Find the derivative of y=(7x^(1/3))/(x^2+9)

To find the derivative of the function y = (7x^(1/3))/(x^2 + 9), we can use the quotient rule. The quotient rule states that if we have a function of the form f(x)/g(x), then the derivative of this function is given by:

(d/dx)(f(x)/g(x)) = (g(x) * f'(x) - f(x) * g'(x)) / (g(x))^2

Let's go step by step:

Step 1: Find the derivative of the numerator, f'(x).
Since the numerator is (7x^(1/3)), we can apply the power rule. The power rule states that if we have a function of the form x^n, then its derivative is given by:

d/dx(x^n) = n * x^(n-1)

Using the power rule, the derivative of f(x) = 7x^(1/3) is:
f'(x) = (1/3) * 7 * x^(1/3 - 1) = (7/3)x^(-2/3)

Step 2: Find the derivative of the denominator, g'(x).
The denominator is (x^2 + 9), and since it is a sum, the derivative will be the sum of the derivatives of its individual terms. The derivative of x^2 is 2x, and the derivative of a constant (in this case, 9) is zero. Therefore, g'(x) is 2x.

Step 3: Apply the quotient rule.
Using the quotient rule, the derivative of y = (7x^(1/3))/(x^2 + 9) is:
y' = [(x^2 + 9) * (7/3)x^(-2/3) - (7x^(1/3)) * (2x)] / (x^2 + 9)^2

Simplifying the expression further, we get:
y' = (7x^(2/3) + 63x^(-2/3) - 14x^(4/3)) / 3(x^2 + 9)^2

Therefore, the derivative of y = (7x^(1/3))/(x^2 + 9) is (7x^(2/3) + 63x^(-2/3) - 14x^(4/3)) / 3(x^2 + 9)^2.