/2) [using the power rule for differentiation]
= (1/2) * (1/sqrt(x))
= 1 / (2 * sqrt(x))
= 1 / (2sqrt(x))
g'(x) = 3 [since the derivative of 3x is 3 and the derivative of 1 is 0]
Now, we can substitute these values back into the quotient rule:
(f'(x) * g(x) - f(x) * g'(x)) / [g(x)]^2
= [1 / (2sqrt(x))] * (3x + 1) - (sqrt(x)) * 3 / [(3x + 1)]^2
Simplifying further:
= (3x + 1) / (2 * 2sqrt(x) * (3x + 1)) - 3sqrt(x) / (3x + 1)^2
= (3x + 1) / (4sqrt(x) * (3x + 1)) - 3sqrt(x) / (3x + 1)^2
= (3x + 1 - 12sqrt(x)) / (4sqrt(x) * (3x + 1))
Therefore, the derivative of (sqrt(x))/(3x+1) is (3x + 1 - 12sqrt(x)) / (4sqrt(x) * (3x + 1)).