Find the derivative of y=x^2/ln3x.
x^2 / ln 3x
getting the derivative,
[(2x)(ln 3x) + (x^2)(1/(3x))(3)] / [(ln 3x)^2]
simplifying,
[ (2x)ln 3x + x ]/[(ln 3x)^2]
hope this helps~ :)
To find the derivative of y = x^2/ln(3x), we can use the quotient rule. The quotient rule states that if we have a function of the form y = u/v, where u and v are functions of x, then the derivative of y with respect to x, denoted as dy/dx, is given by the formula:
dy/dx = (v * du/dx - u * dv/dx) / v^2
In our case, u = x^2 and v = ln(3x). Let's find the derivatives of u and v first:
du/dx = 2x (by applying the power rule for derivatives)
dv/dx = (1/x) * 3x (by applying the chain rule and the derivative of ln(3x))
Simplifying, we get:
dv/dx = 3
Now, we can substitute these values into the quotient rule formula:
dy/dx = (ln(3x) * 2x - x^2 * 3) / (ln(3x))^2
Simplifying further:
dy/dx = (2xln(3x) - 3x^2) / (ln(3x))^2
Thus, the derivative of y = x^2/ln(3x) with respect to x is (2xln(3x) - 3x^2) / (ln(3x))^2.